ncert solved examples

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CHAPTER 2: Structure of Atom

(CLASS: XI)

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NCERT Solved Examples Problem 2.1

Calculate the number of protons, neutrons and electrons in 8035Br .

Solution

In this case, 8035Br , Z = 35, A = 80, species is neutral

Number of protons = number of electrons = Z = 35

Number of neutrons = 80 – 35 = 45

Problem 2.2

The number of electrons, protons and neutrons in a species are equal to 18, 16 and 16 respectively. Assign the proper

symbol to the species.

Solution

The atomic number is equal to number of protons = 16. The element is sulphur (S).

Atomic mass number = number of protons + number of neutrons = 16 + 16 = 32

Species is not neutral as the number of protons is not equal to electrons. It is anion (negatively charged) with charge

equal to excess electrons = 18 – 16 = 2.

Symbol is 32 216 S .

Problem 2.3

The Vividh Bharati station of All India Radio, Delhi, broadcasts on a frequency of 1,368 kHz (kilo hertz). Calculate the

wavelength of the electromagnetic radiation emitted by transmitter. Which part of the electromagnetic spectrum does it

belong to?

Solution

The wavelength, , is equal to c/ , where c is the speed of electromagnetic radiation in vacuum and is the

frequency. Substituting the given values, we have

c 8 13.00 10 m s

1368 kHz

8 1

3 1

3.00 10 m s

1368 10 s= 219.3 m

This is a characteristic radiowave wavelength.

Problem 2.4

The wavelength range of the visible spectrum extends from violet (400 nm) to red (750 nm). Express these

wavelengths in frequencies (Hz). (1 nm = 10–9

m)

Solution

Using equation 2.5, frequency of violet light

8 1

9

c 3.00 10 m s

400 10 m= 7.50 10

14 Hz

Frequency of red light

8 1

9

c 3.00 10 ms

750 10 m = 4.00 10

14 Hz

The range of visible spectrum is from

4.0 1014

to 7.5 1014

Hz in terms of frequency units.



(CLASS: XI)


Problem 2.5

Calculate (a) wavenumber and (b) frequency of yellow radiation having wavelength 5800 Å.

Solution

(a) Calculation of wavenumber ( )

= 5800 Å = 5800 10–8

cm = 5800 10–10

m

10

1 1

5800 10 m= 1.724 10

6 m

–1= 1.724 10

4 cm

–1

(b) Calculation of the frequency ( )

8 1

10

c 3 10 m s

5800 10 m = 5.172 10

14 s

–1

Problem 2.6

Calculate energy of one mole of photons of radiation whose frequency is 5 1014

Hz.

Solution

Energy (E) of one photon is given by the expression

E = h

h = 6.626 10–34

J s; = 5 1014

s–1

(given)

E = (6.626 10–34

J s) (5 1014

s–1

) = 3.313 10–19

J

Energy of one mole of photons = (3.313 10–19

J) (6.022 1023

mol–1

) = 199.51 kJ mol–1

Problem 2.7

A 100 watt bulb emits monochromatic light of wavelength 400 nm. Calculate the number of photons emitted per

second by the bulb.

Solution

Power of the bulb = 100 watt = 100 J s–1

Energy of one photon E = h = hc/ 34 8 1

9

6.626 10 Js 3 10 m s

400 10 m = 4.969 10

−19 J

Number of photons emitted= 1

19

100 J s

4.969 10 J = 2.012 10

20 s

–1

Problem 2.8

When electromagnetic radiation of wavelength 300 nm falls on the surface of sodium, electrons are emitted with a

kinetic energy of 1.68 105 J mol

–1. What is the minimum energy needed to remove an electron from sodium? What is

the maximum wavelength that will cause a photoelectron to be emitted ?

Solution

The energy (E) of a 300 nm photon is given by

h = hc/34 8 1

9

6.626 10 J s 3.0 10 m s

300 10 m= 6.626 10

–19 J

The energy of one mole of photons

= 6.626 10–19

J 6.022 1023

mol–1

= 3.99 105 J mol

–1

The minimum energy needed to remove one mole of electrons from sodium

= (3.99 –1.68) 105 J mol

–1 = 2.31 10

5 J mol

–1

The minimum energy for one electron 5 1

23 1

2.31 10 J mol

6.022 10 electrons mol = 3.84 10

−19 J

This corresponds to the wavelength



(CLASS: XI)


ch

E

34 8 1

19

6.626 10 J s 3.0 10 m s

3.84 10 J = 517 nm

(This corresponds to green light)

Problem 2.9

The threshold frequency 0 for a metal is 7.0 1014

s–1

. Calculate the kinetic energy of an electron emitted when

radiation of frequency = 1.0 1015

s–1

hits the metal.

Solution

According to Einstein’s equation

Kinetic energy = ½ mev2 = h( – 0)

= (6.626 10–34

J s) (1.0 1015

s–1

– 7.0 1014

s–1

)

= (6.626 10–34

J s) (10.0 1014

s–1

– 7.0 1014

s–1

)

= (6.626 10–34

J s) (3.0 1014

s–1

)

= 1.988 10–19

J

Problem 2.10

What are the frequency and wavelength of a photon emitted during a transition from n = 5 state to the n = 2 state in the

hydrogen atom?

Solution

Since ni = 5 and nf = 2, this transition gives rise to a spectral line in the visible region of the Balmer series. From

equation (2.17)

18

2 2

1 12.18 10 J

5 2E = – 4.58 10

–19 J

It is an emission energy

The frequency of the photon (taking energy in terms of magnitude) is given by

E

h

19

34

4.58 10 J

6.626 10 Js= 6.91 10

14 Hz

8 1

14

c 3.0 10 m s

6.91 10 Hz = 434 nm

Problem 2.11

Calculate the energy associated with the first orbit of He+. What is the radius of this orbit?

Solution 18 2

n 2

(2.18 10 J)ZE

n atom

–1

For He+, n = 1, Z = 2

18 2

1 2

(2.18 10 J) (2)E

1 = – 8.72 10

–18 J

The radius of the orbit is given by 2

n

(0.0529 nm)r

n

Z

Since n = 1, and Z = 2 2

n

(0.0529 nm) lr

2 = 0.02645 nm



(CLASS: XI)


Problem 2.12

What will be the wavelength of a ball of mass 0.1 kg moving with a velocity of 10 m s–1

?

Solution

According to de Broglie equation (2.22) 34

1

h (6.626 10 Js)

mv (0.1kg) (10 m s )= 6.626 10

–34 m (J = kg m

2 s

–2)

Problem 2.13

The mass of an electron is 9.1 10–31

kg. If its K.E. is 3.0 10–25

J, calculate its wavelength.

Solution

Since K. E. = ½ mv2

1/21/2 25 2 2

31

2 K.E. 2 3.0 10 kg m sv

m 9.1 10 kg= 812 m s

–1

34

31 1

h 6.626 10 Js

mv (9.1 10 kg) (812 ms )= 8967 10

–10 m = 896.7 nm

Problem 2.14

Calculate the mass of a photon with wavelength 3.6 Å.

Solution

= 3.6 Å = 3.6 10−10

m

Velocity of photon = velocity of light 34

10 8 1

h 6.626 10 Jsm

(3.6 10 m) (3 10 m s )= 6.135 10

–29 kg

Problem 2.15

A microscope using suitable photons is employed to locate an electron in an atom within a distance of 0.1 Å. What is

the uncertainty involved in the measurement of its velocity?

Solution

hx p

4 or

hxm v

4 or

hv

4 xm

34

10 31

6.626 10 Jsv

4 3.14 0.1 10 m 9.11 10 kg= 0.579 10

7 m s

–1 (1J = 1 kg m

2 s

–2) = 5.79 10

6 m s

–1

Problem 2.16

A golf ball has a mass of 40g, and a speed of 45 m/s. If the speed can be measured within accuracy of 2%, calculate

the uncertainty in the position.

Solution

The uncertainty in the speed is 2%, i.e.,

245

100 = 0.9 m s

–1

hx

4 m v

34

3 1 1

6.626 10 Js

4 3.14 40 g 10 kg g (0.9 ms )= 1.46 10

–33 m

This is nearly ~ 1018

times smaller than the diameter of a typical atomic nucleus. As mentioned earlier for large

particles, the uncertainty principle sets no meaningful limit to the precision of measurements.



(CLASS: XI)


Problem 2.17

What is the total number of orbitals associated with the principal quantum number n = 3 ?

Solution

For n = 3, the possible values of l are 0, 1 and 2. Thus there is one 3s orbital (n = 3, l = 0 and ml = 0); there are three

3p orbitals (n = 3, l = 1 and ml = –1, 0, +1); there are five 3d orbitals (n = 3, l = 2 and ml = –2, –1, 0, +1, +2).

Therefore, the total number of orbitals is 1 + 3 + 5 = 9

The same value can also be obtained by using the relation; number of orbitals = n2, i.e. 3

2 = 9.

Problem 2.18

Using s, p, d, f notations, describe the orbital with the following quantum numbers

(a) n = 2, l = 1,

(b) n = 4, l = 0,

(c) n = 5, l = 3,

(d) n = 3, l = 2

Solution

n l orbital

(a) 2 1 2p

(b) 4 0 4s

(c) 5 3 5f

(d) 3 2 3d



(CLASS: XI)


NCERT Exercise

2.1 (i) Calculate the number of electrons which will together weigh one gram.

(ii) Calculate the mass and charge of one mole of electrons.

Ans: (i) 9.1 10–31

kg = 9.1 10–28

kg = Mass of 1 electron

9.1 10–28

g = 1 electron

1 g =28101.9

1 electrons = 1.1 10

–27 electrons

(ii) Mass of 1 electron = 9.1 10–31

kg

Mass of 1 mole electrons = 6.023 1023

9.1 10–31

kg

= 5.48 10–7

kg

Charge on 1 electron = – 1.6 10–19

C

Charge on 1 mole electrons = 6.023 1023

(– 1.6 10–19

) C

= – 96368 C

2.2 (i) Calculate the total number of electrons present in one mole of methane.

(ii) Find (a) the total number and (b) the total mass of neutrons in 7 mg of 14

C.

(Assume that mass of a neutron = 1.675 10–27

kg).

(iii) Find (a) the total number and (b) the total mass of protons in 34 mg of NH3 at STP.

Will the answer change if the temperature and pressure are changed?

Ans: (i) Total electrons in 1 molecule of CH4 = 10 electrons

Electrons in 1 mole of CH4 = 10 mole electrons

= 10 6.023 1023

electrons

= 6.023 1024

electrons

(ii) (a) 1 molecule of 14

C has = (14 – 6) = 8 neutrons

1 mole of 14

C has = 8 mol neutrons

Now, 1 mol of 14

C = 14 g 14

C

14 g of 14

C = 8 mol neutrons

1 g of 14

C = 8

14mol neutrons

7 mg of 14C = 0.007 g of 14

C = 8

0.00714

mol neutrons

= 0.004 mol neutrons

= 0.004 6.023 1023

neutrons

= 2.41 1021

neutrons

(b) Mass of 1 neutron = 1.67 10–27

kg

Total mass of neutrons = 1.67 10–27

2.41 1021

kg

= 4.02 10–6

kg

2.3 How many neutrons and protons are there in the following nuclei?

13 16 24 56 88

6 8 12 26 38C, O, Mg, Fe, Sr

Ans: Neutrons Protons 136 C 7 6

168 O 8 8



(CLASS: XI)


2412 Mg 12 12

5626Fe 30 26

8838Sr 50 38

2.4 Write the complete symbol for the atom with the given atomic number (Z) and atomic mass (A)

(i) Z = 17 , A = 35.

(ii) Z = 92 , A = 233.

(iii) Z = 4 , A = 9.

Ans: (i) 3517 Cl (ii) 233

92 U (iii) 94Be

2.5 Yellow light emitted from a sodium lamp has a wavelength ( ) of 580 nm. Calculate the frequency ( )

and wavenumber ( ) of the yellow light.

Ans: = 580 nm = 580 10–9

m

v = c or 9

8

10580

103c = 5.17 10

14 Hz

Wave number = 910580

11= 1.72 10

6 m

–1

2.6 Find energy of each of the photons which

(i) correspond to light of frequency 3 1015

Hz.

(ii) have wavelength of 0.50 Å.

Ans: (i) E = hv = 6.6 10–34

3 1015

= 1.98 10–18

J

(ii) 10

834

1050.0

103106.6hc E = 3.96 10

–15 J

2.7 Calculate the wavelength, frequency and wavenumber of a light wave whose period is 2.0 10–10

s.

Ans: Frequency, v =10102

1

period time

1 = 5 10

9 Hz

v = c or =9

8

105

103

v

c = 0.06 m

Wave number = 06.0

11= 16.67 m

–1

2.8 What is the number of photons of light with a wavelength of 4000 pm that provide 1J of energy?

Ans: nhc

E

or 34 8

12

n 6.6 10 3 101

4000 10

or n = 2.02 1016

photons



(CLASS: XI)


2.9 A photon of wavelength 4 10–7

m strikes on metal surface, the work function of the metal being 2.13

eV. Calculate (i) the energy of the photon (eV), (ii) the kinetic energy of the emission, and (iii) the

velocity of the photoelectron (1 eV = 1.6020 10–19

J).

Ans: = 4 10–7

m

Work function = 2.13 eV

(i) Energy of photon,

E = hv = 34 8

7

hc 6.6 10 3 10

4 10

J

= 4.95 10–19

J

19

19

4.95 10

1.6 10

eV

= 3.1 eV

(ii) Kinetic energy = Energy of photon – Threshold energy

= (3.1 – 2.13) eV = 0.97 eV

(iii) K.E. = 1

2mv

2, where m = 9.1 10

–31 kg (mass of electron)

K.E. = 0.97 eV = 0.97 1.6 10–19

J = 1.55 10–19

J

1.55 10–19

= 1

2(9.1 10

–31) (v

2)

or v2 =

19

31

2 1.55 10

9.1 10

= 3.4 10

11

or v = 113.4 10 = 5.83 105 m/s

2.10 Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the sodium atom.

Calculate the ionisation energy of sodium in kJ mol–1

.

Ans: Wavelength of radiation, = 242 nm = 242 10–9

m

K.E. = 0

hv = hv0 = ionization energy

Ionisation energy= hv = 34 8

9

hc 6.6 10 3 10

242 10

= 8.2 10

–19 J

Ionisation energy in kJ/mol = 198.2 10

1000

6.023 10

23

= 493.88 kJ/mol

2.11 A 25 watt bulb emits monochromatic yellow light of wavelength of 0.57 m. Calculate the rate of

emission of quanta per second.

Ans: Power = 25 watt

Power = Energy

Time (Here, time = 1 s)

25 = E

1 or E = 25 J

Now, E = nhv = nhc

Given, = 0.57 m = 0.57 10–6

m ; n = ?



(CLASS: XI)


25 = 34 8

6

n 6.6 10 3 10

0.57 10

6

34 8

25 0.57 10n

6.6 10 3 10

= 7.2 10

19 photons (quanta) per second

2.12 Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of

wavelength 6800 Å. Calculate threshold frequency ( 0) and work function (W0) of the metal.

Ans: Velocity, v = zero

Wavelength, = 6800 Å = 6800 10–10

m

Frequency, 8

10

c 3 10v

6800 10

= 4.4 1014

Hz

Now, hv = hv0 + 1

2mv

2

As v = 0 : hv = hv0

or v = v0 = 4.4 1014

Hz

Threshold frequency, v0 = 4.4 1014

Hz

Work function = hv0 = 6.6 10–34

4.4 1014

= 2.904 10–19

J

2.13 What is the wavelength of light emitted when the electron in a hydrogen atom undergoes transition

from an energy level with n = 4 to an energy level with n = 2?

Ans: 2 21 2

1 1v R

n n

z2 = 109677

2 2

1 1

2 4

(1)

2 cm

–1

= 109677 1 1

4 16

cm

–1

= 109677 3

16cm

–1

Now, 1

v

or = 1 16

v 3 109677

cm

= 4.86 10–5

cm

= 486 10–7

cm

= 486 nm

2.14 How much energy is required to ionise a H atom if the electron occupies n = 5 orbit? Compare your

answer with the ionization enthalpy of H atom (energy required to remove the electron from n = 1

orbit).

Ans: Energy required = 2 2

2 2

13.6 z 13.6 (1)

n (5)

= 0.544 eV

Ionisation energy of H–atom = 13.6 eV

Comparison : 13.6

0.544 = 25 times

Ionisation energy of H–atom is 25 times the energy required to remove electron from 5th orbit.



(CLASS: XI)


2.15 What is the maximum number of emission lines when the excited electron of a H atom in n = 6 drops

to the ground state?

Ans: Maximum number of emission lines 2 1 2 1(n n )(n n 1) (6 1)(6 1 1)

2 2

= 15 lines

2.16 (i) The energy associated with the first orbit in the hydrogen atom is –2.18 10–18

J atom–1

. What

is the energy associated with the fifth orbit?

(ii) Calculate the radius of Bohr’s fifth orbit for hydrogen atom.

Ans: (i) En = 18 2

2

2.18 10 z

n

J atom

–1

E5 = 18 2

2

2.18 10 (1)

(5)

= – 8.72 10

–20 J atom

–1

(ii) 2

n

0.529 nr Å

z

20.529 (5)

1

= 13.225 Å

2.17 Calculate the wavenumber for the longest wavelength transition in the Balmer series of atomic

hydrogen.

Ans: Longest wavelength transition corresponds to first line of Balmer series. (n1 = 2 to n2 = 3)

2

2 21 2

1 1v R (z)

n n

= 109677 2 2

1 1

2 3

= 109677

5

36

cm–1

or v 15232.9 cm–1

2.18 What is the energy in joules, required to shift the electron of the hydrogen atom from the first Bohr

orbit to the fifth Bohr orbit and what is the wavelength of the light emitted when the electron returns

to the ground state? The ground state electron energy is –2.18 10–11

ergs.

Ans: Energy required to shift electron from 1st Bohr orbit to fifth Bohr orbit, E = E5 – E1

E = 18 18

2 2

2.18 10 2.18 10

(5) (1)

= 2.18 10–18

1

125

= 2.18 10–18

24

25

= 2.09 10–18

J

or 2.09 10–11

erg (1 J = 107 erg)

Now, for calculating wavelength,

hc

E

or hc

E

34 8

18

6.6 10 3 10

2.09 10

= 9.47 10

–8 m

= 947 10–10

m

= 947 Å



(CLASS: XI)


2.19 The electron energy in hydrogen atom is given by En = (–2.18 10–18

)/n2 J. Calculate the energy

required to remove an electron completely from the n = 2 orbit. What is the longest wavelength of

light in cm that can be used to cause this transition?

Ans: Energy required to remove electron from n = 2 is 18 2

2

2.18 10 z

n

18

2

2.18 10

(2)

= 0.59 10

–18 J atom

–1

Now, E = hc

or 34 8

19

hc 6.6 10 3 10

E 5.9 10

= 3.35 10

–7 m

= 3.35 10–5

cm

= 3350 10–8

cm

= 3350 Å

2.20 Calculate the wavelength of an electron moving with a velocity of 2.05 107 m s

–1.

Ans: = ? ; Mass of electron, m = 9.1 10–31

kg ;

v = 2.05 107 m/s

34

31 7

h 6.6 10

mv 9.1 10 2.05 10

= 3.53 10

–11 m

2.21 The mass of an electron is 9.1 10–31

kg. If its K.E. is 3.0 10–25

J, calculate its wavelength.

Ans: h

mv

Now, K.E. = 21

mv2

or 2 2KE

vm

or 2KE

vm

Now, 22KE 2m KE

mv m 2mKEm m

h h

mv 2mKE

m = 9.1 10–31

kg ; KE = 3.0 10–25

J

34

31 25

6.6 10

2 9.1 10 3.0 10

= 8.93 10–7

m

= 8930 10–10

m

= 8930 Å

2.22 Which of the following are isoelectronic species i.e., those having the same number of electrons?

Na+, K

+, Mg

2+, Ca

2+, S

2–, Ar.

Ans: Na+, Mg

2+ : iso–electronic having 10 electrons each

K+, Ca

2+, S

2–, Ar : iso–electronic having 18 electrons each



(CLASS: XI)


2.23 (i) Write the electronic configurations of the following ions: (a) H– (b) Na

+ (c) O

2– (d) F

–

(ii) What are the atomic numbers of elements whose outermost electrons are represented by (a) 3s1

(b) 2p3 and (c) 3p

5 ?

(iii) Which atoms are indicated by the following configurations ?

(a) [He] 2s1 (b) [Ne] 3s

2 3p

3 (c) [Ar] 4s

2 3d

1.

Ans: (i) (a) H– : 1s

2 (b) Na

+ : 1s

22s

22p

6

(c) O2–

: 1s22s

22p

6 (d) F

– : 1s

22s

22p

6

(ii) (a) Z = 11 (b) Z = 7 (c) Z = 17

(iii) (a) Li (b) P (c) Sc

2.24 What is the lowest value of n that allows g orbitals to exist?

Ans: n = 5

2.25 An electron is in one of the 3d orbitals. Give the possible values of n, l and ml for this electron.

Ans: n = 3, l = 2, ml = – 2, – 1, 0, +1, +2

2.26 An atom of an element contains 29 electrons and 35 neutrons. Deduce (i) the number of protons and

(ii) the electronic configuration of the element.

Ans: (i) Number of protons = No. of electrons = 29

(ii) Electronic configuration: 1s22s

22p

63s

23p

63d

104s

1

2.27 Give the number of electrons in the species 2 2H , H and

2O

Ans: H2+ : 1 electron, H2 : 2 electrons, O2

+ : 15 electrons

2.28 (i) An atomic orbital has n = 3. What are the possible values of l and ml ?

(ii) List the quantum numbers (ml and l) of electrons for 3d orbital.

(iii) Which of the following orbitals are possible?

1p, 2s, 2p and 3f

Ans: (i) n = 3, l = 0 ml = 0

l = 1 ml = – 1, 0, +1

l = 2 ml = – 2, – 1, 0, +1, +2

(ii) n = 3, l = 2, ml = – 2, – 1, 0, +1, +2

(iii) 2s, 2p

2.29 Using s, p, d notations, describe the orbital with the following quantum numbers.

(a) n = 1, l = 0; (b) n = 3; l = 1

(c) n = 4; l = 2; (d) n = 4; l = 3.

Ans: (a) 1s (b) 3p (c) 4d (d) 4f

2.30 Explain, giving reasons, which of the following sets of quantum numbers are not possible.

(a) n = 0, l = 0, ml = 0, ms = + ½

(b) n = 1, l = 0, ml = 0, ms = – ½

(c) n = 1, l = 1, ml = 0, ms = + ½

(d) n = 2, l = 1, ml = 0, ms = – ½

(e) n = 3, l = 3, ml = –3, ms = + ½

(f) n = 3, l = 1, ml = 0, ms = + ½



(CLASS: XI)


Ans: (a) Not possible because ‘n’ can not be zero.

(b) Possible

(c) Not possible because ‘n’ and ‘l’ can not be zero.

(d) Possible

(e) Not possible because ‘n’ and ‘l’ can not be equal

(f) Possible

2.31 How many electrons in an atom may have the following quantum numbers?

(a) n = 4, ms = – ½ (b) n = 3, l = 0

Ans: (a) 16 electrons (b) 2 electrons

2.32 Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de

Broglie wavelength associated with the electron revolving around the orbit.

Ans: According to Bohr’s model

nhmvr

2

or nh

mv2 r

…(1)

According to de–Broglie’s principle

h

mv …(2)

Putting the value of ‘mv’ from (1) in (2)

h

2 rnh

or 2 r

n

or 2 r = n

Hence proved

2.33 What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition

n = 4 to n = 2 of He+ spectrum?

Ans: Let transition in hydrogen spectrum be n1 to n2

H 2 21 2

1 1v R

n n

z2 = R

2 21 2

1 1

n n

…(1)

Hl 2 2

1 1v R

2 4

(2)

2 = R

1 1

4 16

4 = R

4 4

4 16

or Hl

1 1v R

1 4

…(2)

Given that (1) = (2)

2 21 2

1 1 1 1R R

1 4n n

or 2 21 2

1 1 1 1

1 4n n



(CLASS: XI)


or 21n 1 or n1 = 1

22n 4 or n2 = 2

2.34 Calculate the energy required for the process

He+ (g) He

2+ (g) + e

–

The ionization energy for the H atom in the ground state is 2.18 10–18

J atom–1

Ans: He+ He

2+ + e

–

Ionization energy of helium = 18 2

2

2.18 10 Z

n

18 2

2

2.18 10 (2)

(1)

= 8.72 10

–18 J

2.35 If the diameter of a carbon atom is 0.15 nm, calculate the number of carbon atoms which can be

placed side by side in a straight line across length of scale of length 20 cm long.

Ans: Number of carbon atoms = 20 cm

0.15 nm 7

20 cm

0.15 10 cm

= 1.33 10

9

2.36 2 108 atoms of carbon are arranged side by side. Calculate the radius of carbon atom if the length

of this arrangement is 2.4 cm.

Ans: Diameter of 1 carbon atom = 8

2.4

2 10 = 1.2 10

–8 cm

Radius of 1 carbon atom = 81.2 10

2

= 6 10

–9 cm = 0.06 nm

2.37 The diameter of zinc atom is 2.6 Å. Calculate (a) radius of zinc atom in pm and (b) number of atoms

present in a length of 1.6 cm if the zinc atoms are arranged side by side lengthwise.

Ans: Diameter = 2.6 Å

(a) Radius = 2.6

2 = 1.3 Å = 1.3 10

–10 m = 130 10

–12 m = 130 pm

(b) No. of atoms in a length of 1.6 cm 8

1.6 cm 1.6 cm

2.6 Å 2.6 10 cm

= 6.15 10

7 atoms

2.38 A certain particle carries 2.5 10–16

C of static electric charge. Calculate the number of electrons

present in it.

Ans: Charge on 1 electron = 1.6 10–19

C

or 1.6 10–19

C = 1 electron

1C = 19

1

1.6 10 electrons

2.5 10–16

C = 19

1

1.6 10 2.5 10

–16 electrons = 1563 electrons



(CLASS: XI)


2.39 In Milikan’s experiment, static electric charge on the oil drops has been obtained by shining X-rays. If

the static electric charge on the oil drop is –1.282 10–18

C, calculate the number of electrons present

on it.

Ans: Charge on 1 electron = 1.6 10–19

C

1.6 10–19

C = 1 electron

1 C = 19

1

1.6 10 electrons

1.282 10–18

C = 19

1

1.6 10 1.282 10

–18 electrons = 8 electrons

2.40 In Rutherford’s experiment, generally the thin foil of heavy atoms, like gold, platinum etc. have been

used to be bombarded by the -particles. If the thin foil of light atoms like aluminium etc. is used,

what difference would be observed from the above results?

Ans: Smaller number of –particles will be deflected, because of lesser number of protons in lighter nucleus.

2.41 Symbols 79

35Br and 79

Br can be written, whereas symbols 79

35Br and 35

Br are not acceptable. Answer

briefly.

Ans: Try yourself.

2.42 An element with mass number 81 contains 31.7% more neutrons as compared to protons. Assign the

atomic symbol.

Ans: Let number of protons be x

Number of neutrons = x + 31.7

100x

Mass number = No. of protons + No. of neutrons = 81

or x + x + 31.7x

100 = 81

31.7x2x 81

100

231.7x = 8100

or 8100

x 35231.7

No. of protons = Atomic number = 35

Element = Br

Symbol : 8135Br

2.43 An ion with mass number 37 possesses one unit of negative charge. If the ion contains 11.1% more

neutrons than the electrons, find the symbol of the ion.

Ans: Mass no. = 37

Let no. of electrons be x

No. of neutrons = x + 11.1

100x

No. of protons = x – 1

Mass no. = No. of protons + No. of neutrons = 37



(CLASS: XI)


or x – 1 + x + 11.1

100x = 37

211.1x – 100 = 3700

211.1x = 3600

x = 3600

211.1 = 17

No. of protons = x – 1 = 17 – 1 = 16

2.44 An ion with mass number 56 contains 3 units of positive charge and 30.4% more neutrons than

electrons. Assign the symbol to this ion.

Ans: Let no. of electrons be x

No. of neutrons = x + 30.4

x100

No. of protons = x + 3

Mass No. = 56 = No. of protons + No. of neutrons

x + 3 + x + 30.4

x 56100

230.4x + 300 = 5600

230.4x = 5300

x = 5300

230.4 = 23

No. of protons = x – 3 = 23 + 3 = 26

Symbol : 56 326Fe

2.45 Arrange the following type of radiations in increasing order of frequency: (a) radiation from

microwave oven (b) amber light from traffic signal (c) radiation from FM radio (d) cosmic rays from

outer space and (e) X-rays.

Ans: Radiation from FM radio < radiation from microwave oven < amber light from traffic signal < X–rays < cosmic

rays

2.46 Nitrogen laser produces a radiation at a wavelength of 337.1 nm. If the number of photons emitted is

5.6 1024

, calculate the power of this laser.

Ans: = 337.1 nm = 337.1 10–9

m

34 8

9

hc 6.6 10 3 10E

337.1 10

= 5.87 10

–19 J

Power = Energy

Time

(Here time is not given, so we assume t = 1 s)

Power = 195.87 10

1

= 5.87 10

–19 watt



(CLASS: XI)


2.47 Neon gas is generally used in the sign boards. If it emits strongly at 616 nm, calculate (a) the

frequency of emission, (b) distance traveled by this radiation in 30 s (c) energy of quantum and (d)

number of quanta present if it produces 2 J of energy.

Ans: = 616 nm = 616 10–9

m

(a) Frequency of emission, 8

9

C 3 10v

616 10

= 4.87 1014

Hz

(b) Distance = velocity time

velocity of light = 3 108 m/s; time = 30 s

Distance = 3 108 30 = 9 10

9 m

(c) Energy of quantum, 34 8

9

hc 6.6 10 3 10E

616 10

J = 3.21 10

–19 J

(d) nhc

E

or 9

34 8

E 2 616 10n

hc 6.6 10 3 10

= 6.2 10

18 photons

2.48 In astronomical observations, signals observed from the distant stars are generally weak. If the

photon detector receives a total of 3.15 10–18

J from the radiations of 600 nm, calculate the number

of photons received by the detector.

Ans: nhc

E

or 3.15 10–18

= 34 8

9

n 6.6 10 3 10

600 10

or n = (3.15 10–18

600 10–9

)/6.6 10–34

3 108 = 9.54 10 photons

2.49 Lifetimes of the molecules in the excited states are often measured by using pulsed radiation source

of duration nearly in the nano second range. If the radiation source has the duration of 2 ns and the

number of photons emitted during the pulse source is 2.5 1015

, calculate the energy of the source.

Ans: 9

1 1 1v

time period 2ns 2 10

= 5 10

8 Hz

n = 2.5 1015

Energy, E = nhv = 2.5 1015

6.6 10–34

5 108 = 8.25 10

–10 J

2.50 The longest wavelength doublet absorption transition is observed at 589 and 589.6 nm. Calculate the

frequency of each transition and energy difference between two excited states.

Ans: 8

589 9

C 3 10v

589 10

= 5.093 1014

Hz

8

589.6 9

C 3 10v

589.6 10

= 5.088 1014

Hz

Energy difference = hv589.6 – hv589 = h (v589.6 – v589)

= 6.6 10–34

(5.093 1014

– 5.088 1014

)

= 6.6 10–34

1014

(5.093 – 5.088)

= 3.3 10–22

J



(CLASS: XI)


2.51 The work function for caesium atom is 1.9 eV. Calculate (a) the threshold wavelength and (b) the

threshold frequency of the radiation. If the caesium element is irradiated with a wavelength 500 nm,

calculate the kinetic energy and the velocity of the ejected photoelectron.

Ans: Work function = 1.9 eV = 1.9 1.6 10–19

J = 3.04 10–19

J

(a) Threshold wavelength : 0

hcE

or 3.04 10–19

= 34 8

0

6.6 10 3 10

or 34 8

0 19

6.6 10 3 10

3.04 10

= 6.51 10

–7 m = 651 nm

(b) Threshold frequency

8

0 70

c 3 10v

6.51 10

= 4.61 1014

Hz

Wavelength of incident radiation = 500 nm = 500 10–9

m

Frequency of incident radiation, 8

19

c 3 10v

500 10

= 6 1014

Hz

Now, hv = hv0 + K.E.

K.E. = hv – hv0 = h (v – v0)

= 6.6 10–34

(6 1014

– 4.61 1014

)

= 6.6 10–34

1014

(6 – 4.61)

= 6.6 10–20

1.39 = 9.2 10–20

J

2.53 The ejection of the photoelectron from the silver metal in the photoelectric effect experiment can be

stopped by applying the voltage of 0.35 V when the radiation 256.7 nm is used. Calculate the work

function for silver metal.

Ans: K.E. = eV = 1.6 10–19

0.35 = 0.56 10–19

J

= 256.7 nm = 256.7 10–9

m

8

9

c 3 10

256.7 10

= 1.17 10

15 Hz

hv = hv0 + K.E. or hv0 = hv – K.E.= (6.6 10–34

1.17 1015

) – (0.56 10–19

)

= (7.72 10–19

) – (0.56 10–19

)

= 10–19

(7.72 – 0.56)

= 7.16 10–19

J

19

19

7.16 10

1.6 10

= 4.3 eV

2.54 If the photon of the wavelength 150 pm strikes an atom and one of its inner bound electrons is

ejected out with a velocity of 1.5 107 m s

–1, calculate the energy with which it is bound to the

nucleus.

Ans: = 150 pm = 150 10–12

m

v = 1.5 107 m/s

Binding energy or work function, hv0 = ?



(CLASS: XI)


hv = hv0 + 1

2mv

2

or hv0 = hv – 1

2mv

2 =

hc 1

2

mv

2

or hv0 = 34 8

12

6.6 10 3 10

150 10

31 7 219.1 10 (1.5 10 )

2

= [1.32 10–15

] – [0.1 10–15

] = 1.22 10–15

J

2.55 Emission transitions in the Paschen series end at orbit n = 3 and start from orbit n and can be

represented as = 3.29 1015

(Hz) [1/32 – 1/n

2]

Calculate the value of n if the transition is observed at 1285 nm. Find the region of the spectrum.

Ans: = 1285 nm = 1285 10–9

m,

v = 8

9

c 3 10

1285 10

= 2.33 10

14 Hz

2.33 1014

= 3.29 105

2

1 1

9 n

or 14

2 15

1 1 2.33 10

9 n 3.29 10

= 0.0708

2

1 1

9n – 0.0708 = 0.11 – 0.0708 = 0.0392

or n2 =

1

0.0392 = 25.57

or n = (25.57)1/2

5

2.56 Calculate the wavelength for the emission transition if it starts from the orbit having radius 1.3225 nm

and ends at 211.6 pm. Name the series to which this transition belongs and the region of the

spectrum.

Ans: r1 = 1.3225 nm = 1322.5 pm

r = 2 252.9 n 52.9 n

z 1

= 1322.5

or n2 =

1322.5

52.9 = 25 or n = 5

r2 = 211.6 pm

52.9 n2 = 211.6

n2 =

211.6

52.9 = 4 or n = 2

2

2 21 2

1 1v R Z

n n

= 109677 2 2

1 1

2 5

= 109677

21

100cm

–1

1 100

v 21 109677

= 4.34 10

–5 cm

The line belongs to Balmer series and lies in visible region of spectrum.



(CLASS: XI)


2.57 Dual behaviour of matter proposed by de Broglie led to the discovery of electron microscope often

used for the highly magnified images of biological molecules and other type of material. If the

velocity of the electron in this microscope is 1.6 106 ms

–1, calculate de Broglie wavelength

associated with this electron.

Ans: = ?, v = 1.6 106 m/s ; m = 9.1 10

–31 kg

34

6 31

h 6.6 10

mv 1.6 10 9.1 10

= 4.5 10

–10 m = 450 pm

2.58 Similar to electron diffraction, neutron diffraction microscope is also used for the determination of

the structure of molecules. If the wavelength used here is 800 pm, calculate the characteristic

velocity associated with the neutron.

Ans: Mass of neutron, m = 1.67 10–27

kg

= 800 pm = 800 10–12

m, v = ?

= h

mv or v =

34

27 12

h 6.6 10

m 1.67 10 800 10

= 494 m/s

2.59 If the velocity of the electron in Bohr’s first orbit is 2.19 106 ms

–1, calculate the de Broglie

wavelength associated with it.

Ans: 34

31 6

h 6.6 10

mv 9.1 10 2.19 10

= 3.31 10

–10 m = 331 pm

2.60 The velocity associated with a proton moving in a potential difference of 1000 V is 4.37 105 ms

–1. If

the hockey ball of mass 0.1 kg is moving with this velocity, calculate the wavelength associated with

this velocity.

Ans: 34

5

h 6.6 10

mv 0.1 4.37 10

= 1.51 10

–38 m

2.61 If the position of the electron is measured within an accuracy of 0.002 nm, calculate the uncertainty

in the momentum of the electron. Suppose the momentum of the electron is h/4 m 0.05 nm, is there

any problem in defining this value.

Ans: x = 0.002 nm = 0.002 10–9

m, p = ?

x . p = h

4

or 34

9

h 6.6 10p

4 x 4 3.14 0.002 10

= 2.62 10

–23

2.62 The quantum numbers of six electrons are given below. Arrange them in order of increasing

energies. If any of these combination(s) has/have the same energy lists:

1. n = 4, l = 2, ml = –2 , ms = –1/2

2. n = 3, l = 2, ml = 1 , ms = +1/2

3. n = 4, l = 1, ml = 0 , ms = +1/2

4. n = 3, l = 2, ml = –2 , ms = –1/2

5. n = 3, l = 1, ml = –1 , ms = +1/2

6. n = 4, l = 1, ml = 0 , ms = +1/2

Ans: (1) > (6) = (3) > (2) = (4) > (5)



(CLASS: XI)


2.63 The bromine atom possesses 35 electrons. It contains 6 electrons in 2p orbital, 6 electrons in 3p

orbital and 5 electron in 4p orbital. Which of these electron experiences the lowest effective nuclear

charge?

Ans: 4p orbital

2.64 Among the following pairs of orbitals which orbital will experience the larger effective nuclear

charge?

(i) 2s and 3s, (ii) 4d and 4f, (iii) 3d and 3p.

Ans: (i) 2s (ii) 4d (iii) 3p

2.65 The unpaired electrons in Al and Si are present in 3p orbital. Which electrons will experience more

effective nuclear charge from the nucleus?

Ans: Si

2.66 Indicate the number of unpaired electrons in : (a) P, (b) Si, (c) Cr, (d) Fe and (e) Kr.

Ans: (a) P (Z = 15) : 1s22s

22p

63s

23p

3, No. of unpaired electrons = 3

(b) Si (Z = 14) : 1s22s

22p

63s

23p


(c) Cr (Z = 24) : [Ar] 3d54s


(d) Fe (Z = 26) : [Ar] 3d64s


(e) Kr : No. of unpaired electrons in noble gases = 0

2.67 (a) How many sub-shells are associated with n = 4?

(b) How many electrons will be present in the sub-shells having ms value of –1/2 for n = 4?

Ans: (a) No. of sub–shells = n2 = 4

2 = 16

(b) 16 electrons

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