structure of atom - n@[email protected]/.../5/5895441/iit_class_xi_chem_structure_of_ato… ·...

��

IIT-JEE

STRUCTUREOF ATOM

C H E M I S T R YS T U D Y M A T E R I A L

PHASE - I

NARAYANA INSTITUTE OF CORRESPONDENCE COURSESF N S H O U S E , 6 3 K A L U S A R A I M A R K E TS A R V A P R I Y A V I H A R , N E W D E L H I - 1 1 0 0 1 6PH.: (011) 32001131/32/50 • FAX : (011) 41828320Websi te : w w w . n a r a y a n a i c c . c o mE-mai l : i n f o @ n a r a y a n a i c c . c o m

2004 NARAYANA GROUP

This study material is a part of NARAYANA INSTITUTE OF CORRESPONDENCE COURSES for IIT-JEE, 2008-09. This is meant

for the personal use of those students who are enrolled with NARAYANA INSTITUTE OF CORRESPONDENCE COURSES, FNS

House, 63, Kalu Sarai Market, New Delhi-110016, Ph.: 32001131/32/50. All rights to the contents of the Package rest with

NARAYANA INSTITUTE. No other Institute or individual is authorized to reproduce, translate or distribute this material in any form,

without prior information and written permission of the institute.

PREFACEDear Student,

Heartiest congratulations on making up your mind and deciding to be an engineer to serve the society.

As you are planning to take various Engineering Entrance Examinations, we are sure that this STUDY PACKAGE isgoing to be of immense help to you.

At NARAYANA we have taken special care to design this package according to the Latest Pattern of IIT-JEE, whichwill not only help but also guide you to compete for IIT-JEE, AIEEE & other State Level Engineering EntranceExaminations.

The salient features of this package include :

! Power packed division of units and chapters in a scientific way, with a correlation being there.

! Sufficient number of solved examples in Physics, Chemistry & Mathematics in all the chapters to motivate thestudents attempt all the questions.

! All the chapters are followed by various types of exercises, including Objective - Single Choice Questions,Objective - Multiple Choice Questions, Comprehension Type Questions, Match the Following, Assertion-Reasoning& Subjective Questions.

These exercises are followed by answers in the last section of the chapter including Hints & Solutions whereverrequired. This package will help you to know what to study, how to study, time management, your weaknesses andimprove your performance.

We, at NARAYANA, strongly believe that quality of our package is such that the students who are not fortunateenough to attend to our Regular Classroom Programs, can still get the best of our quality through these packages.

We feel that there is always a scope for improvement. We would welcome your suggestions & feedback.

Wish you success in your future endeavours.

THE NARAYANA TEAM

ACKNOWLEDGEMENTWhile preparing the study package, it has become a wonderful feeling for the NARAYANA TEAM to get thewholehearted support of our Staff Members including our Designers. They have made our job really easy throughtheir untiring efforts and constant help at every stage.

We are thankful to all of them.

THE NARAYANA TEAM

CONTENTSCONTENTSCONTENTSCONTENTSCONTENTS

STRUCTURE OF ATOM1. Theory

2. Solved Problems

(i) Subjective Type Problems

(ii) Single Choice Problems

(iii) Multiple Choice Problems

(iv) Miscellaneous Problems

• Comprehension Type Problems

• Matching Type Problems

• Assertion-Reason Type Problems

3. Assignments

(i) Subjective Questions

(ii) Single Choice Questions

(iii) Multiple Choice Questions

(iv) Miscellaneous Questions

• Comprehension Type Questions

• Matching Type Questions

• Assertion-Reason Type Questions

(v) Problems Asked in IIT-JEE

4. AnswersCO

NT

EN

TS

CO

NT

EN

TS

CO

NT

EN

TS

CO

NT

EN

TS

CO

NT

EN

TS

STRUCTURE OF ATOM

IIT-JEE Syllabus Electron, proton and neutron; Constitution of nucleus, Properties of alpha, beta and gamma rays; Rutherford’s scattering experiment; Bohr’s atomic model (mathematical details excluded); Quantum numbers; Pauli’s exclusion principle; Hund’s rule; Aufbau principle; Electronic configuration of elements (upto atomic number 36); Shapes of s, p, and d–orbitals.

CONTENTS

Dalton’s Theory of atom

Subatomic particles

Various atomic models

Photoelectric effect

Particles & wave nature of electron

Heisenberg uncertainty principle

Schrodinger wave equation

Quantum numbers

Rules for filling electrons

Shapes and size of orbitals

Quantum mechanical treatment of H–atom

INTRODUCTION

The classical view of atomic structure was constructed with the body of knowledge accumulated in physics over several centuries. We begin with the focus on atomic nucleus with its protons and neutron. This includes obtaining atomic masses and atomic number. Electrons are at the heart of our modern view of atomic structure.

This chapter, firstly provides experimental evidence for the picture of atom. After that it deals with nature of light and finally explains the behaviour of electrons in atoms leading to the modern view of atomic structure.

FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 • Ph.: (011) 32001131/32 Fax : (011) 41828320

1

NARAYANA INSTITUTE OF CORRESPONDENCE COURSES

Chemistry : Structure of Atom

1. DALTON’S THEORY OF ATOM In 1807, John Dalton developed his atomic theory. According to this theory :

• An atom cannot be subdivided. • Atoms are neither created nor destroyed during chemical reactions. • Atoms of the same element are alike; in particular all atoms of an element have the

same mass. • Atoms of different elements are not alike; in particular, their masses are different.

1.1 REASON FOR FAILURE OF DALTON’S THEORY

In 1833, Michael Faraday showed that there is a relationship between matter and electricity. This was the first major break through to suggest that atom was not a simple indivisible particle of all matter but was made up of small particles. Discovery of electrons, protons and neutrons discarded the indivisible nature of atom proposed by John Dalton.

2. SUBATOMIC PARTICLES The researches done by various eminent scientists and the discovery of radioactivity have

established beyond doubt, that atom is not the smallest indivisible particle but had a complex structure of its own and was made up of still smaller particles like electrons, protons, neutrons etc. At present about 35 different subatomic particles are known but the three particles namely electron, proton and neutron are regarded as the fundamental particles.

We shall now take up the brief study of these fundamental particles.

2.1 CATHODE RAYS (DISCOVERY OF ELECTRON) Gases are normally poor conductor of electricity at ordinary or high pressure. However

when a tube filled with a gas is evacuated to a pressure of 0.01 mm Hg or lower and an electric potential is applied across a pair of electrodes into the tube discharge takes place between the electrodes during which a stream of rays moves from cathode to anode, William Crookes (1879) called these rays as cathode rays.

2.1a PROPERTIES OF CATHODE RAYS

1. Cathode rays travel in straight lines with a high speed approaching to that of light. Their linear propagation is shown by the fact that they caste shadows of the solid objects placed in their path.

2. They produce temperature rise in any object they strike. 3. They pass through thin films of metals but are stopped by thicker foils. 4. Cathode rays can produce mechanical effect e.g. when allowed to fall on a small

paddle wheel they cause rotation of wheel. This experiment shows that cathode rays consist of material particles.


2



5. The cathode rays can be deflected by electric and magnetic field. The direction of deflection is always such that these particles bear a negative charge.

6. Finally no matter what the nature of the cathode or the gas in the discharge tube, the negatively charged particles are always the same as reflected by the same e/m ratio by J.J. Thomson.

These particles constituting cathode rays have been named electron by J.J. Thomson. The fact that electrons are independent of the nature of source from which they come suggest that they are constituent of all elements.

Thomson determined e/m ratio of electrons by applying electric and magnetic field

2.1b METHOD OF EXPERIMENTAL DETERMINATION OF E/M

Anode +(metal plugs with slits)

Tube was evacuatedthrough here and thensealed off.

Cathode – Cathode ray

–

Scale pasted on outside of tubeto measure deflection of cathode ray

3 Magnetic only1 Both or neither2 Electric only

Aluminum plates(5 cm × 2 cm, 1.5 cm apart)

Determination of ratio of electronem

em

of electron = 1.7588 × = 1.7588 × 1110 C/ kg 810 C / g

2.1C DETERMINATION OF CHARGE OF ELECTRON

An American physicist R. A. Millikan determined charge on electron by oil drop experiment Charge on electron = 1. coulombs, from the data obtained from the above two experiments, mass of electron can be calculated.

–196022 10×

–28em 9.1 10 gem

= = ×

Eyepiece of ‘‘telescope’’with three equally spacedcrosshairs

‘‘Telescope’’ (low-power magnifier)

Brass plates 22 cm indiameter and 1.6 cm apart

Drop under observation

Oil was sprayed infine drops.

AirPinhole with cover

X-rays

Determination of charge of electron


3



2.2 POSITIVE RAYS (ANODE RAYS) OR CANAL RAYS Since electron is an essential constituent of atom and since atom as a whole is electrically

neutral. it follows that an equal magnitude of positive charge must be present in an atom. In 1886 Goldstein discovered that not only cathode rays composed of electrons moving towards anode are present but positive rays moving in opposite direction are also present. By perforating the cathode these positive rays can be made to pass through the hole.

2.2a PROPERTIES OF ANODE RAYS

(i) They are deflected in electric and magnetic field just as cathode rays but in the opposite direction showing that they consist of positively charged particles.

(ii) the charge and mass of the particles constituting anode rays vary and depend upon the nature of gas contained in the discharge tube.

(iii) Positive rays are not emitted by anode but originate between the electrodes from the ionization of gas atoms.

The mass of positively charged particle is virtually the same as that of the atoms from which they are derived, Wein (1898) determined the value of e/m for positive particles and found that it was different for different gases. When hydrogen gas was taken in the discharge tube, the lightest positively charge particle was found. It’s mass was nearly the same as that of H atom and carried a positive charge exactly equal and opposite to that of an electron. This positively charged particle was named proton.

–24(1.673 10 g)×

2.3 DISCOVERY OF NEUTRON

The electrically neutral charge particle, neutron was discovered by James Chadwick by bombarding boron or beryllium with α -particles.

9 4 124 2 6Be He C n+ → + 1

0

2.4 CHARACTERISTICS OF THE THREE FUNDAMENTAL PARTICLES ARE Electron Proton Neutron

Symbol e or –1e P n

Approximate relative mass 1/1836 1 1 Approximate relative charge –1 +1 No charge Mass in kg –319.109 10× –271.673 10× –271.675 10× Mass in amu –45.485 10× 1.007 1.008

Actual charge (coulomb) –191.602 10× –191.602 10× 0

Actual charge (e.s.u.) –104.8 10× –104.8 10× 0


4



3. VARIOUS ATOMIC MODELS

3.1 THOMSON’S MODEL

Based on these discoveries J.J. Thomson proposed the Plum Pudding Model. This model proposed that atoms are blobs of a positively charged jelly like material, with electrons suspended in it like raisins in a pudding.

3.2 RUTHERFORD’S MODEL

Rutherford allowed a narrow beam of α–particles to fall on a very thin gold foil (thickness = 4 × 10–4 cm) and

he observed (i) Majority of the -particles pass straight through the gold strip with little or no

deflection. α

(ii) Some -particles are deflected from their path and diverge. (iii) Very few -particles are deflected backwards through angles greater than 90°. (iv) Some were even scattered in the opposite direction at an angle of 180° [Rutherford was

very much surprised by it and remarked that "It was as incredible as if you fired a 15 inch shell at a piece of tissue paper and it came back and hit you"].

3.2a CONCLUSIONS 1. The fact that most of the particles passed straight through the metal foil indicates the

most part of the atom is empty. α

2. The fact that few particles are deflected at large angles indicates the presence of a heavy positively charged body i.e., for such large deflections to occur -particles must have come closer to or collided with a massive positively charged body, and he named it nucleus.

αα

3. The fact that one in 20,000 have deflected at 180° backwards indicates that volume occupied by this heavy positively charged body is very small in comparison to total volume of the atom.


5



3.2b ATOMIC MODEL

Rutherford proposed an atomic model as follows. (i) All the protons (+ve charge) and the neutrons (neutral charge) i.e. nearly the total mass

of an atom is present in a very small region at the centre of the atom. The atom's central core is called nucleus.

(ii) The size of the nucleus is very small in comparison to the size of the atom. Diameter of the nucleus is about 10 while the atom has a diameter of the order 1 of cm. So, the size of atom is 10 times more than that of nucleus.

–13

5

–80

(iii) Most of the space outside the nucleus is empty. (iv) The electrons, equal in number to the net nuclear positive charge, revolve around the

nucleus with high speed in various circular orbits. (v) The centrifugal force arising due to the high speed of an electron balances the

columbic force of attraction of the nucleus and the electron remains stable in its path. Thus according to him atom consists of two parts (a) nucleus and (b) extra nuclear part.

3.2C DEFECTS OF RUTHERFORD'S ATOMIC MODEL 1. Position of electrons : The exact positions of the electrons from the nucleus are not

mentioned. 2. Stability of the atom : Neils Bohr pointed out that Rutherford's atom should be highly

unstable. According to the law of electro-dynamics, the electron should therefore, continuously emit radiation and lose energy. As a result of this a moving electron will come closer and closer to the nucleus and after passing through a spiral path, it should ultimately fall into the nucleus.

It was calculated that the electron should fall into the nucleus in less than10 sec. But it is known that electrons keep moving outsided the nucleus.

–8

To solve this problem Neils Bohr proposed an improved form of Rutherford's atomic model. Before going into the details of Neils Bohr model we would like to introduce you some

important atomic terms.

3.3 SOME IMPORTANT CHARACTERISTICS OF A ELECTROMAGNETIC WAVE

A wave is a sort of disturbance which originates from some vibrating source and travels outward as a continuous sequence of alternating crests and troughs. Every wave has five important characteristics, namely, wavelength (λ), frequency (ν), velocity (c), wave number ( )ν and amplitude (a).

Magnetic fieldcomponent

Directionof

travel

Electro magnetic wave


6



Ordinary light rays, X–rays,γ–rays, etc. are called electromagnetic radiations because similar waves can be produced by moving a charged body in a magnetic field or a magnet in an electric field. These radiations have wave characteristics and do not require any medium for their propagation. (i) Wave length (λ) : The distance between two neighbouring troughs or crests is known

as wavelength. It is denoted by λ and is expressed in cm, m, nanometers (1nm=10–9m) or Angstrom (1Å=10–10m).

(ii) Frequency (ν) : The frequency of a wave is the number of times a wave passes through a given point in a medium in one second. It is denoted by ν(nu) and is expressed in cycles per second (cps) or hertz (Hz) 1Hz = 1cps.

The frequency of a wave is inversely proportional to its wave length (λ)

ν ∝ 1λ

or ν = cλ

(iii) Velocity : The distance travelled by the wave in one second is called its velocity. It is denoted by c and is expressed in cm sec–1.

c = νλ or λ = cν

(iv) Wave number ( : It is defined as number of wavelengths per cm. It is denoted by )ν

ν and is expressed in cm–1.

ν = 1λ

(or) ν = cν

(v) Amplitude: It is the height of the crest or depth of the trough of a wave and is denoted by a. It determines the intensity or brightness of the beam of light.

If all the components of Electromagnetic Radiation (EMR) are arranged in order of decreasing or increasing wavelengths or frequencies, the pattern obtained is known as Electromagnetic Spectrum. The following table shows all the components of light.

S.No. Name Wavelength Frequency(Hz) Source 1. Radio wave 14 73 10 – 3 10× × 5 91 10 –1 10× × Alternating current of

high frequency 2. Microwave 7 63 10 – 6 10× × 9 11 10 – 5 10× × 1 Klystron tube 3. Infrared (IR) 66 10 – 7600× 11 165 10 – 3.95 10× × Incandescent objects 4. Visible 7600–3800 16 143.95 10 – 7.9 10× × Electric bulbs, sun rays 5. Ultraviolet(UV) 3800–150 14 167.9 10 – 2 10× × Sun rays, arc lamps with

mercury vapours 6. X-Rays 150–0.1 16 192 10 – 3 10× × Cathode rays striking

metal plate 7. γ -Rays 0.1–0.01 19 203 10 – 3 10× × Secondary effect of

radioactive decay 8. Cosmic Rays 0.01–zero 203 10× –Infinity Outer space


7



3.4 ATOMIC SPECTRUM Electromagnetic spectrum consist of radiations of different wavelength and frequencies. An

instrument used to separate the radiations of different wavelength is called spectroscope or spectrograph. It consist of a prism or diffraction grating for dispersion of radiations and a telescope to examine the emergent radiations. In spectroscope the telescope is replaced by photographic film. 1. Emission Spectrum: When the radiation emitted from some source e.g. from the sun

or by passing electric discharge to high temperature is passed through prism and then received on photographic plate the spectrum obtained is known as emission spectrum.

Emission spectra are as two types: (a) Continuous Spectra : When white light from any source such as sun a bulb or hot

glowing body is analysed by passing through the prism. It is observed that it splits up into seven different wide band of colours. These colours are so continuous that each merges into the next. Hence spectrum is continuous spectrum.

(b) Line Spectra : When some volatile salt like NaCl is placed in Bunsen Burner or an electric discharge is passed through a gas. Light is emitted if this light is resolved and passed in a spectroscope. It is found that no continuous spectrum is obtained but isolated lines are obtained. This spectrum is called line spectrum. Each line corresponds to a wavelength.

(2) Absorption Spectra : When white light from any source is first passed through the solution or vapours of a chemical substance and than analysed by spectroscope. It is observed that some dark lines are obtained. These dark lines are supposed to result from the fact that when white light is passed through chemical substance radiations of certain wavelength are absorbed. Dark lines are obtained at same place where coloured lines were obtained in emission spectra.

3.4a HYDROGEN SPECTRUM

If an electric discharge is passed through hydrogen gas taken in a discharge tube under low pressure, and the emitted radiation is analysed with the help of spectrograph, it is found to consist of a series of sharp lines in the UV, visible and IR regions. This series of lines is known as line or atomic spectrum of hydrogen. The lines in the visible region can be directly seen on the photographic film.

Each line of the spectrum corresponds to a light of definite wavelength. The entire spectrum consists of six series of lines each series, known after their discoverer as the Balmer, Paschen, Lyman, Brackett, Pfund and Humphrey series. The wavelength of all these series can be expressed by a single formula.

2 21 1

1 1R –n n

= ν = λ

1

ν = wave number = wave length λ R = Rydberg constant (109678 ) –1cm and have integral values as follows 1n 2n


8



Series 1n 2n Main spectral lines Lyman 1 2, 3, 4, etc Ultra-vio Balmer 2 3, 4, 5 etc Visible Paschen 3 4, 5, 6 etc Infra-red Brackett 4 5, 6, 7 etc Infra-red Pfund 5 6, 7, 8, etc Infra-red

Note : All lines in the visible region are of Balmer series but reverse is not true, i.e., all Balmer lines will not fall in visible region

The pattern of lines in atomic spectrum is characteristic of hydrogen.

3.5 PLANCK’S QUANTUM THEORY OF RADIATION In order to explain the spectral distribution of energy radiated by a back body. Max Planck

(1901) put forward quantum theory of radiation. According to this theory. (i) Radiant energy is emitted or absorbed discontinuously in the form of tiny bundles or

packets known as quanta or photon. (ii) The energy of a quantum is directly proportional to the frequency of the radiation

E ∝ v E = hv v = frequency of radiation h = planck’s constant

(iii) A body can emit or absorb energy only in whole number multiples of quantum i.e 1hv, 2hv, etc.

Energy in fractions of a quantum cannot be lost or absorbed. this is known as quantization of energy.

Illustration 1: Calculate the frequency and wave number of radiation with wavelength 480 nm.

Solution : Given λ = 480 nm = 480 × 10–9 m [1 nm = 10–9 m] c = 3 × 108 m/s

Frequency, 8 1

14 19

c 3 10 ms 6.248 10

−−

−

×= = = ×

λ ×v = 6.25 × 105 10 s 14 Hz.

Illustration 2: Find the quantum numbers of the excited state of electrons in He+ ion which on transition to ground state and first excited state emit two photons of wavelengths. 30.4 nm and 108.5 nm respectively. (RH = 1.09678 × 107 m–1).

Solution : 2H 2 2

1 2

1 1R Zn n

= − λ

1

n1 = 1, n2 = ?


9



79 2

1 2

1.09678 10 430.4 10 1 n−

= × × −

2

11 1×

∴n 2 = 2

n1= 2 (first excited state), n2 = ?

79 2

2

1.09678 10 4108.5 10 2 n−

= × × −×

2

1

1 1 n2 = 5

Illustration 3: In the measurement of the quantum efficiency of photosynthesis in green plants, it was found that 8 quanta of red light at 6850 Å were needed to release one molecule of O2. The average energy storage in the photosynthesis process is 112 kcal/mol O2 released. What is the energy conversion efficiency in the photosynthetic process ?

Solution : Energy possessed by one quanta, 1

hcE =λ

= 34 8

1910

625 10 3 10 2.9 10 J / molecule6850 10

−−

−

× × ×= ×

×6.

Average energy storage in photosynthesis = 112 kcal/mol

= 3

1923

10 4.2 7.81 10 J / molecule6.023 10

−× ×= ×

×112

Energy conversion efficiency = Average energy storage 100Energy of 8 quanta

×

= 19

19

81 10 100 33.66%8 2.9 10

−

−

×× =

× ×7.

EXERCISE – 1 1. Calculate the energy associated with photon of light having a wavelength 6000 Å.

2. How many photons of light having a wavelength 4000 Å are necessary to provide 1.00 J of energy?

3. How many moles of photon would contain sufficient energy to raise the temperature of 225 g of water 21°C to 96°C? Specific heat of water is 4.18 J g–1 K–1 and frequency of light radiation used is 2.45 × 109 s–1.

3.6 BOHR’S ATOMIC MODEL Bohr developed a model for hydrogen and hydrogen like atoms one-electron species

(hydrogenic species). He applied quantum theory in considering the energy of an electron bond to the nucleus.


10



3.6a IMPORTANT POSTULATES

An atom consists of a dense nucleus situated at the center with the electron revolving around it in circular orbits without emitting any energy. The force of attraction between the nucleus and an electron is equal to the centrifugal force of the moving electron.

Of the finite number of circular orbits possible around the nucleus, and electron can revolve only in those orbits whose angular momentum (mvr) is an integral multiple of factor h/ . 2π

nhmvr2

=π

where, m = mass of the electron v = velocity of the electron

n = orbit number in which electron is present

r = radius of the orbit

As long as an electron is revolving in an orbit it neither loses nor gains energy. Hence these orbits are called stationary states. Each stationary state is associated with a definite amount of energy and it is also known as energy levels. The greater the distance of the energy level from the nucleus, the more is the energy associated with it. The different energy levels are numbered as 1, 2, 3, 4, (from nucleus onwards) or K, L, M,N etc.

Ordinarily an electron continues to move in a particular stationary state without losing energy. Such a stable state of the atom is called as ground state or normal state.

If energy is supplied to an electron, it may jump (excite) instantaneously from lower energy (say 1) to higher energy level (say 2, 3, 4, etc) by absorbing one quantum of energy. This new state of electron is called as excited state. The quantum of energy absorbed is equal to the difference in energies of the two concerned levels.

Since the excited state is less stable, atom will lose it’s energy and come back to the ground state.

Energy absorbed or released in an electron jump, is given by ( E)∆

2 1E E – E hv∆ = =

Where E and are the energies of the electron in the first and second energy levels, and v is the frequency of radiation absorbed or emitted.

2 1E

Note: If the energy supplied to hydrogen atom is less than 13.6 eV, it will accept or absorb only those quanta which can take it to a certain higher energy level i.e., all those photons having energy less than or more than a particular energy level will not be absorbed by hydrogen atom. But if energy supplied to hydrogen atom is more than 13.6 eV then all photons are absorbed and excess energy appear as kinetic energy of emitted photo electron.

3.6b MERITS OF BOHR’S THEORY (i) The experimental value of radii and energies in hydrogen atom are in good agreement

with that calculated on the basis of Bohr’s theory.


11



(ii) Bohr’s concept of stationary state of electron explains the emission and absorption spectra of hydrogen like atoms.

(iii) The experimental values of the spectral lines of the hydrogen spectrum are in close agreement with the calculated by Bohr’s theory.

3.6C LIMITATIONS OF BOHR’S THEORY (i) It does not explain the spectra of atoms or ions having more than one electron. (ii) Bohr’s atomic model failed to account for the effect of magnetic field (Zeeman effect)

or electric field (Stark effect) on the spectra of atoms or ions. It was observed that when the source of a spectrum is placed in a strong magnetic or electric field, each spectral line further splits into a number of lines. This observation could not be explained on the basis of Bohr’s model.

(iii) de-Broglie suggested that electrons like light have dual character. It has particle and wave character. Bohr treated the electron only as particle.

(iv) Another objection to Bohr’s theory came from Heisenberg’s Uncertainty Principle. According to this principle “it is impossible to determine simultaneously the exact position and momentum of a small moving particle like an electron”. The postulate of Bohr, that electrons revolve in well defined orbits around the nucleus with well defined velocities is thus not attainable.

3.6d BY BOHR’S THEORY (i) Radius and Energy levels of hydrogen atom: Consider an electron of mass `m’ and

charge `e’ revolving around a nucleus of charge Ze (where, Z = atomic number and e is the charge of the proton) with a tangential velocity v.r is the radius of the orbit in which electron is revolving.

By Coulomb’s Law, the electrostatic force of attraction between the moving electron

and nucleus is Coulombic force 2

2

KZer

=

0

1K4

=πε

(where is permitivity of free space) 0ε

9 2K 9 10 Nm C= × –2

2 In C.G.S. units, value of K = 1 dyne 2 –cm (esu)

The centrifugal force acting on the electron is 2mv

r

Since the electrostatic force balance the centrifugal force, for the stable electron orbit.

2 2

2

mv KZer r

= … (i)

(or) 2

2 KZemr

v = … (ii)

According to Bohr’s postulate of angular momentum quantization, we have


12



nhmvr2

=π

nhv2 mr

=π

2 2

22 2 2

n hv4 m r

=π

… (iii)

Equating (2) and (3)

2 2 2

2 2 2

KZe n hmr 4 m r

=π

Solving for r we get 2 2

2 2

n hr4 mKZe

=π

where n = 1, 2, 3, … ∞ Hence only certain orbits whose radii are given by the above equation are available for

the electron. The greater the value of n, i.e., farther the level from the nucleus the greater is the radius.

The radius of the smallest orbit (n = 1) for hydrogen atom (Z = 1) is . 0r

2 2 2 –34 2

–110 2 2 2 –31 –19 2 9

n h 1 (6.626 10 )r 54 me K 4 (3.14) 9 10 (1.6 10 ) 9 10

× ×= = = ×

π × × × × × × ×.29 10 m

0r 0.5629Å=

Radius of nth orbit for an atom with atomic number Z is simply written as

2 2

n 0n nr r 0.529 Åz z

= = ×

(ii) Energy level of Hydrogen atom: The total energy, E of the electron is the sum of kinetic energy and potential energy. Kinetic energy of the electron = 21

2 mv

Potential energy 2Ze

r–K

=

Total energy 2

2 KZe–r

12 mv= … (iv)

From equation (1) we know that

2 2

2

mv KZer r

=

2

212

KZemv2r

∴ =

Substituting this in equation (iv)

Total energy (e) 2 2KZe KZe KZe– –

2r r 2r= =

2


13



1KE – PE, KE –TE2

= =

Substituting for r, gives us

2 2 4

2 2

2 mZe e KEn h

π=

2

where n = 1, 2, 3, …

This expression shows that only certain energies are allowed to the electron. Since this energy expression consists so many fundamental constant, we are giving you the following simplified expressions.

2

–122

ZE –21.8 10 ergn

= × × per atom.

2

–192

Z– 21.8 10n

= × × J per atom = –2

2

Z6n

×13. eV per atom

2

n 2

zE –13.6n

= eV per atom

(1eV = 3. –2383 10 Kcal)×

(1eV = 1. erg) –12602 10×

(1eV = 1. J) –19602 10×

2

2

ZE –313.6n

= × kcal/mole (1 cal = 4.18 J)

The energies are negative since the energy of the electron in the atom is less than the energy of a free electron (i.e., the electron is at infinite distance from the nucleus) which is taken as zero. The lowest energy level of the atom corresponds to n = 1, and as the quantum number increases, E become less negative.

When n = , E = 0 which corresponds to an ionized atom i.e., the electron and nucleus are infinitely separated.

∞

(ionization). –H H+→ + e(iii) Velocity of electron

We know that, mvr nh nh; v2 2

= =π πmr

By substituting for r we are getting

22 KZev

nhπ

=

Where excepting n and z all are constant, v = 8 Z2.18 10 cm / sec.n

×

Further application of Bohr’s work was made, to other one electron species (Hydrogenic ion) such as and . In each case of this kind, Bohr’s prediction of the spectrum was correct.

He+ 2Li +


14



(iv) Explanation for hydrogen spectrum by Bohr’s theory: According to the Bohr’s theory electron neither emits nor absorbs energy as long as it stays in a particular orbit. However, when an atom is subjected to electric discharge or high temperature, and electron in the atom may jump from the normal energy level, i.e., ground state to some higher energy level i.e., excited state. Since the life time of the electron in excited state is short, it returns to the ground state in one or more jumps.

During each jumps, energy is emitted in the form of a photon of light of definite wavelength or frequency. The frequency of the photon of light thus emitted depends upon the energy difference of the two energy levels concerned ( ) and is given by 1 2n , n

2 4 2

2 1 2 22 1

–2 mZ e K 1 1hv E – E –h n

2 π= =

2n

2 2 4 2

3 21 2

2 mZ e K 1 1v –h n

π=

2n

The frequencies of the spectral lines calculated with the help of above equation are found to be in good agreement with the experimental values. Thus, Bohr’s theory elegantly explains the line spectrum of hydrogen and hydrogenic species.

Bohr had calculated Rydberg constant from the above equation.

2 2 4 2

3 21 2

C 2 mZ e K 1 1–h n n

πν = = λ

2

2 2 4 2

3 21 2

1 2 mZ e K 1 –h c n n

π= ν = λ

2

1

where 2 4 2

–7 –1 –13

e K 1.097 10 m or 109678 cmh c

π= ×

2 m

i.e. Rydberg constant (R) 22 21 2

1 1RZ –n n

= = λ

1∴ ν

ν = wave number. Illustration 4: Calculate the ratio of the radius of Li+2 ion in 3rd energy level to that of He+

ion in 2nd energy level.

Solution: 2n

zr ∝

2

1 1

2 2

zr n z

=

2

1

3

r n

n 3 1 =

n 2 2 =

z (for Li ) 1 = 2+


15



z (for He ) 2 = 2 +

2

1

2

2r 2 3

= × =

r 3 32

Illustration 5: How far from the nucleus is the electron in a hydrogen atom if it has energy of –0.850 eV.

Solution: 2

–19n 2

Z– 21.69 10n

= × ×E

2

–19 –192 21.69 10 –0.85 1.6 10

n= × ×

Z– × ×

Illustration 6: Find the wavelengths of the first line of He+ ion spectral series whose

interval between extreme lines is 4 –

1 2

1 1– 2.7451 10= × cmλ λ

1

Solution: Extreme lines means first and last.

2 22 2 2 2

2 1 1 1

1 1 1 1 1 1– RZ – – R –n n (n

= λ λ ∞ + n)

2

21

RZ(n 1)

=+

2

42

1

109677.76 22.7451 10(n 1)

×× =

+

1(n 1) 4+ =

1n 3=

Wavelength of first line,

22 2

1 1109677.76 2 –3 4

= × × λ 1

–84689 10 cm 4689Åλ = × =

Illustration 7 : The velocity of electron in a certain Bohr's orbit of hydrogen atom bears the ratio of 1 : 275 to the velocity of light

(a) What is the quantum number (n) of orbit? (b) Calculate the wave number of radiations emitted when electron jumps

from (n+1) state to ground state

Solution : Velocity of electron = velocity light × 1275

= 83 10 m/sec

275× = 1.09 × 106 m/sec

Vn = 22 ZKe

nhπ =

62.18 10 Zn

× × m/sec


16



For hydrogen Vn = 62.18 10

n× m/sec

∴ 1.09 × 106 = 62.18 10

n× ⇒ n = 2

When electron jumps from (n+1) i.e., 3 to ground state

1ν =

λ= RH = 2 2

1 11 3

−

⇒ 1λ

= 1.09678 × 107 1 11 9

− = 9.75 × 102 m–1

Illustration 8: Two hydrogen atoms collide head on and end up with zero kinetic energy. Each then emits a photon of wave length 121.6 nm which transition leads to this wavelength? How fast were the hydrogen atoms travelling before collision?

RH = 1.09678 × 107 m–1 MH = 1.672 × 10–27 Kg Solution: As wavelength of emitted radiation is 121.6 nm, which falls in UV region of

spectrum Therefore, n1 = 1

For hydrogen atom H 2 2

1 1R1 n

= − λ

1

∴ 9

1121.6 10−×

= 1.09678 × 107 2

1 11 n

−

n = 2 Energy released is due to collision and all the kinetic energy released in form

of radiant energy

12

mv2 = hcλ

⇒ 12

1.67 × 10–27 × V2 = 34 8

9

6.625 10 3 10121.6 10

−

−

× × ××

∴ V = 4.42 × 104 m/sec

EXERCISE –2

1. Calculate the radii of the first two Bohr orbits of Li2+.

2. Ionisation energy of H-atom is 13.6 eV. Calculate ionization energy of He+, Li2+ and Be3+ (all isoelectronic of H).

3. If the electron in a hydrogen atom were in the energy level with n = 3, how much energy would be required to ionize the atom ?


17



4. PHOTOELECTRIC EFFECT Emission of electrons from a metal surface when exposed to light radiations of appropriate

wavelength is called photoelectric effect. The emitted electrons are called photoelectrons. Work function or threshold energy may be defined as the minimum amount of energy

required to eject electrons from a metal surface. According to Einstien,

Maximum kinetic energy of the ejected electron = absorbed energy – work function

1 22 max 0

0

1 1mv hv hv hc

= − = − λ λ

where v0 and λ0 are threshold frequency and threshold wavelength.

4.1 LAWS OF PHOTOELECTRIC EFFECT (i) Rate of emission of photoelectrons from a metal surface is directly proportional to the

intensity of incident light. (ii) The maximum kinetic energy of photoelectrons is directly proportional to the

frequency of incident radiation; moreover, it is independent of the intensity of light used.

(iii) There is no time lag between incidence of light and emission of photoelectrons. (iv) For emission of photoelectrons, the frequency of incident light must be equal to or

greater than the threshold frequency. Illustration 9 : Photo - electrons are liberated by ultraviolet light of wavelength 3000Å from

a metallic surface for which the photoelectric threshold is 4000Å. Calculate de-Broglie wavelength of electrons emitted with maximum kinetic energy

Solution: As, ν = hν° + KE ⇒ o

hc hc K.E= +λ λ

∴ .E. = hc o

o

1 1 hc(λ − λ − = λ λ λ × λ

) =

34 8 10 10

10 10

625 10 3 10 (4000 10 3000 10 )3000 10 4000 10

− −

− −

× × × × − ×× × ×

6.=

−

6565 × 10–19 Joule

Also 1 mν2

2 = 1.6565 × 10–19

⇒ m2v2 = 2 × 1.65665 × 10–19 × m

⇒ v2 = 2 × 1.6565 × 10–19 × 9.109 × 10–31 ⇒ mν = 5.49 × 10–25

= 34

25

6.625 10mv 5.49 10

−

−

×=

×h = 1.2 × 10–9 m


18



Illustration 10 : Photo - electrons are liberated by ultraviolet light of wavelength 3000Å from a metallic surface for which the photoelectric threshold is 4000Å. Calculate de-Broglie wavelength of electrons emitted with maximum kinetic energy

Solution: As, hν = hν° + KE ⇒ o

hc hc K.E= +λ λ

∴ K.E. = hc o

o

1 1 hc(λ − λ− =

λ λ λ × λ

) =

= 34 8 10 10

10 10

625 10 3 10 (4000 10 3000 10 )3000 10 4000 10

− −

− −

× × × × − ×× × ×

6. −

=1.6565 × 10–19 Joule

Also 1 mν2

2 = 1.6565 × 10–19

⇒ m2v2 = 2 × 1.65665 × 10–19 × m ⇒ m2v2 = 2 × 1.6565 × 10–19 × 9.109 × 10–31

⇒ mν = 5.49 × 10–25

λ = 34

25

h 6.625 10mv 5.49 10

−

−

×=

× = 1.2 × 10–9 m

5. PARTICLE AND WAVE NATURE OF ELECTRON In 1924, de Broglie proposed that an electron, like light, behaves both as a material particle

and as a wave. This proposal gave birth to a new theory known as wave mechanical theory of matter. According to this theory, the electrons, protons and even atoms, when in motion, possess wave properties.

de Broglie derived an expression for calculating the wavelength of the wave associated with the electron.

According to Planck’s equation

cE hv h= = ⋅λ

…(i)

The energy of a photon on the basis of Einstein’s mass–energy relationship is E = mc2 …(ii)

where c is the velocity of the electron. Equating both the equations, we get

2ch m=λ

c

h hmc p

λ = =


19



Momentum of the moving electron is inversely proportional to its wavelength. Let kinetic energy of the particle of mass ‘m’ is E.

E = ½ mv2 2Em = m2v2

2Em = mv = p (momentum)

h hp 2Em

λ = =

5.1 DAVISSON AND GERMER MADE THE FOLLOWING MODIFICATION IN DE BROGLIE EQUATION

Let a charged particle say an electron be accelerated with a potential of V; then the kinetic energy may be given as :

½mv2 = eV m2v2 = 2eVm

mv 2eVm p= = h

2eVmλ =

de Broglie waves are not radiated into space, i.e. they are always associated with electron. The wavelength decreases if the value of mass (m) increases, i.e., in the case of heavier particles, the wavelength is too small to be measured. de Broglie equation is applicable in the case of smaller particles like electron and has no significance for larger particles.

5.2 BOHR’S THEORY VERSUS DE BROGLIE EQUATION

One of the postulates of Bohr’s theory is that angular momentum of an electron is an

integral multiple of h2π

. This postulate can be derived with the help of de Broglie concept

of wave nature of electron.

Consider an electron moving in a circular orbit around nucleus. The wave train would be associated with the circular orbit as shown in figure.

(a) (b)

(a) The number of waves around the orbit is a whole number (6) times the wavelength.

(b) The number of waves around the orbit is not a whole number multiple of the wavelength.

When the wave goes around the orbit many times, it cancles itself out.


20



If the two ends of the electron wave meet to give a regular series of crests and troughs, the electron wave is said to be in phase, i.e., the circumference of Bohr’s orbit is equal to whole number multiple of the wavelength of the electron wave.

So, 2πr = n λ

or 2 rnπ

λ =

From de Broglie equation,

hmv

λ =

Thus, h 2mv n

π=

r

or hmvr n2

= ⋅π

[v = velocity of electron and r = radii of the orbit]

i.e., Angular momentum = hn2

⋅π

This proves that the de Broglie and Bohr–concepts are in perfect agreement with each other. Illustration 11 : What is the mass of a photon of sodium light with a wavelength of 5890 Å.

Solution: hmc

λ =

or hc

=λ

m

So, 27

8

6.63 105890 10 3 10

−

−

×× × × 10m = = 3.752 × 10–33 g

Illustration 12 : An electron beam can undergo diffraction by crystals. Through what potential should a beam of electrons be accelerated so that its wavelength becomes equal to 1.54 Å

Solution : If an electron accelerated by potential V, has velocity 'u' and associated de Broglie's wavelength λ.

1/2 mν2 = eV … (i)

λ = hmu

… (ii)

From equation (ii) hmλ

µ =


21



∴ 1/2 m 2h

mλ

= e × V ⇒ V =

2

2

h2m cλ

= 34 2

31 10 2 19

(6.625 10 )2 9.108 10 (1.54 10 ) 1.602 10

−

− −

×× × × × × × − = 63.3 volt

Illustration 13 : As the speed of a particle of mass m increases, its de Borglie wavelength decreases. What is the rate of change of λ with the kinetic energy, E for an

electron with = 1 A ? λ

λ

Solution: According to de Broglie.

Matter wavelength, h hλ =

mv p=

We have, kinetic energy

2 2

22

1 p hE mv2 2m (2

= = =λ m)

∴ h2mE

λ =

The rate of change of de Broglie wavelength with the kinetic energy E, λ

ddE

λ =1/ 2h d(E)

dE2m

−

= 3

2

1 h2 h2mE E

λ− × = −

×m

10 3 31

6 -34 2

at 14

d (1 10 ) 9.1 10 2.07×10 m JdE (6.62 10 )

− −

−λ =

λ × × × = − = × 1

EXERCISE – 3 1. What accelerating potential must be imparted to a proton beam to give it an effective

wavelength of 0.050Å ? 2. A 1.0 g projectile is shot from a gun with velocity of 100 m s–1. What is the de Broglie

wavelength ? 3. Two particles A and B are in motion; if wavelength of A is 5 × 10–8 m, calculate

wavelength of B so that its momentum is double of A.

6. HEISENBERG UNCERTAINTY PRINCIPLE For heavier objects, the velocity and their position can be measured accurately at any instant

of time but same is not the case for subatomic particles. As a consequence of dual nature of matter, Heisenberg in 1927 gave a principle about the uncertainties in simultaneous measurement of position and momentum of small particles.

This principle states that “It is impossible to measure simultaneously the position and momentum of a small

microscopic moving particle with absolute accuracy or certainty i.e. if an attempt is made to


22



measure any one of these two quantities with higher accuracy the other becomes less accurate”

Mathematical expression for the Heisenberg uncertainty principle is simply written as

x . p h / 4∆ ∆ ≥ π

where ∆ = p m v∆

Explaination of Heisenberg uncertainty Principle : Suppose we attempt to measure both the position and momentum of an electron, to pinpoint the position of the electron we have to use light so that the photon of light strikes the electron and the reflected photon is seen in the microscope. As a result of the hitting, the position as well as the velocity of the electron are disturbed. The accuracy with which the position of the particle can be measured depends upon the wavelength of light used the uncertainty in position is ±λ . The shorter the wavelength, the greater is the accuracy. But shorter wavelength means higher frequency and hence higher energy. So if position is fixed velocity will not be fixed.

Illustration 14 : Why electron cannot exist inside the nucleus according to Heisenberg’s uncertainty principle?

Solution : Diameter of the atomic nucleus is of the order of 10–15m

The maximum uncertainty in the position of electron is 10–15 m.

Mass of electron = 9.1 ×10–31 kg.

∆x. ∆p = h4π

∆x × (m.∆v) = h/4π

∆v = h 14 x.

×π ∆ m

= 346.63 10

2247

−×

× × 15 31

110 9.1 10− −× ×

∆v = 5.80 × 1010 ms–1

This value is much higher than the velocity of light and hence not possible.

EXERCISE – 4 1. Calculate the wavelength of a soft ball of mass of 100 g traveling at a velocity of

35 ms–1. 2. If uncertainties in the measurement of position and momentum are equal, calculate

uncertainty in the measurement of velocity. 3. Calculate the uncertainty in the velocity of wagon of mass 2000 kg whose position is

known to an accuracy of ± 10 m.


23



7. SCHRODINGER WAVE EQUATION

2

22

8 m (E U) 0hπ

∇ ψ + − ψ =

This equation implies in essence that a body of mass m, velocity v, potential energy U and total energy E has a wave associated with it of an amplitude given by wave function Ψ. The schrodinger equation has an infinite many solution. For any physical situation Ψ must be finite, single valued and continuous. The particular values of Ψ which yield satisfactory solution of above equation are called wave functions. Wave function for an electron is called an atomic orbital. Each solution is identified by three quantum numbers. A set of three quantum numbers is needed to describe an electron because electrons in atoms are moving in three dimensional space.

2ψSchrodinger equation

Wave functionψ

SolveProbability of finding electron

for electron this wave function is known as atomic orbital

The wave function Ψ has got no meaning but the square of the wave function Ψ2 is a

mathematical expression of how the probability of finding an electron in a small volume varies from place to place. Square of wave function is like the square of amplitude of an electromagnetic wave. The intensity of electromagnetic radiation is proportional to square of amplitude.

8. QUANTUM NUMBERS To understand the concept of Quantum Numbers, we must know the meaning of some terms

clearly so as to avoid any confusion.

Energy Level: The non-radiating energy paths around the nucleus are called as Energy Levels of Shells. These are specified by numbers having values 1, 2, 3, 4, ... or K, L, M, N, ... in order of increasing energies. The energy of a particular energy level is fixed.

Sub-Energy Level: The phenomenon of splitting of spectral lines in electric and magnetic fields reveals that there must be extra energy levels within a definite energy level. These were called as Sub-Energy Levels or Sub-Shells. There are four types of sub-shells namely; s, p, d, f.

First energy level (K or ) has one sub-shell designated as 1s, the second energy level (L or 2) has two sub-shell as 2s & 2p, the third energy level (M or 3) has three sub shell as 3s, 3p and 3d, and the fourth energy level (N or 4) has four sub-shells as 4s, 4p, 4d and 4f. The energy of sub-shell increases roughly in the order: s < p < d <f.

Orbital: Each sub-energy level (sub-shell) is composed of one or more orbitals. These orbitals belonging to a particular sub-shell have equal energies and are called as degenerate orbitals. s-sub-shell has one orbital, p has three orbitals, d have five orbitals and f has seven orbitals.


24



To describe or to characterize the electrons around the nucleus in an atom, a set of four numbers is used, called as Quantum Numbers. These are specified such that the states available to the electrons should follows the laws of quantum mechanics or wave mechanics.

8.1 PRINCIPAL QUANTUM NUMBER: (n)

This quantum number represents the main energy levels (principal energy levels) designated as n = 1, 2, 3, ... or the corresponding shells are named as K, L, M, N, ... respectively. It gives an idea of position and energy of an electron. The energy level n = 1 corresponds to minimum energy and subsequently n = 2, 3, 4, ..., are arranged in order of increasing energy.

Higher is the value of n, greater is its distance from the nucleus, greater is its size and also greater is its energy.

It also gives the total electrons that may be accommodated in each shell, the capacity of each shell is given by the formula , where n : principal quantum number. 22n

8.2 AZIMUTHAL QUANTUM NUMBER: (l)

This number determines the energy associated with the angular momentum of the electron about the nucleus. It is also called as the angular momentum quantum number. It accounts for the appearance of groups of closely packed spectral lines in electric field.

It can assume all integral values from 0 to n–1. The possible values of l are :

0, 1, 2, 3, ..., n–1. Each value of l describes a particular sub-shell in the main energy level and determines the

shape of the electron cloud. When n = 1, l = 0, i.e., its energy level contains one sub-shell which is called as a s-sub-

shell. So for l = 0, the corresponding sub-shell is a s-sub-shell. Similarly when l = 1, 2, 3, the sub-shells are called p, d, f sub-shells respectively.

As you known for n = 1, l = 0, there is only one sub-shell. It is represented by 1s. Now for n = 2, l can take two values (the total number of values taken by l is equal to the value of n in a particular energy level). The possible values of l are 0, 1. The two sub-shell representing the IInd energy level are 2s, 2p. In the same manner, for n = 3, three sub-shells are designated as 3s, 3p, 3d corresponding to l = 0, 1, 2, and for n = 4, four sub-shells are designated as 4s, 4p, 4d, 4f corresponding to l = 0, 1, 2, 3.

The orbital Angular momentum of electron = h( 1)2

+π

.

Note that its value does not depend upon value of n.

8.3 MAGNETIC QUANTUM NUMBER (m) An electron with angular momentum can be thought as an electric current circulating in a

loop. A magnetic field due to this current is observed. This induced magnetism is determined by the magnetic quantum number. Under the influence of magnetic field, the electrons in a given sub-energy level prefer to orient themselves in certain specific regions


25



in space around the nucleus. The number of possible orientations for a sub-energy level is determined by possible values of m corresponds to the number of orbitals in a given sub-energy level).

m can have any integral values between –l to +l including 0, i.e., m = –l, 0 +, l, …, 0, 1, 2, 3, 4, . . ., l–1 + l. We can say that a total of (2l + 1) values of m are there for a given value of l– 2, –1, 0, 1, 2, 3.

In s sub-shell there is only one orbital [l = 0, ⇒ m = (2l +1) = 1]. In p sub-shell there are three orbitals corresponding to three values of m : –1, 0 +1. [l = 1

m = (2l +1) = 3]. These three orbitals are represented as along X, Y, Z axes perpendicular to each other. ⇒ x yp , p , pz

In d sub-shell, there are five orbitals corresponding to –2, –1, 0 +1, +2, [l = 2 m = ( . These five orbitals are represented as . ⇒ 2 2 1) 5]× + = 2 2

2xy yz zx x –y

d ,d ,d ,d ,dz

In f sub-shell there are seven orbitals corresponding to –3, –2, –1, 0, +1, +2, +3 [l = 3 m = . ⇒ (2 3 1) 7]× + =

8.4 SPIN QUANTUM NUMBER (s) When an electron rotates around a nucleus it also spins about its axis. If the spin is

clockwise, its spin quantum number is +1/2 and is represented as ↑ . If the spin is anti-clockwise, its value is –1/2 and is represented as ↓ . If the value of s is +1/2, then by convention, we take that electron as the first electron in that orbital and if the value of s is –1/2, it is taken as second electron.

Illustration 15: Write down the values of quantum numbers of all the electrons present in the outermost orbit of argon (At. No. 18)

Solution: The electronic configuration of argon is 1 2 2 6 2 2 2 2x y zs , 2s 2p ,3s 3p 3p 3p

Values of quantum numbers are: n 1 m s

23s 3 0 0 +1/2, –1/2 2x3p 3 1 ± 1 +1/2, –1/2

2y3p 3 1 ± 1 +1/2, –1/2

2z3p 3 1 0 +1/2, –1/2

Illustration 16: (a) An electron is in 5f-orbital. What possible values of quantum numbers n, l, m and s can it have?

(b) What designation is given to an orbital having (i) n = 2, l =1, and (ii) n = 3, l = 0?

Solution: (a) For an electron in 5f-orbital, quantum number are:

n = 5, l = 3; m = –3, –2, –1, 0, +1, +2, +3 and s = either + 1 1or –2 2

(b) (i) 2p, (ii)3s


26



EXERCISE – 5

1. Give the possible values for the missing quantum numbers : (a) n = 3, l = 0, ml = ? (b) n = 3, l = ?, ml

= –1 (c) n = ?, l = 1, m1 = + 1 (d) n = ?, l = 2, ml = ? 2. Write appropriate values of n, l and ml quantum numbers for each of the following

orbital designations : (a) 4s (b) 3p (c) 5f (d) 6g (e) 3d (f) 7s Also arrange them in increasing energy. 3. Which of the following sets of quantum numbers is/are not allowable ? Why not ? (a) n = 2, l = 1, ml = 0 (b) n = 2, l = 2, ml = –1 (c) n = 3, l = 0, ml = 0 (d) n = 3, l = 1, ml = 1 (e) n = 2, l = 0, ml = –1 (f) n = 2, l = 2, ml = –1 4. What type of electron orbital is designated for : (a) n = 2, l = 1, ml = –1 (b) n = 2, l = 2, ml = –1 (c) n = 5, l = 0, ml = 0

9. RULES FOR FILLING ELECTRONS

9.1 AUFBAU PRINCIPLE The aufbau principle states that in the ground state of an atom, the orbital with a lower energy is filled up first before the

filling for the orbital with a higher energy commences. In other words : The electrons enter the various orbitals in the order of increasing energy.

1s

2s

3s

4s

5s

6s

7s

2p

3p

4p

5p

6p

3d

4d

5d

4f


27

This order of orbitals is only a guidance and exceptions are known in few cases of atoms.



9.2 PAULI'S EXCLUSION PRINCIPLE This an important generalisation given by Wolfgang Pauli, in 1925, which determines the

maximum number of electrons that an energy level can accommodate. Pauli's exclusion principle states that It is impossible for any two electrons in the same atom to have all the four quantum numbers

identical. Thus, in the same atom, any two electrons may have three quantum numbers identical but the

fourth must be different. This principle is very useful in determining the maximum number of electrons that can occur in any shell/sub shell. For K-shell, for instance, since n=1, can have only one value (viz., 0) and m can also have only one value (viz., 0). Hence s can be either +1/2 or -1/2. Thus, there are two combinations of the quantum number, as shown below:

n =1 = o m = 0 s = +1/2 n =1 = o m = 0 s = -1/2 This shows that in the K-shell, there is only one subshell = 0 and there are only two

electrons of opposite spins. For L-shell, since n = 2, can have two values ( 0 and 1,) m can have one value of = 0 and

three values of =1 (-1,0 and +1) and s can have two values (=+1/2 and-1/2) for each value of m. These possibilities would give rise to eight combinations of the four quantum numbers, keeping in view the exclusion principle . The maximum number of electrons in a shell is given by 2n2.

9.3 HUND'S RULE OF MAXIMUM MULTIPLICITY This rule states that Electron pairing in orbitals of same energy (degenerate orbitals) will not take place unless

all the available orbitals of a given subshell contain one electron each with parallel spin. For example, we know that there are three p orbitals (px, py and pz) of the p-subshell in a

principle energy level. According to Hund's rule, each of the three p orbitals must get one electron each of parallel spin before any one of them gets the second electron of opposite spin as shown below

9.4 RULE OF EXTRA STABILITY OF COMPLETELY HALF FILLED OR COMPLETELY FILLED ORBITAL

When the energy of a wave function is evaluated by using the appropriate Schrodinger equation, two types of energies, viz., Colulomb and exchange energies, are involved. The coulomb energy is a measure of the mean repulsion of two electrons occupying different orbitals, the exchange energy is the lowering of energy for a set of two electrons of parallel spin which arises as a result of in distinguish ability of the electron coordinates. For a system containing more than two electrons of parallel spin, the lowering of energy due to exchange of electrons is given by

∆E = N × K


28



where N is the total number of possible exchanges of coordinates between sets of two electrons of parallel spin and K is the average exchange energy per set of electrons of parallel spin. For a system containing n electrons of parallel spin. For a system containing n electrons of parallel spin, N is given by

n!N2(n 2)!

=−

The exchange energies for the two possible configurations of Cr are as follows: 3d4, 4s2 n = 5 ∆E = 10K 3d5, 4s1 n = 6 ∆E = 15K that is, the exchange energy stabilization of 3d5, 4s1 configuration is larger than 3d4, 4s2

configuration. Besides this, since the two electrons with parallel spins remain farther apart than two electrons with opposed spins, the Columbic repulsion energy is also smaller in case of two parallel spins. Thus the configuration 3d54s1, which contains more number of parallel spins, is more stable than 3d54s2. The stability of the 3d104s1 configuration for the copper atom may be explained in the same way.

Thus, it may be concluded that the unusual stability of electronic configurations involving half-filled and completely filled orbitals is due to the larger exchange energy and lesser columbic repulsion.

The simplified explanation of the above illustration is given below : Exchange Energy : This stabilizing effect arises whenever two or more electrons with the

same spin are present in the degenerate orbitals of a subshell. These electrons tend to exchange their positions and the energy released due to this exchange is called exchange energy. The number of exchanges that can take place is maximum when the subshell is either half filled or completely filled (Figure given below). As a result the exchange energy is maximum and so is the stability.

1 2 3

4

4 exchange byelectron 1




Fig. Possible exchanges for a d5 configuration.

9.5 ELECTRONIC CONFIGURATIONS OF ELEMENTS Based on the rules, we can easily determine the electronic configurations of most elements.

We just need to know the atomic number of an element, the order in which orbitals are to be


29



filled and the maximum number of electrons in a shell, sub-shell or orbital. The configuration so obtained can be represented in two ways. As an illustration, let us consider fluorine (Z = 9):

F (Z = 9) 1s2 2s2 2p 2p 2 2p 1 or 2x y z ↑ ↓ ↑ ↓

2pz2py2px2s1s

↑ ↓ ↑ ↓ ↑

Importance of knowing the exact electronic configuration of an element lies in the fact that

the chemical properties of an element are dependent on the behaviour and relative arrangement of its electrons.

Electronic configurations of heavier elements (beyond Z = 56) deviate a little from the order mentioned previously. These are brief listed below

a) Lanthanides La (Z = 57) : [Xe] 6s25d1 (not 4f1) Ce (Z = 58) : [Xe] 6s25d14f1 Pr (Z = 59) : [Xe] 6s25d14f2 b) Actinides Ac (Z = 89) : [Rn] 7s26d1 (not 5f1) Th (Z = 90) : [Rn] 7s26d15f1 Pa (Z = 91) : [Rn] 7s26d15f2 c) Beyond Z = 103 Z = 104 : [Rn] 5f146d27s2 Z = 105 : [Rn] 5f146d37s2 Z = 106 : [Rn] 5f146d47s2 Z = 112 : [Rn] 5f146d107s2

EXERCISE – 6 1. State the basic ideas that are violated by each of the following electronic configuration

and replace each by the correct configuration : (a) B5 – 1s2 2s2 (b) Na11 – 1s2 2p6 2p6 2d1 (c) K19 – [Ar] 3d1 (d) Ti22 – [Ar] 4s2 4p2 (e) Hg80

– [Xe] 4f10 5d10 6s2 6p2 2. Which of the following will be coloured ion ? (a) Fe2+ (b) Cu+

(c) Sc3+ (d) Mn2+ Which has the maximum magnetic moment ? 3. Magnetic moment of X3+ ion of 3d series is 35 B.M. What is atomic number of X3+ ?

10. SHAPES AND SIZES OF ORBITALS By shape of an orbital we mean the shape of the region in space in which there is probability

of finding an electron. It is basically determined by the azimuthal quantum number ( ), while the orientation of an orbital depends on the magnetic quantum number (m). The size of an orbital increases with the value of the principal quantum number (n). Shapes of orbitals in the various sub-shells are as discussed below:

The place where probability of finding an electron is 0 (Ψ = 0) is known as node and a plane passing through node is known as nodal plane.


30



Total number of nodal planes = There are two types of nodes:

(i) Radial node: These are the points at some distance from the nucleus where there is zero probability of finding the electrons. S-orbital being spherically symmetric and hence non-directional, can’t have angular node. (i.e. points where ) 2 24 r 0π Ψ =

(ii) Angular nodes: These are directional in nature so these are associated with p and d orbitals.

For a particular quantum number n Total nodes = n – 1 Radial nodes = n – – 1

Angular nodes =

10.1 s – ORBITALS s - orbitals ( = 0) : The orbitals are spherical and

symmetrical about the nucleus. There is vacant space between two successive s-orbitals, known as ‘radial node’. Due to nucleus symmetricity, these are non-directional.

��

��

Z radial node

2s1sx

y

nucleus

10.2 p – ORBITALS p - orbitals ( = 1): These have a dumb - bell shape, having three possible mutually

perpendicular orientations. As they are not symmetrical, they have a directional character. The two lobes of p-orbitals are separated by a nodal plane, where the probability of finding electron is zero.

The three 2p orbitals and Nodal Planes

zy

x

p orbitalx

yz plane

xz plane

z

y

x

p orbitaly p orbitalz

z

y

x

xy plane

10.3 d – ORBITALS d - orbitals ( = 2): They have relatively complex geometry. Out of the five d-orbitals, the

first two ( lie along the axes, while the other three (d)22 2z x yd

−d & xy, dzx, dyz) project in

between the axis.


31



The five d orbitals

d orbitalxy

d orbitalx – y 2 2

d orbitalxz d orbitalyz

d orbitalz2

z

y

x

z

y

x

z

y

x

z

y

x

z

y

x

10.4 NODES AND NODAL PLANES FOR d-ORBITALS

Nodal cone

z

y

x

z

y

x

Two nodal surface forming two nodal cones about z axis for orbital.2Zd xz - nodal plane

xy - nodalplane

y

z

z x

y

z

Two nodal planes xy and xz for d orbitalyz


32



yz - nodal plane

xy - nodal plane

y

xx

y

z

Two nodal planes yz and xy for d orbitalxz

z

z yz - nodal plane

xz - nodal plane

y

x

z

y

x

Two nodal planes xz and yz for d orbitalxy

z

y

x

z

y

x

Nodal planes for at angle 45° to x and y axis2 2x yd

−

Illustration 17 : Name the orbitals corresponding to given set of quantum numbers

(a) n = 3, = 2, m = ± 2 (b) n = 4, = 0, m = 0 (c) n = 3, = 1, m ± 1 (d) n = 2, = 1, m = 1 Solution: (a) (b) 4s 2 2x y

3d−

(c) 3 or (d) 2pxp 3py x or 2py


33



Illustration 18 : (a) Find the orbital angular momentum of an electron in the following orbital

(i) 3p (ii) 3d (iii) 3s (b) Arrange the electrons represented by the following sets of quantum

number in decreasing order of energy

(i) n = 4, l = 0, m = 0, ms = ±1/2 (ii) n = 3, l = 2, m = 0, ms = ±1/2 (iii) n = 3, l = 1, m = 1,ms = –1/2 (iv) n = 3, l = 0, m = 0,ms = –1/2

Solution : (a) µ = ( 1+ ) h2π

= ( 1)+

(i) For 3p, = 1 µ = 1(1 1) 2+ =

(ii) For 3d , = 2 µ = h2(2 1) 62

+ =π

(iii) For 3s, = 0, µl = 0

(b) For the same value of n, higher the value of (n + ), higher is the energy.

(c) For the same value of n + , higher the value of n, higher will be the energy.

For (i) ( n + ) = 4 for 4s orbital

For (ii), (n+ ) = 4, for 3d orbital

For (iii) ( n + ) = 4 for 3p orbital

For (iv), (n+ ) = 3 3s orbital

Decreasing order of energy = (ii) > (i) > (iii) > (iv) 11. QUANTUM MECHANICAL TREATMENT OF HYDROGEN ATOM Hydrogen atom is the simplest system consisting of one electron and one proton.

Schrodinger equation is written in terms of Cartesian coordinates (x, y, z). If electron moves at a distance from the stationary nucleus and atom is spherical in shape. If the position of electron is expressed in polar coordinates (r, θ, φ), the schrodinger equation can be solved easily.

The Cartesian coordinates can be converted to polar coordinates by following relation. x = r sin θ cos φ y = r sin θ sin φ


34



z = r cos θ and also, x2 + y2 + z2 = r2 When Schrodinger equation is solved in polar coordinates, for hydrogen atom. It gives the

possible energy states and corresponding wave function [Ψ (r, θ, φ)], called atomic orbitals. The schrodinger equation cannot be exactly solved for a multi electron atom.

Atomic Orbitals and their Pictorial Representation : An atomic orbital is an electron wave function, Ψ (r, θ, φ) obtained from the solution of the schrodinger equation. Wave function Ψ is a mathematical function of the three coordinates of electron (r, θ, φ). This wave functionΨ can be written into three separate parts each of which is a function of only one coordinate.

Ψ = f(r, θ, φ) Ψ (r, θ, φ) = R(r), α(θ), ω(φ) R is function of r, α is function of θ and ω is function of φ. Where, R is radial function and α and ω are angular function. The radial function depends

on quantum numbers n and . The angular function depends on and m.

Significance of Wave Function Ψ : The orbital wave function Ψ has no physical significance. It is the square of the absolute value of wave function |Ψ|2, which has a physical significance. It measure the electron probability density at a point in an atom.

If the variations of Ψ or |Ψ|2 with r, θ, φ are to be observed graphically. It would need a four dimensional graph in space but on the plane of paper we can draw only two dimensional graph to overcome this difficult.

We draw separate diagrams for (i) Variation of radial function (ii) Variation of Angular function R is known as radial wave function R2 is known as radial probability density and 4π2r2R2 is known as radial probability.

11.1 FUNCTION VARIATION OF RADIAL FUNCTION (i) Plot of Radial Wave Function (R) vs r: In all cases R approaches zero as r

approaches infinity. There is a node (where the probability of finding electron is zero) in 2s orbital radial function. At the node the value of radial function changes from positive to negative for 2s orbital. In general ns orbitals have (n – 1) nodes.

The plots of the radial wave function R Vs. distance r

Node


35



(ii) Radial Probability Density (R2)vs r: The square of radial wave function R2 for an orbital gives the radial density. The radial density gives the probability density of finding the electron at a point along a particular line.

The plots of the radial probability density R Vs. distance r2

Node

(iii) Radial Probability Functions (4π2r2R2) vs r: Radial probability function gives the

probability of finding the electron at a distance r from the nucleus regardless of direction.

The plots of the radial density function 4 r R Vs. distance rπ22


36

FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 ! Ph.: (011) 32001131/32 Fax : (011) 41828320

37

Chemistry : Structure of AtomINSTITUTE OF CORRESPONDENCE COURSES

NARAYANA

SECTION - ISUBJECTIVE TYPE PROBLEMS

Problem 1 : Estimate the wavelength of second line in Balmer series.Solution : T147he transition responsible for second Balmer line is 4 → 2.

∆E = 13.6 Z2 ( )22

1 2

1 1n n

−

= 2.55 eV

Now λ = 19 8

19

hc 6.63 10 3 10E 2.55 1.6 10

−

−

× × ×=∆ × ×

⇒ λ = 4.862 × 10�7 m 4862 Å

Problem 2 : A spectral line in the spectrum of H-atom has a wave number of 82200 cm–1 .What transition is responsible for this radiation? (Rydberg constantRH=1.096 × 107 m-1).

Solution : λ = 1 / ν = 1/8220000⇒ it lies in UV region or in Lyman seriesHence, n1 = 1Using the relation for wave umber for H-atom:

v = 1/λ = RH 2 21 2

1 1n n

−

82200 = 1.096 × 105 2 21 2

1 1n n

−

⇒ 1 � 22

1n

= 34

⇒ n2 = 2

Problem 3 : Calculate the wavelength of light radiation that would be emitted, when an electron inthe fourth Bohr’s orbit of He+ ion falls to the second Bohr’s orbit. To what transitiondoes this light radiation correspond in the H-atom ?

Solution : Let us calculate ∆E first.

∆E = 21.7 × 10�19 Z2 2 21 2

1 1n n

−

Substituting n1 = 2 and n2 = 4, Z = 2 we get ; ∆E = 1.632 × 10�8 J

Now λ = hcE∆ = 1.218 × 10�7 m = 1218 Å

The value of λ = 1218 Å implies that, in H-atom this transition would lie in Lyman Series.Hence our aim is now to find the transition : n2 → 1. (1st orbit)

⇒ ∆ E = 21.7 × 10�18 = 21.7 × 10-19 22

11n

−

⇒ n2 = 2

Hence the corresponding transition in H-atom is 2 → 1.



38

INSTITUTE OF CORRESPONDENCE COURSES

NARAYANA

Problem 4 : Find the wavelength of radiation required to excite the electron in ground level of Li++

(Z = 3) to third energy level. Also find the ionisation energy of Li2+ .R = 1.097 × 107 m-1

Solution : ν = RZ2 2 21 2

1 1n n

−

Putting the values : n1 = 1, n2 = 3, Z = 3We get ν = 8.77 × 107 m�1

⇒ λ = 1ν = 113.9 Å

Ionisation energy is the energy required to remove the electron from ground state to infinityi.e. corresponding transition responsible is 1 → ∞

i.e. ∆E = 13.6 × 32 2 2

1 11

− ∞ ∆E = 122.4 eV = 1.95 × 10�7 J = ionisation energy

Problem 5 : Calculate the uncertainty in position assuming uncertainty in momentum within 0.1 %for :(a) a tennis ball weighing 0.2 kg and moving with a velocity of 10 m/s.(b) a electron moving in an atom with a velocity of 2 × 106 m/s.

Solution : Using Uncertainty Principle,

∆x ∆p = h4π

(a) p = mv = 0.2 × 10 = 2.0 kgm/s∆p = 0.1 % of p = 2 × 10�3

⇒ ∆ x = 34

3

h 6.63 104 p 4 3.14 2 10

−

−

×=π∆ × × ×

= 2.135 × 10�32 m.(b) For an electron, p = mv

= 9.1 × 10�31 × 2 10�6

= 1.82 × 10�24 Kgm/s∆p = 0.1 % of p = 1.82 × 10�27 Kgm/s

∆x = 34

21

h 6.63 104 p 2 3.14 1.82 10

−

−

×=π∆ × × ×

⇒ ∆ x = 2.89 × 10�8 m

Problem 6 : Hydrogen when subjected to photo-dissociation, yields one normal atom and one atompossessing 1.97 eV more energy than normal atom. The bond dissociation energy ofhydrogen molecule into normal atoms is 103 kcals mol–1 . Compute the wave length ofeffective photon for photo dissociation of hydrogen molecule in the given case.

Solution : H2 → H + H*where H is normal H-atom and H* is excited H-atom.So the energy required to dissociate H2 in this matter will be greater than the usual bondenergy of H2 molecule.E(absorbed) = dissociation energy of H2 + extra energy of excited atomenergy required to dissociate in normal manner = 103 × 103 cal (given)


39


NARAYANA

= ( )3

1923

103 10 4.187.17 10 J / atom

6 10−× ×

= ××

extra energy possessed by excited atom is 1.97 eV = 1.97 × 1.6 × 10�19 J= 3.152 × 10�19

E (abosorbed)= 7.17 × 10�19 + 3.152 × 10�19 = 1.032 × 10�18 JNow calculate the wavelength of photon corresponding to this energy.

λ = 34 8hc 6.63 10 34 10

E 1.032 10 J

−× × ×=×

= 1.926 × 10�7 m = 1926 Å

Problem 7 : Find out the number of waves made by a Bohr electron in one complete revolution inits 3rd orbit.

Solution : For an electron to be in a particular energy level of radius r, with de Bronglie wavelength λ:2πr = nλ

(where n = number of waves in one revolution)

λ = h

mv ⇒ n = 2 r mv

hπ

for third orbit ;r = 0.53 × 10�10 (3)2 m ; v = 2.165 × 106 (1/3)

substituting the value of r, v, m and h, we get,n = 3

Problem 8 : With what velocity should an α α α α α particle travel towwards the nucleus of a copper atomso as to arrive at a distance 10–13 metre from the nucleus of the copper atom?

Solution : To arrive at a distance �d� from the nuceleus, kinetic energy of α - particle equals electrostaticpotential energy.

KE = PE

21 m V2 α α = ( )( )cuK q q

dα

qα = 2 × 1.6 × 10�19 C d = 10�23 mqα = 29 × 1.6 × 10�19 C

mα = 4 × 1.67 × 10�27 kgK = 1/4πε0 = 9 × 109 N-m/C2

⇒ Vα = cu2k q qm d

α

α

Substitute the given values,Vα = 6.325 × 106 m/s.

Problem 9 : Find the energy released (in ergs), when 2.0 gm atom of Hydrogen undergo transitiongiving spectral line of lowest energy in visible region of its atomic spectrum.

Solution : ∆E = 2.178 × 10�18 (Z2) 2 21 2

1 1n n

−

J / atom

For visible photon, n1 = 2



40


NARAYANA

For lowest energy transition : n2 = 3

⇒ ∆ E = 2.178 × 10�18 (12) 2 2

1 12 3

− J / atom

For 2.0 gm atom,

∆E = (2 × 6.023 × 1023 ) 2.178 ×10�18 5

36 Joules. = 3.63 × 105 J

= 3.63 × 1012 ergs

Problem 10 : Find the ratio of frequencies of violet light (λ λ λ λ λ = 4.10 × 10–5) to that of red light(λ λ λ λ λ = 6.56 × 10–5 cm). Also determine the ratio of energies carried by them.

Solution : C = νλ ;c : speed of light, ν : frequency & λ : wavelength

1

2

νν = 2

1

λλ 1 : red & 2 : violet

⇒ 1

2

νν =

5

5

6.56 10 1.64.10 10

−

−

× =×

Now the energy associated with electromagnetic radiation is given by E = hν.

⇒1

2

EE = 1 2

2 1

1.6ν λ= =ν λ

Hence the ration of energies is same as that of frequencies.


41


NARAYANA

SECTION - IISINGLE CHOICE PROBLEMS

Problem 1 : A photon of wavelength 300 nm is absorbed by a gas and then reemitted as two photons.One photon is red with wavelength of 760 nm. The wave number of the second photon willbe(a) 2.02 × 106 m�1 (b) 3.02 × 106 m�1

(c) 1.02 × 106 m�1 (d) 2.2 × 106 m�1

Solution : E = hcλ ...(i)

E1 = 1

hcλ ...(ii)

E2 = 2

hcλ ...(iii)

E = E1 + E2 = 1 2

1 1hc +λ λ

1λ

= 1 2

1 1+λ λ

-91

(300×10 ) = -9 2

1 1+λ(760×10 )

2

1λ = 91 1- ×10

300 760

= (0.00333 � 0.00131) × 109 = 2.02 × 106 m�1

(a)

Problem 2 : A certain laser transition emits 6.37 × 1015 quanta per second per square meter. The poweroutput in joules per square meter per second would be (l = 632.8 nm).(a) 5 × 10�5 J m�2 sec�1 (b) 2 × 10�3 J m�2 sec�1

(c) 9.2 × 10�3 J m�2 sec�1 (d) none of the above

Solution : Energy falling per square meter per sec.

= No. of quanta falling per square meter per second

Energy of one qunta = 15 -34 8

-196.37 ×10 × 6.626 ×10 × 3×10

632.8×10= 2 × 10�3 J m�2 sec�1

(b)

Problem 3 : The uncertainity in the location of circulating electron is equal to its de Broglie wavelength.The minimum percent error in its measurement of velocity under this circumstance will beapproximately(a) 4 (b) 8(c) 18 (d) 22

Solution : Let the electron be moving with momentum, p, its wavelength will be equal to h/p.



42


NARAYANA

hxp

∆ =

From Heisenberg�s uncertainity principle,hx p4π

∆ × ∆ =

h × pp4π× h

∆ =

∆pp =

14π

Minimum percent error in measuring its velocity would be∆v100 ×v =

∆p ×100p =

1004π =

1004 × 3.14 = 7.96 » 8

(b)

Problem 4 : A mono electronic species in energy level with energy X was provided with excess of energyso that it jumps to higher energy level with energy Y. If it can emit 6 wavelenths originatedfrom all possible transition between these group levels, then which of the following relation iscorrect?

(a) X 3= 1+Y n

(b)X n=Y 6

(c) 2X = (n -1)Y

(d)X 3= 1 +Y n

(where n is the principal quantum number of energy level X).

Solution : If n is the principal quantum number of energy level corresponding to energy X, the principalquantum number of energy level Y is (n + 3) as it emit 6 wavelengths.

XY

=2

2(n + 3)

n

XY

= 31 +n

(a)

Problem 5 : If we plot a graph of stopping potential of photoelectrons against the frequency of radiationthat has emitted it, then its sope would be equal to(a) 6.625 × 10�34 (b) 3.656 × 1012

(c) 2.414 × 1014 (d) 5.796 × 1018

Solution : From Fintein�s equation for photoelectric effect, we havehv = hv0 + 1/2 mv2

max;where v0 is threshold frequencyIf Vstop be the stopping potential of photoelectrons then its maximum possible kinetic energymust be equal to the opposing work i.e. eVstop.

hv = hv0+ eVstop


43


NARAYANA

v = v0 + stope Vh

So the given graph will be a straight line with slope equal to

eh

= 19

341.6×10

6.626×10

−

−

= 2.414×1014

(c)

Problem 6 : The radius of hydrogen atom in its ground state is 5.3 × 10�11 m. After collision with anelectron it is found to have a radius of 21.2 × 10�11m. The principal quantum number of thefinal state of the atom is(a) 2 (b) 3(c) 4 (d) 5

Solution : Radius of H� atoms in its ground state,r1 = 5.3 × 10�11 meters

Radius of Bohr atom in its nth orbit,

rn = 5.3 × 10�11 2n

zmeters

For H-atom, z = 1 and rn = 2.2 × 10�11 meters

n = 1/211

1121.2×105.3×10

−

−

= 2

(a)

Problem 7 : Suppose 10�17 J of energy is needed by theinterior of human eye to see an object. How manyphotons of green light (l = 550 nm) are needed to generate this minimum amount of energy?(a) 14 (b) 28(c) 39 (d) 42

Solution : Let the number of photons required be n.nhcλ = 10�17

n = 1710hc

− ×λ = 17 9

34 810 550 10

6.626 10 3 10

− −

−× ×

× × × = 27.6 » 28 photons

(b)

Problem 8 : Which element has a hydrogen like spectrum whose lines have wavelength one fourth ofatomic hydrogen?(a) He+ (b) Li2+

(c) Be3+ (d) B4+



44


NARAYANA

Solution :H

1λ

= 22 21 2

1 1R 1n n

× −

...(i)

X

1λ = 2

2 21 2

1 1RZn n

−

...(ii)

Dividing equation (i) by equation (ii), we get x

H

λλ = 2

1

Z

14

= 21

Z

Z = 2 (He+) (a)

Problem 9 : Ionisation potential of hydrogen is 13.6 eV. Hydrogen atom in the ground state are excitedby monochromatic light of energy 12.1 eV. The spectral lines emitted by hydrogen accordingto Bohr�s theory(a) One (b) Two(c) Three (d) Four

Solution : The electron in H atom is excited to III shell after absorbing 12.1 eV. The possible transitionstate = (3-1) = 3. (c)

Problems 10 : Which of the following statements is not correct?(a) Special stability of half-filled and fully-filled atomic configurations amongst s- and

p-block elements is reflected in ionization potential trends along a period(b) Special stability of half-filled and fully-filled atomic configurations amongst s- and

p-block elements is reflected in electron affinity trends along a period(c) Aufbau order is not obeyed in cases where energy difference between ns and

(n � 1)d subshell is large.(d) Special stability of half-filled subshell is attributed to higher exchange energy of

stabilizationSolution : Aufbau order is obeyed in cases where energy difference between ns and (n�1)d subshell

is large.


45


NARAYANA

SECTION - IIIMULTIPLE CHOICE PROBLEMS

Problem 1: In hydrogen like sample two different types of photons A and B are produced by electronictransition. Photon B has it�s wavelength in the infrared region if photon A has more energythan B, then the photon A may belong to the region(a) ultraviolet (b) visible(c) infrared (d) none

Solution : (a, b, c)Since B is in infrared region and A has more energy than B hence it will have lesser wavelength i.e. ultra violet, visible or infrared region.

Problem 2 : A hydrogen like atom in ground state absorbs �n� photons having the same energy and itemits exactly �n� photons when electronic transition takes place. Then the energy of theabsorbed photon may be(a) 91.8 eV (b) 40.8 eV(c) 48.4 eV (d) 54.4 eV

Solution : (a, b)Since it absorbs �n� photons and it also emits exactly n-photons therefore transition musthave taken place from 1 to 2.∴ Energy of photon = 10.2 Z2

where Z = 1, 2, 3, 4

Problem 3: Identify the correct statement(s)(a) |Ψ| is the probability of finding the electron in an orbital(b) p-orbital is directional in nature

(c) 2 2x yd

− has dumb bell shape along x-and y-axis

(d) 2zd has dumb bell shape along x-axis and y-axis.

Solution : (b, c)It is obvious from the fact.

Problem 4 : If uncertainty in momentum is twice the uncertainty in position of an electron then uncertainty

in velocity is : h

2 = π !

(a)1

2m! (b) 4 mπ

!

(c)1

4m! (d)

1m!

Solution : (d)

hP. x4

∆ ∆ ≥π

∵ 2∆x = � ∆P (given)



46


NARAYANA

∴2P h

2 4∆ ≥

π∴

2 2 hm ( V) { P m V}2

∆ ≥ ∆ = ∆π

∵

2P h2 4

∆ ≥π

∴ 2

hV2m

∆ ≥π

∴1 h 1V or Vm 2 m

∆ ≥ ∆ ≥π

!

Problem 5 : In a hydrogen like sample electron is in 2nd excited state, the binding energy of 4th state ofthis sample is 13.6 eV, then

(a) A 25 eV photon can set free the electron from the second excited state of this sample

(b) 3 different types of photon will be observed if electrons make transition up to groundstate from the second excited state

(c) If 23 eV photon is used then K.E. of the ejected electron is 1eV

(d) 2nd line of Balmer series of this sample has same energy value as 1st excitation energyof H-atoms

Solution : (a, b)

B.E. of 4th state = 2 2

2 2

z z13.6 13.6 13.6 z 4n 4

⇒ = ⇒ =

sample is Be3+ ∴ energy of elecron in 3rd statestate = 1.5 × 42 = 24 eV

therefore 25 eV photon will cause ionisation.

Problem 6 : The orbitals which have same number of nodes are(a) 2s, 2p (b) 3p, 3d(c) 2s, 3p (d) 3s, 4d

Solution : (a, b)2s has one spherical node and 2p has one planar node, 3p has one planar and one sphericalnode whereas 3d has two planar nodes but no spherical node.

Problem 7 : The probability of finding the electron in the px orbital is(a) Zero at the nucleus(b) Maximum on two opposite sides of the nucleus along X-axis(c) Zero on Z-axis(d) Same on all sides around the nucleus.

Solution : (a, b, c) Fact.


47


NARAYANA

Problem 8 : The possible correct sets of quantum numbers for the unpaired electron of oxygen atom inthe orbitals of p-subshel are(a) n = 2, l = 1, m = 0 (b) n = 2, l = 1, m = �1(c) n = 2, l = 1, m = + 1 (d) n = 2, l = 1, m = � 2

Solution : (a, b, c)Option (d) is not possible because if l = 1, m has three values i.e. �1, 0, + 1.

Problem 9 : The diagram illustrates a possible electronic configuration of which of the following species

↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↓ ↓(a) Cl+ ion (b) S atom(c) Ar+2 ion (d) P�2 ion

Solution : (a, b)Total number of electrons = 16Cl+ ion and S atom have 16 electrons.

Problem 10 : Which of the following statements are wrong?(a) Heisenberg�s uncertainity principle is applicable only to microscopic particles(b) de Broglie concept of dual nature is applicable to all material particles(c) The electronic configuration of palladium (Z = 46) is [Kr] 4d85s2

(d) dxy, dyz and dxz orbitals have identical shapes.Solution : (a, c)

Heisenberg�s uncertainity principle is applicable to microscopic as well as macroscopicparticles but has significance only for microscopic particles. Hence, (a) is wrong. Electronicconfiguration of Pd is [Kr] 4d105s0. Hence, (c) is wrong.



48


NARAYANA

MISCELLANEOUS PROBLEMSSECTION - IV

COMPREHENSION TYPE PROBLEMS

Write up – IIn the Rutherford�s experiment, α-particles were bombarded towards the copper atoms so as to arriveat a distance of 10�13 metre from the nucleus of copper and then getting either deflected or traversingback. The α-particles did not move further closer.

Problem 1 : The velocity of the α-particles must be(a) 8.32 × 108 cm/sec (b) 6.32 × 108 cm/sec(c) 6.32 × 108 m/sec (d) 6.32 × 108 km/sec

Solution : (b)

At the distance of closest approach 2

0 0

2Ze 2e 1v4 r m

= ×πε ×

Hence, substituting the value of m = 4, z = 29, e = 1.66 × 10�19, r0 = 10�13 mts.

v = 6.32 × 108 cm/secHence, choice (b) is correct while (a), (c), (d) are incorrect as from above calculation.

Problem 2 : Which of the following metals can be used instead of gold in α-scattering experiment(a) Pt (b) Na(c) K (d) Cs

Solution : (a)It is because �Pt� is highly malleable and ductile and least ractive.(b), (c) and (d) are soft metals and highly reactive, therefore, cannot be used.

Problem 3 : From the Rutherford�s α-particle scattering, it can be concluded that

(a) 4N sin2θα (b)

1Nsin4α

θ

(c)1N

sin / 24αθ (d) N sin

2θ=

Where N = Number of α-particles scattered by an angle θ.

Solution : It is observed that N = no. of particles of scattering 4

1sin ( / 2)

∝θ

Hence, choice (c) is correct while choice (a), (b) and (d) are incorrect.


49


NARAYANA

Write up – IIThe branch of science which deals with dual behaviour of mater is said to be quantum mechanics. Thefundamental equation of quantum mechanics is Schrodinger wave equation. The important features ofquantum mechanics are(i) The energy of electron is quantized as a result of wave like properties of electrons.(ii) The position and momentum of an electron cannot be determined simultaneously. The path of

electron cannot be determined. We can only talk of probability of finding electron.(iii) An atomic orbitals is represented by wave function �Ψ� for an electron. �R� is radial wave function

whose value varies with distance from the nucleus Ψ2 determines total probability (angular andradial) whereas R2 is called raidal probability. All information about electron in an atom is stored inorbital wave function Ψ. �R� is radial wave function.

(iv) R can be +ve or �ve but R2 is always +ve. Similarly, Ψ can be +ve or � ve but Ψ2 is always +ve Ψ2

can be equal to zero.

Problem 4 : The regions or space where Ψ2 = 0 are called(a) nodes (b) antinodes(c) orbitals (d) energy levels

Solution : (a).Nodes are region where there is no probability of finding elctron, i.e., Ψ2 = 0.

Problem 5 : Number of nodal planes in f-orbitals are(a) 0 (b) 1(c) 2 (d) 3

Solution : (d)There are three nodal planes in f-orbitalsNo. of nodal planes = l = 3

Problem 6 : Which of the following probability distribution curves represents 2s orbital for H-atom

(a)

r

R2 (b)

r

R2

(c)

r

R2 (d)

r

R2

Solution : (b)It is because 2s orbital has one node where R2 = 0.



50


NARAYANA

MATCHING TYPE PROBLEM

7. Column (I) Column (II)(a) Binding energy of He+ atom in an excited state (p) infrared region(b) 7 → 3 transition in hydrogen atom (q) 3.4 eV(c) 5 → 1 transition in hydrogen atom (r) 13.6 eV(d) Series limit of Balmer series in hydrogen atom (s) 10 spectral lines observed

Sol. (a) B.E. of He+ atom = 2

2

13.6 2n×

n = 1, 2, 3 .....Hence it can be 13.6 eV, 3.4 eV both

(b) In 7 → 3 transition ∆n = 7 � 3 = 4

∴ Maximum no. of spectral line observed = 4(4 1) 10

2+ =

It is line of Paschen series Hence infrared region(c) 5 → 1 : 10 lines(d) Series limit of Balmer series is the last line i.e. 3.4 eV energy.(a) � (q, r) (b) � (p, s)(c) � (s) (d) � (q)

ASSERTION-REASON TYPE PROBLEMS

The question given below consist of an ASSERTION and the REASON. Use the following key for theappropriate answers

(a) If both Assertion and Reason are correct and Reason is the correct explanation for Assertion(b) If both Assertion and Reason are correct and Reason is not the correct explanation for

Assertion(c) If Assertion is correct but Reason is not correct.(d) If Assertion is incorrect but Reason is correct.

8. Assertion : In third energy level there is no f-subshell.Reason : For n = 3, the possible values of l are 0,1 and 2.Solution : Both assertion and reason are true and reason is a correct explanation for assertion Hence

Ans. (a)

9. Assertion : Heisenberg�s uncertainty principle holds good for all objects but it is of significance only formicroscopic particles.

Reason : Heisenberg�s uncertainity principle is due to the limitation of the measuring instruments tomeasure accurately the position and velocity of microscopic particles.

Solution : Heisenberg uncertainity principle is not due to limitation of the measuring instrument.Ans. (c)


51


NARAYANA

10.Assertion : Electronic configurations of Fe3+ (containing 23 electrons) is not same as that of Vanadium(Z = 23), i.e. isoelectronic species do not have the same electronic configuration.

Reason : Here, Fe3+ follows Aufbau principle where as Vanadium does not.

Solution : Assertion is correct but reason is false.

Electronic configuration of Fe3+ : [Ar]3d5

Electronic configuration of V : [Ar]3d34s2

Ans. (c)



52


NARAYANA

ASSIGNMENTS

SECTION - ISUBJECTIVE QUESTIONS

LEVEL - I

1. Magnetic moment of X3+ ion of 3d series is 35 BM. What is the atomic number of X3+ ?

2. The electromagnetic radiation of wave length 3186 Å is just sufficient to ionise the outermost electronof a gaseous Cs atom. Calculate the ionization energy of Cs in kJ/mole.

3. Calculate the energy emitted when electrons 1 g atom of hydrogen undergo transition giving spectrallines of lowest energy in the visible region of its atomic spectrum.

4. Calculate the wave number for the shortest wavelength transition in the Balmer series of atomic hydrogen.5. Calculate the de Broglie wavelength of the electron in the ground state of hydrogen atom, given that its

kinetic energy is 13.6 eV. (1eV = 1.602 × 10–19 J).6. With what velocity must an electron travel so that its momentum is equal to that of a photon of wavelength

560 nm?7. A laser emits monochromatic radiation of wavelength 663 nm. If it emits 1015 quanta per second per

square metre, calculate the power output of the laser in joule per square metre per second.8. If astronomical observations, signals observed from the distant starts are generally weak. If the photon

detector receives a total of 3.15 × 10–8 J from the radiations of 600 nm, calculate the number of photonsreceived by the detector.

9. The energy of an α-particle is 6.8 × 10–18 J. What will be the wavelength associated with it?10. The ionization energy of hydrogen in excited state is + 0.85 eV. What will be the energy of the photon

emited when it returns to the ground state?

LEVEL - II

1. Electromagnetic radiation of wavelength 242 MM is just sufficient to ionise the sodium atom. Calculatethe ionisation energy in kJ mol–1.

2. Find the number of quanta of radiations of frequency 4.67 × 1013 s–1 that must be absorbed in order tomelt 5g of ice. The energy required to melt 1 g of ice is 333 J.

3. Calculate the wavelength and energy of radiation emitted for the electronic transition from infinite tostationary state of hydrogen atom.

4. If the energy difference between two electronic states is 46.12 kcal mol–1, what will be the frequencyof the light emitted when the electrons drop from higher to lower states?

5. Calculate the ratio of the velocity of light and the velocity of electron in the first orbit of a hydrogenatom.

6. The wavelength of a certain line in Balmer series is observed to be 4341 Å. To what value of ‘n’ doesthis correspond?


53


NARAYANA

7. The uncertainty in position and velocity of a particle are 10–10 m and 5.27 × 10–24 m s–1 respectively.Calculate the mass of the particle.

8. Calculate the momentum of a particle which has a de Broglie wavelength of 2.5 × 10–10 m.

9. Find the number of waves made by a Bohr electron in one complete revolution in the 3rd orbit.10. The ionisation energy of hydrogen atom is 13.6 eV. What will be the ionisation energy of He+.

LEVEL - III

1. Helium can be excited to 1 11s 2p configuration by supplying 1933.12 10−× J of energy. The lowest excited

singlet state 1 11s 2s lies 5000 cm-1 below 1 11s 2p state. What should be the average He–X bond energy

so that HeX2 could form non-endothermically ? Assume the compound would be formed from lowestexcited singlet state of He+. Heat of atomization of X = 200 kJ mol–1 .

2. A uranyl actinometer containing oxalic acid is irradiated with light of wavelength 4350 Å for 15 minutes.At the end of this time, it was found that oxalic acid equivalent to 12 ml of 0.001 M KMnO4 solution hasbeen decomposed by light. At this wavelength, the quantum efficiency of actinometer is 0.58. Find thepower of radiation source in 10–4 watt assuming that one photon decomposes only one molecule ofoxalic acid.

3. The reaction between H2 and Br2 to form HBr in the presence of light is initiated by the photodecompostionof Br2 into free bromine atoms (free radicals) by absorption of light. The bond dissociation energy ofBr2 is 192 kJ/mole. What is the longest wavelength of the photon that would inititate the reaction?

4. A hydrogen like atom (described by the Bohr’s model) is observed to emit six wavelength, originatingfrom all possible transitions between a group of levels. These levels have energies between – 0.85 eVand – 0.544 eV including both these values).(a) Find the atomic number of the atom.(b) Calculate the smallest wavelength emitted in these transitions.

(use hc = 1240 eV-nm, E1H = – 13.6 eV)5. Electrons in hydrogen like atom (Z = 3) make transition from the fifth to the fourth orbit and from the

fourth to the third orbit. The resulting radiations are incident normally on a metal plate and ejects photo-electrons. The stopping potential for the photo-electrons ejected by the shorter wavelength is 3.95 eV.Calculate the work function of the metal and stopping potential for the photo-electrons jected by thelonger wavelength. (RH = 1.094 × 107 ms–1).

6. A particle of charge equal to that of an electron and mass 208 times the mass of electron(i.e., mu-meson) moves in a circular orbit around a nucleus of charge + 3e. (Take the mass of thenucleus to be infinite). Assuming that the Bohr model of the atom is applicable to this system.(RH = 1.097 × 107 m–1).(i) Derive an expression for the radius of Bohr orbit.(ii) Find the value of n for which the radius of the orbital is approximately the same as that of the first

Bohr orbit for H-atom.(iii) Find the wavelength of the radiation emitted when the mu-meson jumps from the third orbit to the

first orbit.



54


NARAYANA

7. Estimate the difference in energy between 1st and 2nd Bohr’s orbit for a H atom. At whatminimum atomic number a transition from n = 2 to n = 1 energy level would result in the emission ofX-rays with l = 3.0 × 10–8 m? Which hydrogen atom like species does this atomic number correspondsto?

8. A series of lines in the spectrum of atomic H lies at wavelengths 656.46 , 480.27 , 434.17 , 410.29 nm.What is the wavelength of next line in this series?

9. When a beam of 10.6 eV photons of intensity 2.0 W/m2 falls on a platinum surface ofarea 1 × 10 -4 m2 and work function 5.6 eV , 0.53 % of the incident photons eject photo electrons. Findthe number of photo electrons emitted per second and their mimimum energies in (eV). Take1 eV = 1.6 × 10 -19 J .

10. Assume that the de Broglie wave associated with an electron can form a standing wave between theatoms arranged in a one dimensional array with nodes at each of the atomic sites . It is found that onesuch standing wave is formed if the distance d between the atoms of the array is 2 Å . A similiarstanding wave is again formed if d is increased to 2.5 Å but not for any intermediate value of d . Findthe energy of the electrons in electron volts and the least value of d for which the standing wave of thetype described can be formed .


55


NARAYANA

SECTION - IISINGLE CHOICE QUESTIONS

1. If the speed of electron in the Bohr’s first orbit of hydrogen atom be x, then speed of the electron in the3rd orbit is(a) x/9 (b) x/3(c) 3x (d) 9x

2. Which one of the following set of quantum numbers is not possible for 4p electron?(a) n = 4, l = 1, m = – 1, ms = + 1/2 (b) n = 4, l = 1, m = 0, ms = + 1/2(c) n = 4, l = 1, m = 2, ms = + 1/2 (d) n = 4, l = 1, m = – 1, ms = – 1/2

3. The orbital angular momentum of an electron in 2s orbitals is

(a)1 h2 2

+ ⋅π

(b) zero

(c)h

2π(d)

h22

⋅π

4. The uncertainity in the momentum of an electron is 1.0 × 10–5 kg ms–1. The uncertainity in its positionwill be (h = 6.62 × 10–34 kg m2 s–1)(a) 1.05 × 10–28 m (b) 1.05 × 10–26 m(c) 5.27 × 10–30 m (d) 5.25 × 10–28 m

5. Naturally occuring boron is 20% 5B10 and 80% 5B

11. The atomic weight of boron is(a) 10.50 (b) 11.0(c) 10.80 (d) 10.20

6. A particle A moving with a certain velocity has a de Broglie wavelength of 1 Å. If particle B has mass25% of that A and velocity 75% of that of A, the de Broglie wavelength of B will be approximately(a) 1 Å (b) 5.3 Å(c) 3 Å (d) 0.2 Å

7. What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition,n = 4 to n = 2 in the He+ spectrum.(a) n = 4 to n = 1 (b) n = 3 to n = 2(c) n = 3 to n = 1 (d) n = 2 to n = 1

8. Which of the following is violation of Pauli’s exclusion principle?

(a) ↑↓ ↑↓ (b) ↑↑ ↑ ↑ ↑

(c) ↑↓ ↑ ↓ ↑ (d) ↑ ↑ ↑ ↑

9. The probability of finding an electron residing in a px orbital is zero(a) in the yz plane (b) in the xy plane(c) in the y-direction (d) in he z-direction



56


NARAYANA

10. Which of the following has more unpaired d-electrons?(a) Zn+ (b) Fe2+

(c) Ni3+ (d) Cu+

11. 4000 Å proton is used to break the iodine molecule, then the % of energy converted to the KE of iodineatoms if bond dissociation energy of I2 molecule is 246.5 kJ/mole(a) 8% (b) 12%(c) 17% (d) 25%

12. What will be the energy emitted when electrons of 1.0 g atom of hydrogen undergo transition giving thespectral line of lowest energy in the visible region of its atomic spectrum.(a) 18.25 × 104 J (b) 11.91 × 103 J(c) 52.18 × 105 J (d) 63.12 × 104 J

13. Probability of finding the electron Ψ2 of ‘s’ orbital doesn’t depend upon :(a) Distance from nucleus (r) (b) Energy of ‘s’ orbital(c) Principal quantum number (d) Azimuthal quantum number

14. Consider the following statements. The d-orbitals have1. four lobes and two nodes 2. four lobes and one node3. same sign in the opposite lobes 4. opposite sign in the opposite lobesWhich of the above statements are correct?(a) 1 alone (b) 1 and 3(c) 2 and 3 (d) 1 and 4

15. The maximum number of electrons in a subshell having the same value of spin quantum number is givenby(a) l + 2 (b) 2l + 1

(c) l (l + 1) (d) ( 1)+l l

16. Potential energy of Li2+ electron is

(a)2

0

3e4 r−π∈ (b)

2

0

3e4 rπ∈

(c)2

20

2e4 r

−π∈ (d)

2

0

e4 r

−π∈

17. The energy of an electron in the second Bohr orbit of H-atom is –E. The energy of the electron in theBohr’s first orbit is(a) – E/4 (b) – 4E(c) 4E (d) – 2 E

18. The wave number of the shortest wavelength of absorption spectrum of H-atom is (Rydbergconstant = 109700 cm–1)(a) 109700 cm–1 (b) 3/4 × 109700 cm–1

(c) 1/2 × 109700 cm–1 (d) 9 × 10 × 109700 cm–1

19. In the Bohr’s orbit, what is the ratio of total kinetic energy and total energy of the electron?(a) – 1 (b) – 2(c) + 1 (d) + 2

20. In two H atoms A and B the electrons move around the nucleus in circular orbits of radius r and 4rrespectively. The ratio of the times taken by them to complete one revolution is(a) 1 : 4 (b) 1 : 2(c) 1 : 8 (d) 2 : 1


57


NARAYANA

SECTION - IIIMULTIPLE CHOICE QUESTIONS

1. Which of the following quantum numbers is/are not allowed

(a) n = 3, l = 2, m = 0 (b) n = 2, l = 2, m = –1

(c) n = 3, l = 0, m = 1 (d) n = 5, l = 2, m = –1

2. For radial probability distribution curves, which of the following is/are correct.

(a) The number of maxima in 2s orbital are two

(b) The number of spherical or radical nodes is equal to n – l – 1

(c) The number of angular nodes are ‘l’

(d) 3dz2 has two angular nodes.

3 ‘g’ orbital is possible if

(a) n = 5, l = 4 (b) It will have 18 electrons

(c) It will have 9 types of orbitals (d) It will have 22 electrons

4. Which of the following transitions are allowed in the normal electronic emission spectrum of an atom?

(a) 2s → 1s (b) 2p → 1s

(c) 3d → 2p (d) 5p → 3s

5. Choose the correct relations on the basis of Bohr’s theory.

(a) Velocity of electron 1n

∝ (b) Frequency of revolution 31n

∝

(c) Radius of orbits 2n Z∝ (d) Force on electron 41n

∝

6. The magnitude of the spin angular momentum of an electron is given by

(a)hS s(s 1)

2= +

π(b)

hS s2

=π

(c)3 hS

2 2= ×

π(d)

1 hS2 2

= ± ×π

7. The ground-state electronic configuration of the nitrogen atom can be represented as

(a) ↑↓ ↑↓ ↑ ↑ ↑ (b) ↑↓ ↑↓ ↑ ↓ ↑

(c) ↑↓ ↑↓ ↑ ↓ ↓ (d) ↑↓ ↑↓ ↓ ↓ ↓



58


NARAYANA

8. Which of the following statements relating to the spectrum of hydrogen atom are false?

(a) The lines can be defined by quantum number

(b) The line of longest wavelength in the Balmer series corresponds to the transition between n = 3 andn = 2 levels

(c) The spectral lines are closer together at long wavelengths

(d) A continuum occurs at n = ∞

9. Which of the following statements are correct?

(a) Angular momentum of 3s electron is equal to that of 2s electron

(b) Energy levels of H-atom are not equally spaced

(c) Energy of electron is zero only when it is stationary

(d) Humphrey series lies in the far infrared region

10. Which of the following statements are not correct?

(a) An orbital containing an electron having quantum number n = 2, l = 0, s = + 1/2 is spherical

(b) All photons have same energy

(c) the frequency of X-rays is less than radiowaves

(d) As intensity of light increases, the frequency increases.


59


NARAYANA

MISCELLANEOUS QUESTIONSSECTION - IV

COMPREHENSION TYPE QUESTIONS

Write up - IThe wave numbers of the spectral lines of Lyman Series may be calculated using the equation givenbelow as derived by Niels Bohr for an atom or ion free from electron-electron repulsion.

2H 2

1R Z 1n

ν = ⋅ − where n = 2, 3, 4 5….¥

where2 2 4

eH 3

2 m k eR

Chπ= = Rydberg constant

me is the mass of electron but for actual calculation we will have to consider the nuclear motion and thenme will have to be replaced by reduced mass (m) where

N e

N e

m mm m

⋅µ =

+ , mN = mass of nucleus

All the other symbols have their usual significances.

1. If proton in the nucleus of hydrogen atom is replaced by positron (+1e0) having the charge of proton but

mass that of electron, the wave number of the lowest energy transition in the Lyman series of theabove designated H-atom considering the nuclear motion will be equal to

(a) H3 R4

(b) H3 R8

(c) HR2

(d) H3 R2

2. What will be the ratio of the wave number of the 1st line to that of the 5th line of Lyman series ofHe+ ion?

(a)2532

(b)2735

(c)10032 (d)

10835

3. The 3rd IP of Be (Z = 4) per atom/ion (RH = 109677 cm–1 and h = 6.625 × 10–34J.s,C = 3 × 108 ms–1 and 1.6 × 10–19 J = 1 eV) is equal to(a) 217.6 eV(b) 163.2 eV(c) 123.0 eV(d) Can’t be calculated using Bohr’s expression



60


NARAYANA

Write up - II

In the quantum mechanics there is operator ˆ(A) for every physically observable property like momentum(linear or angular), position, K.E., total energy etc., and whenever the function (ψ) is operated upon bythe operator of any property , we get back ψ multiplied by a constant (a) which is the value of theproperty under consideration. The function ψ obeying the condition is called eigenfunction of the operator

A and the constant “a” is called eigen-value. Mathematically

A aψ = ⋅ψThis equation is called eigen-value equation.The Schrodinger wave equation is the energy eigen value equation. The operator for total energy iscalled Hamiltonion operator (H) defined as

2 2 2 2

2 2 2 2hH V

8 m x y z ∂ ∂ ∂= − + + + π ∂ ∂ ∂

Where V = PE, m = mass of microparticle.Solving Schrodinger wave equation for H-atom after putting proper value of V, we get wave functions(ψ) and corresponding energy values. Each wave function denotes an orbital. ψ2 measures the probabilityof finding the electron and ψ2dx gives the probability of finding the electron in the region x and x + dx.4πr2ψ2(r)dr gives the probability of finding the electron in spherical shell of thickness dr at a distancer from the nucleus. A plot of radial distribution function [4πr2ψ2(r)] vs. distance from the nucleus (r)gives the radial probability distribution curve of electron. For 1s and 2s orbitals each containing only oneelectron, these curves are as given below

2 24 ( )r rπ ψ

r

1s 2s

r

2 24 ( )r rπ ψ

The wave functions of 1s and 2s orbitals of hydrogenic atom or ion are given below:

ψ1s = 0

1/ 2Zr / a

0

Z2 ea

−

ψ2s = 0

1/ 2Zr / 2a

0 0

Z Zr2 e2a a

− −

In 2s-orbital there occurs a point at the radius at which curve touches X-axis. Here the probability offinding the electron is zero. This is called radial node.

On the basis of the above write-up, answer the following questions :

4. An electron moves along x-axis and restricted to move only between x = 0 and x = L. Hence

(a)L

2

0

dx 1ψ =∫ (b)L

2

0

dx 0ψ =∫

(c)L

2

0

dx 1ψ >∫ (d)L

2

0

dx 1ψ <∫


61


NARAYANA

5. The value of r at which radial node of 2s-electron of H-atom will appear is:(a) r = a0 (b) r = 2a0

(c) 0ar2

= (d) 0ar3

=

6. The value of r at which the maxima of the probability distribution curve of 1s orbital electron of H-atomwill occur is(a) r = a0 (b) r = 2a0

(c) 0ar

3= (d) 0a

r2

=

MULTIPLE MATCHING TYPE QUESTIONS

Match the following

7. Column – I Column – II(a) Angular momentum (p) increases by increasing shell numbers(b) Kinetic energy (q) decreases by decreasing atomic numbers(c) Potential energy (r) increases by decreasing atomic numbers(d) Velocity (s) decreases by increasing shell numbers

8. Column – I Column – II(a) Visible, infrared, ultraviolet – (p) increases

wavelength of light(b) Lyman, Balmer, Paschen – (q) decreases

number of lines in respective spectrums(c) As we move outwards from an atom (r) first increases then decreases

distance between two consecutive shells(d) As we more outwards from an atom (s) Remains same

energy differnce between twoconsecutive shells

ASSERTION-REASON TYPE QUESTIONS

The question given below consist of an ASSERTION and the REASON. Use the following key for theappropriate answers

(a) If both Assertion and Reason are correct and Reason is the correct explanation for Assertion

(b) If both Assertion and Reason are correct and Reason is not the correct explanation forAssertion

(c) If Assertion is correct but Reason is not correct.

(d) If Assertion is incorrect but Reason is correct.



62


NARAYANA

9. Assertion : In Rutherford’s gold foil experiment, very few α-particles are deflected back.

Reason : Nucleus present inside the atom is heavy.

10. Assertion : Limiting line in the Balmer series has a wavelength of 364.4 mm.

Reason : Limiting line is obtained for a jump of electron from n = ∞.

11. Assertion : There are two spherical nodes in 3s-orbital.

Reason : There is no planar node in 3s-orbital

12. Assertion : An orbital cannot have more than two electrons.

Reason : The two electrons in an orbital create opposite magnetic field.

13. Assertion : The configuration of B-atom cannot be 1s22s2.

Reason : Hund’s rule demands that the configuration should display maximum multiplicity.14. Assertion : The plots of radial probability density and radial probability function versus distance r

from the nucleus for any particular orbital are identicalReason : Radial probability density (R2) is along the radial distance where as radial probability

function (4πr2R2) is the probability in a shell of thickness dr.15. Assertion : The 19th elecron in potassium atom enters into 4s-orbital and not the 3d-orbital.

Reason : (n + l) rule is followed for determining the orbital of the lowest energy state.


63


NARAYANA

SECTION - V(PROBLEMS ASKED IN IIT-JEE)

A. Only one option is correct (Objective Questions)1. The energy of an electron in the first Bohr orbital of H atom is – 13.6 eV. The possible energy values(s)

of the excited state(s) for electrons in Bohr orbitals of hydrogen is (are) :(a) –3.4 eV (b) – 4.2 eV (1998)(c) – 6.8 eV (d) + 6.8 eV

2. The electrons, identified by quantum numbers n and l, (i) n = 4, l = 1, (ii n = 4, l = 0, (iii) n = 3, l = 2 and(iv) n = 3, l = 1 can be placed in order of increasing energy, from the lowest to highest, as :

(1999)(a) (iv) < (ii) < (iii) < (i) (b) (ii) < (iv) < (i) < (iii)(c) (i) < (iii) < (ii) < (iv) (d) (iii) < (i) < (iv) < (ii)

3. The electronic configuration of an element is 1s2, 2s2, 2p6, 3s2, 3p6, 3d5 4s1. This represents its :(a) excited state (b) ground state(c) cationic form (d) anionic form (2000)

4. The number of nodal planes in 2 px orbital is : (2001)(a) one (b) two(c) three (d) zero

5. The wavelength associated with a golf ball weighing 200g and moving at a speed of 5 m/h is of theorder: (2001)(a) 10–10 m (b) 10–20 m(c) 10–30 m (d) 10–40 m

6. The quantum numbers 1 1and2 2

+ − for the electron spin represents : (2001)

(a) rotation of the electron in clockwise and anticlockwise direction respectively(b) rotation of the electron in anticlockwise and clockwise direction respectively(c) magnetic moment of the electron pointing up and down respectively(d) two quantum mechanical spin states which have no classical analogue

7. If the nitrogen atom had electronic configuration 1s7, it would have energy lower than that of the normalground state of configuration 1s2 2s2 2p3, because the electrons would be closer to the nucleus, yet 1s7

is not observed because it violates : (2002)(a) Heisenberg uncertainty principle (b) Hund’s rule(c) Pauli’s exclusion principle (d) Bohr’s postulate of stationary orbitals

8. Rutherford’s experiment, which established the nuclear model of the atom, used a beam of :(a) β–particles, which impinged on a metal foil and got absorbed (2002)(b) γ–ray, which impinged on a metal foil and ejected electrons(c) helium atoms, which impinged on a metal foil and got scattered(d) helium nuclei, which impinged on a metal foil and got scattered



64


NARAYANA

9. Which hydrogen like species will have same radius as that of Bohr orbit of hydrogen atom?(a) n = 2, Li2+ (b) n = 2, Be3+ (2004)(c) n = 2, He+ (d) n = 3, Li2+

10. The number of radial nodes of 3s and 2p orbitals are respectively : (2005)(a) 2, 0 (b) 0, 2(c) 1, 2 (d) 2, 11

B. More than one options are correct (Objective Questions)

1. An isotone of 7632Ge is : (1984)

(a) 7732Ge (b) 77

33As

(c) 7734Se (d) 78

34Se

2. Many elements have non–integral atomic masses because : (1984)(a) they have isotopes(b) their isotopes have non–integral masses(c) their isotopes have different masses(d) the constituents, neutrons, protons and electrons, combine to given fractional masses

3. When alpha particles are sent through a thin metal foil, most of them go straight through the foilbecause : (1984)(a) alpha particles are much heavier than electrons(b) alpha particles are positively charged(c) most part of the atom is empty space(d) alpha particle move with high velocity

4. The sum of the number of neutrons and proton in the isotope of hydrogen is : (1986)(a) 6 (b) 5(c) 4 (d) 3

5. The atomic nucleus contains : (1988)(a) protons (b) neutrons(c) electrons (d) photons

6. The energy of an electron in the first Bohr orbit of H atom is – 13.6 eV. The possible energy value (s)of the excited state (s) for electrons in Bohr orbits of hydrogen is (are) : (1998)(a) – 3.4 eV (b) – 4.2 eV(c) – 6.8 eV (d) + 6.8 eV

7. Which of the following statement (s) is (are) correct ? (1998)(a) The electronic configuration of Cr is [Ar]3d5 4s1, (Atomic number of Cr = 24).(b) The magnetic quantum number may have a negative value(c) In silver atom, 23 electrons have a spin of one type and 24 of the opposite type. (Atomic number of

Ag = 47)(d) The oxidation state of nitrogen in NH3 is – 3.


65


NARAYANA

8. Decreases in atomic number is observed during : (1998)(a) alpha emission (b) beta emission(c) positron emission (d) electron capture

9. Ground state electronic configuration of nitrogen atom can be represented by : (1999)

(a) ↑↓ ↑↓ ↑ ↑ ↑ (b) ↑↓ ↑↓ ↑ ↓ ↑

(c) ↑↓ ↑↓ ↑ ↓ ↓ (d) ↑↓ ↑↓ ↑ ↑ ↑

SUBJECTIVE QUESTIONS1. Estimate the difference in energy between 1st & 2nd Bohr’s orbit for a H atom . At what minimum

atomic number a transition from n = 2 to n = 1 energy level would result in the emission of x - rays withl = 3.0 ´ 10 -8 m ? Which hydrogen atom like species does this atomic number corresponds to?[JEE 1993]

2. What transition in the hydrogen spectrum would have the same wavelength as the Balmer transitionn = 4 to n = 2 of He+ spectrum ? [JEE 1993]

3. Find out the number of waves made by a Bohr electron in one complete revolution in its3rd

orbit. [JEE 1994]

4. Iodine molecule dissociates into atoms after absorbing light of 4500 Å . If one quantum of radiation isabsorbed by each molecule , calculate the kinetic energy of iodine atoms . (Bond energy ofI2 = 240 k J mol -1). [JEE 1995]

5. A bulb emits light of l = 4500 Å . The bulb is rated as 150 watt and 8 % of the energy is emitted as light.How many photons are emitted by the bulb per second. [JEE 1995]

6. Consider the hydrogen atom to be a proton embedded in a cavity of radius a0 (Bohr’s radius) whosecharge is neutralized by the addition of an electron to the cavity in vacuum infinitely slowly .

(a) Estimate the average of total energy of an electron in its ground state in a hydrogen atom as thework done in the above neutralization process . Also if the magnitude of the average kinetic energyis half the magnitude of the average potential energy , find the average potential energy.

(b) Also derive the wavelength of the electron when it is a0 from the proton . How the wavelength ofan electron in the ground . State Bohr’s orbit . [JEE 1996]

7. An electron beam can undergo diffraction by crystals . Through what potential should a beam ofelectrons be accelerated so that its wavelength becomes equal to 1.54 Å. [JEE 1997]

8. A compound of vanadium has a magnetic moment of 1.73 BM . Work out the electronic configurationof the vanadium ion in the compound . [JEE 1997]

9. Calculate the energy requird to excite one litre of hydrogen gas at 1 atm and 398 k to the first excitedstate of atomic hydrogen . The energy for the dissociation of H - H bond is 436 k J mol -1 Alsocalculate the minimum frequency of photon to break this bond. [JEE 2000]



66


NARAYANA

ANSWERSEXERCISE - 1

1. 3.312 × 10–12 erg 2. 2.01 × 1018

3. 7.22 × 104 mol

EXERCISE - 21. r1 = 0.1763 Å; r2 = 0.7053 Å

2. He(IE) 54.4 eV+ = (IE) ; 2Li

(IE) 122.4 eV+ = ; 3Be(IE) 217.6 eV+ =

3. H(IE) in 3rd orbit = 1.51 eV

EXERCISE - 3

1. 32.77 eV 2. 6.63 × 10–33 m3. 2.5 × 10–8 m

EXERCISE - 4

1. 1.893 × 10-34 m 2.1 hV

2m∆ =

π

3. ∆V = 2.64 × 10–39 ms–1

EXERCISE - 5

1. (a) ml = 0 (b) l = 1

(c) n = 2 (d) n = 3, ml = – 2 to + 2 (any value)2. 3p < 4s < 3d < 7s < 5f < 6g3. (b), (e) and (f) are not allowed.4. (a) 2p (b) 4d(c) 5s

EXERCISE - 61. (a) 1s2 2s2 2p1 : maximum capacity of 2s orbital is 2.

(b) 1s2 2s2 2p6 3s1 : 2d orbital does not exist.(c) [Ar] 4s1 : 3d is filled only after 3s has been completely filled.(d) [Ar] 4s2 : 3d2 : 3d is filled before 4p.(e) [Xe] 4f14 5d10 6s2 : maximum capacity of 4f suborbit is 14.

2. (a) Fe2+ = [Ar] 3d6

(b) Cu+ = [Ar] 3d10 no unpaired electron in 3d(c) Sc3+ = [Ar] no unpaired electron.(d) Mn2+ = [Ar] 3d5

3. Z = 26


67


NARAYANA

SECTION - I(Subjective Questions)

LEVEL - I

1. 26 2. 375.25 kJ / mol3. 182.5 kJ 4. 27419.25 cm–1

5. 0.3328 nm 6. 1300 ms–1

7. 3 × 10–4 Jm–2 s–1 8. 9.519. 2.31 × 10–16 m 10. 12.75 eV

LEVEL - II1. 493.6 kJ mol–1 2. 5.38 × 1022

3. λ = 9.11 × 10–8 m ; E = 2.17 × 10–18 J 4. 4.84 × 1014 cycle sec–1

5. c/v = 137 6. n = 57. 0.099 kg 8. 2.64 × 10–24 kg m sec–1

9. 3 10. 54.5 eV

LEVEL - III1. 1167.5 kJ mol–1 2. 1583. 6235 Å 4. Li2+, 4060 nm5. w = 3.193 × 10–19 J, Vs = 0.756 V6. (i) r = n2 × 10–14 m (ii) n = 25

(iii) 5.48 × 10–11 m7. 10.2 eV, 2, He+ 8. 397.2 nm9. 6.25 × 1011 10. d = 0.5 Å, E = 150.8eV.

SECTION - II(Single Choice Questions)

1. (b) 2. (c)3. (b) 4. (c)5. (c) 6. (b)7. (d) 8. (b)9. (a) 10. (b)11. (c) 12. (a)13. (d) 14. (b)15. (b) 16. (a)17. (b) 18. (a)19. (a) 20. (c)



68


NARAYANA

SECTION - III(Multiple Choice Questions)

1. (b, c) 2. (a, b, c)3. (a, b, c) 4. (a, b, d)5. (b, c, d) 6. (a, b, d)7. (a, c) 8. (a, d)9. (b, c, d) 10. (b, c, d)

SECTION - IV(Comprehension Type Questions)

1. (b) 2. (b)3. (a) 4. (a)5. (b) 6. (a)

(Multiple Matching type questions)7. (a) (p) (b) (q, s)

(c) (p, q) (d) (q, s)8. (a) (r) (b) (p)

(c) (p) (d) (q)

(Assertion Reason Type Questions)9. (b) 10. (a)11. (b) 12. (b)13. (c) 14. (d)15. (a)

SECTION - V(Problems asked in IIT-JEE)

OBJECTIVE QUESTIONSA. Only one option is correct

1. (a) 2. (a)3. (b) 4. (a)5. (c) 6. (d)7. (c) 8. (d)9. (b) 10. (a)

B. More than one option are correct1. (b, d) 2. (a, c)3. (a, c) 4. (d)5. (a, b) 6. (a)7. (a, b, c) 8. (a)9. (a, d)

SUBJECTIVE QUESTIONS1. 10.2 eV , z = 2 , He+ 2. n1 = 1 , n2 = 23. 3 4. 0.216 ´ 10 -19 J5. 27.2 × 1018 7. 63.3 volt8. 1 s2 , 2 s2 , 2 p6 , 3 s2 3 p6 3 d1 9. 98.19 k J , 10.93 ´ 1014 Hz

structure of atom - n@[email protected]/.../5/5895441/iit_class_xi_chem_structure_of_ato… ·...

Documents