atomic structure · isobars atoms having same mass number but different atomic number. example 15p...

Dalton’s Theory of MatterDalton proposed that matter was made up of extremely small particles called atoms.

Main postulates of Dalton’s atomic theory :

1. Matter is made up of small invisible particles calledatom.

2. Atom is the smallest particle of an element that takes part in a chemical reaction.

3. Atoms of same element are identical.

4. Atoms of different elements are different.

5. Atoms can neither be created nor can be destroyed.

A chemical reaction is just a rearrangement of atoms.

6. Atoms of different elements combine in a fixed ratio of small whole numbers to form compoundatoms, called molecules.

But later scientists established that atom was not the smallest indivisible particle but has a complexstructure of its own, and was made up of still smaller particles like electrons, protons, neutrons etc.

Electron, proton and neutron are regarded as fundamental particles.

Models of Atoms

Thomson’s Model (Plum Pudding Model)

He suggested that atom is a positively charged sphere having electrons embedded uniformly giving anoverall view of plum pudding.

Rutherford’s ModelHe allowed a narrow beam of light to fall on a thin gold foil and determined the path of particles with thehelp of ZnS fluorescent screen.

The ZnS gives off a visible flash of light when struck by anα-particles.

Observation of Rutherford1. Most of the α-particles pass straight through the gold strips with little or no deflection.

2. Some α-particles are deflected from their path and diverge.

3. Few α-particles are deflected backwards through angles greater than 90°.

Atomic Structure

1Unit

Atomic Model Proposed by RutherfordRutherford proposed atomic model as follows:

1. Atom consists of two parts (a) Nucleus and (b) Extra nuclear part

2. All the protons (positive charge) and the neutrons (neutral charge) i.e.,nearly the total mass of anatom is present in very small region at the centre of the atom. The atom’s central core is callednucleus.

3. Diameter of nucleus is about 10 cm13− . Atom has a diameter of order of 10 8− cm. Size of atom is

105 times more than that of nucleus.

4. Most of space outside nucleus is empty.

5. Electrons revolve around the nucleus with fast speed in various circular orbits.

6. The centrifugal force arising due to fast speed of an electron balances coulombic force of attractionof the nucleus and electron remains stable in its path.

Defect’s in Rutherford’s Model1. The exact position of electrons from the nucleus are not mentioned.

2. Neils Bohr found Rutherford’s model defective in respect of stability of atom. According to law ofelectrodynamics the electron should therefore, continuously emit or use energy for moving and getcloser to the nucleus and after passing through a spiral path, ultimately fall into the nucleus.

Neils Bohr proposed an improved form of Rutherford’s atomic model based on the quantum theoryof radiation put forth by Max Planck.

Few Important Terms1. Atomic number Atomic number of an element is the number of protons contained in the

nucleus of the atom of that element. It is represented by Z.

2. Nucleons It is the number of neutrons and protons together present in a nucleus.

3. Mass number The total number of protons and neutrons present in the nucleus. It isrepresented by A.

IUPAC notation of an atom (nuclide)

The element can be represented as ZAX

Where, X = chemical symbol of element.

4. Isotopes Atoms of the element with same atomic number but different mass number.

Example 11H , 1

2H , 13H are three isotopes of hydrogen.

5. Isobars Atoms having same mass number but different atomic number.

Example 15P32 and 16S32 are isobar.

6. Isotones Atoms having same number of neutrons but different number of protons ormass number.

Example 6C14 , 8 C16, 7 N15 are isotones.

7. Isoelectronic Atoms, molecules or ions having same number of electrons.

Example N2, CO, CN

8. Atomic mass unit It is equal to 1/12 of the mass of 6C12 atom.1 u 1.66= × −10 27 kg

2 Indian National Chemistry Olympiad

Characteristics of WavesA wave is a sort of disturbance which originates from some vibrating sources and travels outwards as acontinuous sequence of alternating crests and trough.

Every wave has five important characteristics, i.e.,wavelength ( )λ , frequency ( )ν , velocity ( )c , wave number( )ν and amplitude (a).

1. Wavelength ( )λ The distance between two neighbouring troughs and crests is known as

wavelength. It is expressed in cm, m , nm and Å.

2. Frequency (ν) The frequency of a wave is the number of times a wave passes through a given

point in one second. It is denoted by ν. It is expressed in cycles per second (cps) or hertz. (Hz)1Hz 1 cps= .

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

νλ

∝ 1

or νλ

= c

3. Velocity The distance travelled by the wave in one second. It is denoted by c and is expressedin cm/s.

c = νλ

4. Wave number (ν) It is defined as the number of wavelengths per cm. It is denoted by ν and isexpressed in cm–1.

νλ

= 1, ν ν=

c

5. Amplitude (a) It is the height of the crest or depth of trough of a wave. It is denoted by a. Itdetermines the intensity or brightness of the beam of light.

Radiant EnergyIt is the energy transmitted from one body to another in the form of radiation. Maxwell found that analternating current of high frequency radiates energy in the form of waves which travel in space with thesame speed as light. These waves were called as electromagnetic waves or electromagnetic radiations.

Radiant energy has wave nature and is associated with electric as well as magnetic field. Theseradiations are called electromagnetic radiation.

Various forms of electromagnetic radiations are cosmic rays, γ-rays, X-rays, UV rays, visible rays,infrared rays, microwaves, radar waves and radio waves such asTV, FM, AM rays.

Electromagnetic radiations arranged in the increasing order of wavelength or frequency, calledelectromagnetic spectrum.

Thus, the spectrum of radiations means different radiations arranged according to their wavelength.When a band of visible light is passed through a prism of glass, all the colours of white light getseparated into violet, indigo, blue, green, yellow, orange and red. It is known as spectrum and there is nooverlapping of colours.

E hvhc= =λ

where, ν = Frequency, c = Velocity of radiations, λ = Wavelength, E = Energy of radiations

Atomic Structure 3

Planck’s Quantum TheoryWhen different solids are heated to same temperature they glow and emits different types of radiations.

1. Substances radiate or absorb energy discontinuously in the small packets or bundles of energy.

2. The smallest packet of energy is called quantum. In case of light the quantum is known as photon.

3. The energy of a quantum is directly proportional to the frequency of the radiation.

E ∝ νE h= νν = Frequency of radiation

h = Planck’s constant

h = × −6.626 10 erg s27 –1 = × −6.626 10 34 J s/

4. A body can radiate or absorb energy in whole number multiples of a quantum i.e., h, 2h, 3h, … nhνwhere, n is the positive integer.

Atomic SpectrumSpectrum is the impression produced on a photographic film when the radiation(s) or particularwavelength(s) is (are) analysed through a prism or diffraction grating.

Types of Spectrum

(i) Emission spectrumSpectrum produced by radiation emitted is known as emission spectrum. This spectrum corresponds to

the radiation emitted when an excited electron returns back to the ground state.

(ii) Absorption spectrumSpectrum produced by the absorbed radiation.

Types of Emission Spectra

(i) Continuous spectrumWhen white light from any source is analysed by passing through prism, it splits up into 7 different widebands of colour. These colours are so continuous that each of them merges into next without any gap orbreak.

(ii) Line spectrumWhen an electric discharge is passed through a gas at low pressure light is emitted. If this light isresolved by a spectroscope, it is found that some isolated coloured lines are obtained on a photographicplate separated from each other by dark spaces. Each line in the spectrum corresponds to a particularwavelength each element gives its own characteristic spectrum.

(iii) Band spectrumThe spectrum which consists of bright bands each having sharp edge.

Atomic Spectra of Hydrogen AtomIf an electric discharge is passed through hydrogen gas in atomic state taken in a discharge tube underlow pressure and the emitted radiations is analysed with the help of spectrograph the spectrum doesnot have radiations of all the wavelengths but radiations of only certain wavelengths. The spectrum ofhydrogen is found to consist of a series of sharp lines in the UV visible and IR regions.


The wavelength of all these series can be expressed as :

1 1 1

12

22λ

ν= = −

R

n n

ν = Wave number

λ = Wavelength

R = Rydberg constant (10967 z cm–1)

n1 ; n2 have integral values

Series n1 n2 Main spectral lines

Lyman 1 2,3,4… UV

Balmer 2 3,4,5… Visible

Paschen 3 4,5,6… IR

Brackett 4 5,6,7… IR

Pfund 5 6,7,8… IR

Bohr’s Model and Hydrogen SpectrumMain postulates of Bohr’s Model are as follows :

(i) An atom consists of positively charged nucleus.

(ii) The electrons revolve around the nucleus in certain permitted circular orbits of definite radii.

(iii) The permitted orbits are those for which the angular momentum of an electron is an integralmultiple of h/2π.

h = Planck’s constant

If m is the mass; v is the velocity of an electron in a permitted orbit of radius r then

L mvr nh p n= = =/ ;2 1, 2, 3…

L = Orbital angular momentum

n = Number of orbit

The integer n is called principal quantum. This equation is known as Bohr quantization postulate.

(iv) When electrons move in permitted discrete orbits they does not radiate or loose energy these arestationary or non-radiating orbits. Greater the distance of energy level from the nucleus, the more isthe energy associated with it. The different energy levels were numbered as 1, 2, 3, 4… and called asK, L, M, N, etc.

(v) Ordinarily an electron continues to move in a particular stationary state of orbit. Such a stable state,of atom is ground state when energy is given to the electron it jumps to any higher energy level and issaid to be in excited state. When electrons jump from higher to lower energy state the energy isradiated.

Hydrogen SpectrumAccording to Bohr, the spectrum arises when, the electron and nucleus from any excited stationary orbitof principal quantum number n2 jumps to the ground state or any lower stationary orbit of principalquantum number n1 , so that the difference of the energy associated with these orbits is emitted as aphoton of frequency (ν).

Accordingly, h E Eν = −2 1 = −

2 22 2 4 2

22 2

2 2 2 2

12 2

π πmZ e K

n h

mZ e K

n h

Hence, ν π= −

2 1 12 2 4 2

312

22

mZ e K

h n n(Z = 1 for H atom)

Atomic Structure 5

Since the wavelength, λ of the photon is related to ν by ν λ= c/

Hence,c me K

h n nλπ= −

2 1 12 4 2

312

22

1 2 1 12 4 2

312

22λ

π= −

me K

h c n nor ν

λ1 1 1

12

22

= = −

R

n nH

where, RH is called the Rydberg constant of the hydrogen atom.

Rp mK e

h CH

7 11.097 10 m= = × −2 2 2 4

3

Above equations is called Rydberg equation.

Different types of series lie in hydrogen spectrum and their wave numbers of photons are as follows :

(i) Lyman seriesThe spectral lines correspond to the transition of the electron from higher energy state (n) to the lowerenergy state corresponding to nf = 1

ν R1 1

n= −

H

22 2; n = 2, 3, 4, … ∞

These lines are found in the ultraviolet region of the electromagnetic spectrum.

(ii) Balmer series

The spectral lines correspond to the transition of the electron from some higher energy correspondingto nf = 2.

Thus, ν R= −

H

1

2

12 2n

; n = 3, 4, 5, … ∞.

These lines are found in the visible region.

(iii) Paschen series

Here, ν = −

Rn

H1

3

12 2

; n = 4, 5, 6, …∞

These lines are found near in the infrared region.

(iv) Brackett series

Here, ν = −

Rn

H1

4

12 2

; n = 5, 6, 7 … ∞

These lines are also found in the infrared region.

(v) Pfund series

Here, ν = −

Rn

H1

5

12 2

; n = 6, 7, 8, … ∞

These lines are found in the far infrared region.

Merits of Bohr’s Theory(i) Value of radii and energies in hydrogen atom are in good agreement with that calculated on the

basis of Bohr’s theory.

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

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


Limitations of Bohr’s Theory(i) It doesn’t explain the spectra of atoms having more than one electron.

(ii) Bohr’s atomic model failed to account for the effect of magnetic field (Zeeman effect) or electricfield (Starck effect ) on the spectra of atoms or ions.

(iii) Postulate of Bohr, that electrons revolve in well defined orbit’s around the nucleus with welldefined velocities is thus not valid.

Dual Nature of ElectronEinstein suggested dual character of light wave and particle to explain the phenomenon of diffractionand interference by taking light as waves and phenomenon of black body radiation and photoelectriceffect on the basis of its particle nature.

de-Broglie said that matter can also have a dual character. He gave an expression between wavelength ‘ ’λof particle and its mass ‘ ’m moving with velocity ‘ ’ν i.e., λ ν= h m/ .

de-Broglie EquationThe wavelength of the wave of a particle is calculated by analogy with photon.

Case I Photon having wave character energy,

E h= ν (according to Planck’s quantum theory) …(i)

Case II Photon having particle character energy,

E mc= 2 …(ii)

By Einstein equation, here m is the mass of photon and c is the velocity of light.

By Eqs. (i) and (ii)

h cmν = 2

If the m and c of photon are replaced by the m and n of particle then

λν

= h

m⇒ λ = h

P

As m Pν = the momentum of the particle

Heisenberg’s Uncertainty PrincipleIt is impossible to measure simultaneously the position and momentum of a small microscopic movingparticle with absolute accuracy or certainly.

If an attempt is made to measure any one of these two quantities with higher accuracy, the otherbecomes less accurate.

Its mathematical expression is ∆ ∆x P h× ≥ / .4π

Quantum or Wave Mechanical Model of AtomThis model of atom is based on de-Broglie’s principle, Heisenberg’s uncertainty principle and dualnature of matter. According to this model motion of electron in three dimensional space is representedby Schrodinger wave equation as follows

∂∂

+ ∂∂

+ ∂∂

+ − =2

2

2

2

2

2

280

ψ ψ ψ π ψx y z

m

hE v( )

Atomic Structure 7

where x, y and z are coordinates.

m = Mass of electron

E = Total energy

ν = Potential energy

ψ = Wave function of electron wave

ψ2 = Probability of finding electron in atom.

Features of Quantum, Mechanical Model1. The energy of electron in atom is quantised.

2. Solution of Schrondinger equation gives the energy values of quantised levels.

3. It doesn’t specify the exact position and momentum of electron but focuses on most probableregions — Orbitals.

4. Electron in each orbital has definite amount of energy.

5. Whenever an electron is described by a wave function ψ it means electron occupies an orbital.

6. ψ2 is probability density and always has positive value.

7. The solution of Schrodinger wave equation gives values of n, l and m.

Probability distribution curve (graph) is a plot prepared by plotting probability ( )ψ2 against the distance

from nucleus ( ).r Radial probability distribution curve gives probability of finding electron in a givenvolume.

For 1s-orbital

At mode probability of finding electron is zero.


4π ψr2 2

r

Node

r

4π ψr2 2

r

4π ψr2 2

For 2s-orbital For 2p-orbital

Shape of Orbitals

Photoelectric EffectSir JJ Thomson, observed the photoelectric effect. When a light of certain frequency strikes the surfaceof a metal, electrons are ejected from the metal, and the ejected electrons are called photoelectrons.Only cesium shows this effect in visible light as it has low ionization energy but others show it undermore high UV rays.

Experimental Findings of Photoelectric Effect(i) When light of required energy strikes the surface of metal, electrons come out.(ii) The light of minimum frequency, i.e., threshold frequency is require to eject the electrons.

Atomic Structure 9

z

dyz

y

x

dxy

(a)

z

yx

dxz

(b)

z

y

x

(c)

z

y x

dx2

(d)

z

yx

dz2

(e)

– y2

d-orbital

2px

y

x

z

y

x

p-orbital

2pz

y

x

2pyzz

s-orbital

(iii) Photoelectric current is directly proportional to the intensity of light of same frequency is more willbe the current if the intensity is greater.

Let, the threshold frequency = ν0

If a photon of frequency ν0 strikes the metal it give its energy to the electron = ( )hν0

By getting hν0 energy electron can break away from the atom by overcoming the attraction force ofnucleus. But if the frequency < ν0 then no effect will take place

The more of frequency than ν0 gives more energy to the electron

∴ Given a certain velocity (i.e., kinetic energy ) to the electron

h hν ν= +0 KE

h h mν ν ν2− =01

2⇒ 1

22

0m hν ν ν= −( )

Now, hν0 is the work function denoted by φand it is constant for a particular metal.

The KE of photoelectron and frequency of induct light is linear i.e., a straight line whose slope is equal toPlanck's constant ‘ ’h and intercept is ‘ ’hν0 . Quantum n0 and atomic orbitals. There are 4 quantumnumbers which describe the location energy, orbital, its shape and orientation etc. of the electron.

Principal Quantum Number(a) It is denoted by letter n and can have +ve value e.g.,1, 2, 3, …

(b) It gives angular distance between electrons and nucleus.

(c) It tells the energy of that electron.

(d) The maximum number of electrons present in any principal shell is given by2 2n .

Shell Maximum no. of electrons

nK 1 2L 2 8M 3 18N 4 32

Azimuthal or Angular Quantum Number(a) Denoted by ‘l ’ and represent the subshells present in the main shell.

(b) It can have any integral value 0 to n − 1.

(c) It tells the shape of the shell.

(d) Through it angular momentum can be calculated as, ω π= +l l h( ) /1 2 or l l h( )+ 1

Magnetic Quantum NumberAngular motion of an electron produces an electric field around the nucleus which give rise to magneticfield. Due to which electrons can orient themselves in certain preferred regions of spaces around thenucleus called orbitals. The magnetic quantum number determine the number of preferred orientationof the electron present in the sub-shell.

m has value between −l to +l including 0.

Spin Quantum NumberElectron revolves around the nucleus as well as spins about its own axis either clockwise oranti-clockwise.

∴ It can have two value + 1

2or − 1

2. This helps us to explain magnetic properties of the substances.


Shapes and Size of OrbitalSpace around nucleus where probability of finding an electron is maximum is known as orbital.

s-orbitals (l = 0), the shape of s-orbit symmetrical about the p-orbitals ( l = 1).

The probability of finding the p-electron is maximum intwo lobes on the opposite sides of the nucleus. This giverise to a dumb-bell shape for the p- orbital for p-orbitall = 1. Hence, m = − 1, 0, +1 thus p-orbital have threedifferent orientations. These are designated aspx ,py , andpz depending upon whether the density of electron ismaximum along the x, y and z-axis respectively. They aresymmetrical but have directional characters. The twolobes of p-orbitals are separated by a nodal plane, wherethe probability of finding electron is zero the threep-orbitals belonging to a particular energy shell haveequal energies and are called degenerate orbitalsd-orbitals ( )l = 2 .

For d-orbitals, l = 2 Hence, m = − 2, –1, 0, +1, +2 thus5d-orbitals they have different geometrical shape orbitalsare (d d dxy yz zx, , ) are lie in between the axis and the other two d

z2 and dx y2 2−

lie along the axis.

Aufbau PrincipleThis principle states that the electrons are added in orbitals one by one in the various orbitals in order oftheir increasing energy starting with the orbital of lowest energy.

( )n l+ Rule The energy of an orbital depends upon the sum of the values of the principal quantumnumber (n) and the azimuthal quantum number (l ), this is called ( )n l+ rule. According to this rule, ‘‘Inneutral isolated atom’’ the lower the value of ( )n l+ for an orbital, lower is its energy.

Pauli’s Exclusion PrincipleAccording to this principle ‘‘No two electrons in an atom will have the same value of all the fourquantum numbers.”

Hund’s Rule of Maximum MultiplicityElectron pairing in p, d and f orbitals cannot occur until each orbital of a given sub-shell contains oneelectron each or is singly occupied’’.

Factors responsible for the extra stability of half-filled and completely filled sub-shells :

(i) Symmetrical distribution It is well known fact that symmetry leads to stability. Thus theelectronic configuration in which all the orbitals of the same sub-shell are either completely filledor are exactly half-filled are more stable because of symmetrical disturbance of the electrons.

(ii) Exchange energy The electrons with the parallel spins present in the degenerate orbitals tendto exchange their position. The energy released during this exchange is called exchange energy. Thenumber of exchanges that can take place in maximum when the degenerate orbitals (orbitals ofsame sub-shell having equal energy) are exactly half-filled or completely filled. As a result, theexchange energy is maximum and so is the stability.

Atomic Structure 11

1s

2s

3s

4s

5s

6s

7s

2p

3p3d

4d4f

5d4p

5p

6p

Aufbau principle

Some Important Points1. Some uncommon fundamental particles

Positron +10e (Anderson, 1932)

Neutrino and antineutrino (Jumi, 1934)

Mesons ( , , and )π π π µ+ −0 (Yukawa, 1935, Kemma).

2. General symbol for an atom of element X is ZA X where, A = mass number, Z = atomic number.

A Z n= +

3. Mass of e− at high speed mm

u

c

=

−

0

2

1

where, m0 = rest mass, u is velocity of e− and c is velocity of light. If u c= , then m → ∞

4. Radius of nucleus = ×A

rAvogadro no.

1

4

3

3π

Frequency ν λ= c / ; wave number νλ

= 1

Energy of photon E hhc

hc= = =νλ

ν (Energy is inversely proportional to wavelength).

5. ∆E E E hn n= − =2 1

ν if n n2 1> ; emission spectra if n n2 1< absorption spectra

6. Angular momentum of e nh

m r− = =2π

ν

7. Orbital angular momentum of e− = +l lh

( )12π

8. Spin angular momentum of e− = +s sh

( )12π

9. Velocity of e− in an orbit ue

h

Z

nn =

2 2π = ×

2.178 106 Z

nm per = u Z

n

1

10. Energy of an e− in a shell ( )En

Eme

h

Z

n

E

nn = −

× =2 1

4

2 4

20

1

2

ππε

2

2; E

Ze

rn

n

=2

2

EZ

nn = − × −21.79 10 19

2

2J-atm = ×13 6

2

2.

Z

neV = − ×313.5208

Z

n

2

2kcal mol 1−

11. In CGS1

41

0πε=

12. In MKS1

49 10

0

9

πε= × Nm C2 –2, 4 0πε = permittivity factor of medium

∆En n

= − −

1312

1 1

22

12

kJ mol–1

∆E = energy required to jump from e− or emitted when e− drops from n2 to n1 level.

13. Separation energy E En n→ ∞ − (given)

14. λ =−

hc

E En n2 1

, νλ

= = −

1 1 1

12

22

Rn n

H

15. RH = Rydberg constant =109677 cm–1 for ‘‘H’’ atom

16. RmZ e K

h cH = 2 2 2 4 2

3

π;


17. K =ε

= ×1

49 10

0

9

πNm C2 –2

18. Number of fine lines of a line in atomic spectrum of H n n= ×1 2

19. The possible number of λ given out = Σ∆n = −∆ ( )n n2 1

20. λ = =h

mV

h

p, de-Broglie equation

21. Heisenberg’s uncertainty principle ∆ ∆x ph× ≥

4πor

h

2π

22. Maximum number of lines = −n n( )1

2

23. Einstein photoelectric equation1

2

20mv h h= −ν ν

24. 2π λr n= number of waves in any orbit =circumferences

wavelength= n

25. PE = − Ze

rn

2

26. KE = =

1

2

22

2Ze

rmv

Ze

rn

27. ν = −a Z b( ) ; (Moseley) a,b are constants Z = atomic number

28. Para-magnetic momentum µ = +n n( )2

29. Isorlirs Molecules having same number of atoms and same number of e− for example N O2 and CO2

(3 atoms and 22 e−).

30. Isodiaphers Same difference of neutrons and protons .

31. Isoelectronic Same number of e−.

32. Diamagnetic Doubly occupied (no unpaired e−).

33. Paramagnetic Atoms which consist of at least one of the orbitals singly occupied.

34. Kernel Part remained after H e− moving the outermost shell.

Core charge Number of e− in outer most orbit.

35. Number of orbitals in any sub-shell = +( )2 1l

36. Maximum number of e− in main energy shell = 2 2n

37. Number of sub-shells in main energy shell = n

38. Number of orbitals in a main energy shell = n2

Illustrative Solved ExamplesExample 1. 1.8 g H2 atoms are excited to radiations. The study of spectra indicates that 27% of the atoms

are in 3rd energy level and 15% of atoms in second energy level the rest in ground state. IP of H is 13.6 eV.Calculate

(a) number of atoms present in third and second energy level.

(b) total energy evolved when all the atoms return to ground state.

Solution 1g H contains = H atoms

1.8 g H contains = ×N 1.8atoms = × ×6.023 10 1.823 = ×10.84 1023 atoms

Atomic Structure 13

(a) ∴ Number of atoms in third shell = × ×10.84 10 27

100

23

atoms = ×292.68 1021

∴ Number of atoms in second shell10.84 10 15

100

23

= × × = ×162.6 1021 atoms

Number of atoms in first shell10.84 10 58

100

23

= × × = ×628.72 1021 atoms

(b) When all the atoms return to first shell, then

E E E′ = − × ×( )3 1 292.68 1021 = − +

× × −13.6

913.6 1.602 10 19 292.68 1021× ×

= ×5.668 105 J

E E E′ ′ = − × ×( )2 1 162.6 1021 = − +

× × × ×−13.6

413.6 1.602 10 162.6 119 021

= ×2.657 105 J

∴ E E E= ′ + ′ ′ = × + ×5.668 10 2.657 105 5J = 832.50 kJ

Example 2. A hydrogen like atom (at. no. Z) is in a higher excited state of quantum number n. This excitedatom can make a transition to the first excited state by successively emitting 2 photons of energies10.200 eV and 17.00 eV respectively. Alternatively, the atom from the same excited state can make atransition to the second excited state by successively emitting 2 photons of energy 4.25 eV and 5.95 eVrespectively. Determine the value of n and Z.

Solution Total energy liberated during transition of electron from nth shell to first excited state

= + =10.20 17 27.20 eV

= × × −27.20 1.602 10 12 erg

∴ hcR Z h c

nH 2λ

= × × × −

22

1

2

1…(i)

Similarly, total energy liberated during transition of electron fromnth shell to second excited state

= + =4.25 5.95 10.20 eV.

Dividing Eqs. (i) by (ii)

n = 6

On substituting value of n in Eqs. (i) and (ii)

Z = 3

Example 3. A gas identical H like atoms in the lowest (ground) energy level A and some atoms in aparticular upper (excited) energy level B and there are no atoms in any other energy level. The atoms ofthe gas make to a higher energy level by absorbing monochromatic light of photon energy 2.7 eV.Subsequently the atoms emit radiations of only six different photons energies. Some of the emittedphotons have energy 2.7 eV. Some have more and some have less than 2.7 eV.

(a) Find principal quantum number of initially excited level B.

(b) Find the ionization energy for the gas atoms.

(c) Find the maximum and minimum energies of the emitted photons.

Solution The electrons being present in first shell and another shell n1, There are excited to higher level

n2 by absorbing 2.7 eV and on deexcitation emits six lines and thus excited state n2 comes to be 4.

[b n n= = −Σ∆ Σ( )2 1 ∴ n2 4= ]


Now, ER chn

1 21= ; E

R ch

nn

n1

12

= ; E R chn n=

Since deexcitation leads to different λ having photon energy ≤ 2.7 eV and this absorption of 2.7 eVenergy causing excitation to fourth shell and then reemitting photons of ≤ 2.7 eV is possible only whenn1 2= .

The deexcitation from 4th shell occurs in first, second and third shell.

E E4 2− = 2.7 eV

E E4 3− > 2.7 eV

E E4 1− > 2.7 eV

E ER ch E

n nn

1 2 2 2212

= = =

Since n1 2=

Also E E4 2− = 2.7 eV ∴ − + =E E12

124 2

27. eV

∴ E1 = − 14.4 eV ∴ IP 14.4= eV

E E EE E

max = − = − +4 31

2124 1

= − +14.4

1614.4 = 13.5 eV

E E EE E

min = − = − + =4 312

134 3

0.7 eV

Example 4. Balmer’s empirical formula is1 1

2

12 2λ

= −

Rn

H ; n = 3 4 5, ,

Here, Rm e

h ce

H =ε

=4

02 38

109678/cm

RH = Rydberg constant

me = mass of an electron

(a) Calculate longest wavelength in Å in the Balmer series of singly ionised He+ . (Ignore nuclear motion).

(b) Write the formula analogous to Balmer's formula applies to the series of spectral lines which arise fromtransition from higher energy level to lowest energy level of H atom. And also use it to determine theground state energy of H atoms in eV.

(c) ah

m ee0

02

2= ε =

π0.53 Å

Determine the lowest energy and radius of first Bohr orbit of muonic hydrogen atom.

In ‘muonic hydrogen atom’ electron is replaced by 207 times heavier particle, muon. But the charge issame.

(d) Onsider a spherical shell of radius a0 and thickness 0.001a. Estimate the probability of finding theelectrons in this shell.

Volume of spherical shell of inner radius r and thickness ∆r is given by V r r= 4 2π ∆ .

(e) (i) H H + H2 → (ii) H H + H2+→

Put appropriate channel labels (i) and (ii) in the boxes.

(f) Calculate the energy change in H H– → + e– system. Assuming that the Bohr energy formula is valid

for each electron with nuclear charge Z replaced by Zeff , calculate Zeff for H−.

(g) H– has a 2 electron, atomic.

Atomic Structure 15

Solution (a) Longest wavelength λ , corresponds to n = 3

For He+ ,1

41

2

12 2λ

= −

Rn

H ; λL = 1641.1 Å

(b)1 1

1

12 2λ

= −

Rn

H ; n = 2 3 4, , , . . .

E hcR= = −H 13.6 eV

(c) Lowest energy = − × = −207 13.6 282 keV

Radius of the first Bohr orbit = = × −0.53

2072.6 10 Å3

(d) Probability = ×[ ( )]ψ πa a a02

02

04 0.001 = = ×− −0.004 5.41 10 4e 2

(f) Channel (i) = 4.7 eV; channel (ii) = 17.6 eV

(g) Electron affinity = − − − =13.6 ( 14.3) 0.7 eV

− + =13.6 27.2 0.7;Zeff2 Zeff = 0.7

Example 5. Oxalic acid in presence of

UO (UO )22+

22++ hv → +

[UO ] + H C O UO + CO + CO22+

2 2 4 22+

2− → + H O2

By absorbing light of λ → 250 nm. 450 nm.

The reaction is used in estimating the amount of decomposed acid and quantum efficiency of the process.Here the quantum efficiency is given to be 50%. 1.0 L of 0.005 M oxalic acid solution was irradiated withlight of λ = 310 nm in presence of UO2

2 + ion. 20 mL of irradiated solution on titration required 25 mL of

0.005 N KMnO4 solution. How many photons are absorbed per second if the solution was irradiated for10.5 nm?

Solution Equivalents of KMnO4 that reached with excesss

H C O0.005 25

10001.25 102 2 4

4= × = × −

∴ Equivalents of excess H C O2 2 4 in 20 mL = ×1.25 104

∴ Equivalents of excess H C O2 2 4 in 20 mL = × ×−1.25 10 504 = × −6.25 10 3

∴ Moles of excess H C O2 2 4 in 1 L = × −3.125 10 3

Initial moles of oxalic acid = 0.005

∴ Moles of oxalic acid consumed by (UO )22 + = × −1.875 10 3

∴ Moles of (UO ) 1.875 1022 3+ −= ×

∴ Moles of photons = × −1.875 10 3

Number of photons absorbed/sec = × × × ×× ×

−100

1.875 10 6.023 10

50 10.5 60

3 23

= ×3.58 1018

Example 6. The second ionization potential of Be is 17.98 eV. If the electron is Be+is assumed to move in aspherical orbit with a central field effective nuclear charge (Zeff ) consisting of the nucleus units of chargeis the nucleus sheilded by other electrons? The energy of electron in first Bohr orbit of H is – 13.6 eV. If theextent of shielding the K electrons of Li atom is the same as you have calculated above, find the ionizationpotential of Li.


Solution IE = =13.617.98

Z

n

2

2∴ Zeff = 2.3

Shielding extent = − = − =Z Zeff 4 2.3 1.7

Zeff for lithium = − =3 1.7 1.3

∴ IE of lithium = × =13.6 (1.3)

(2)5.746 eV

2

2

Example 7. A mixture of atoms of ordinary hydrogen 11H and tritium, 1

3H excited and its spectrum is

observed. How far apart in wavelength will be the H2 lines of the 2 kinds of hydrogen ? Rydberg constantfor ordinary H2 atom is1.096 .58 10 m7 1>> × − , proton mass 1.002276= u, mass of atomic 1

3H 3.016049= u,

electron mass 0.000549= u.

Solution For hydrogen,1 1

2

1

31 2 2λ

= ⋅ −

R R = −

36

51

1

1

2R

R

R

ButR

R1

2

1

2

= µµ

, the ratio of the reduced masses

∴ µ11

1

=+

M m

M mand µ 2

2

2

=+

M m

M m

where, m = Electron mass

∴ µµ

1

2

1

2

2

1

= × ++

M

M

M m

M m

We know that M1 100 2 6= > > u, M2 = 3.016049 u and n = 0.000549 u

∴ µµ

1

2

= 0.999637 ∴ ∆λ = × × −36

510 4

R3.63

where, Rm

= ×1.0967758107

∆λ = 2.385 Å

Atomic Structure 17

atomic structure · isobars atoms having same mass number but different atomic number. example 15p...

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