HC Verma Solutions Class 12 Chapter 21 Bohr’s Theory And Physics Of Atom

HC Verma Solutions Vol 2 Chapter 21 Bohr’s Theory and Physics of Atom gives students the opportunity to learn and solve questions related to the unit in a more effective manner. The solutions offer greater insight into all the exercises and the detailed answers provided in the solutions will further help students to take their preparation to a higher level. The exercises are based on topics such as the dimensional formula of acceleration, wavelength, and radius of atoms etc.  Additionally, some of the problems given include;

  • Questions on finding the energy to take out one photon along with finding the intensity and we also have questions on counting the number of photons.
  • There are questions related to finding energies associated with types of radiations.
  • Questions are also based on Rydberg’s Constant, excitation potential, ionization potential, angular momentum, magnetic dipole moment, etc.
  • Finding the frequency of radiation emitted, time period, no. of revolutions, the energy released of atomic particles.

The HC Verma Solutions will prove to useful for Class 12 students who are preparing for competitive exams like JEE. With these solutions, students will also develop better skills to tackle different types of questions and prepare efficiently.

Download This Solution As PDF: HC Verma Solutions Chapter 21 PDF

Key Concepts Related To Bohr’s Theory And Physics Of Atom

Some of the important topics covered in this chapter are;

  • Early Atomic Models
  • Hydrogen Spectra
  • Difficulties With Rutherford’s Model
  • The Wave Function of an Electron
  • Quantum Mechanics of Hydrogen Atom
  • Nomenclature in Atomic Physics
  • Laser

Class 12 Important Questions In Chapter 21

1. It is said that the numerical value of ionization energy in eV is equal to the ionization potential in volts. If these quantities are measured in some other units what will happen to the equal status?

2. Why is the Balmer series normally observed and analysed before the other series?   

3. The numerical value of ionization energy in eV equals the ionization potential in volts. Does the equality hold if these quantities are measured in other units?

4. Which parameters will be the same for all hydrogen-like atoms and ions in their ground states?

(a) the energy of the atom (b) orbital angular momentum of the electron (c) radius of the orbit (d) speed of the electron

5. What happens when a photon stimulates the emission of another photon?

(a) the two photons will have the same wavelength (b) same phase (c) same energy (d) same direction

HC Verma Solutions Vol 2 Bohr’s Theory And Physics of Atom Chapter 21

Question 1: The Bohr radius is given by a0 = (Єoh2)/(πme2) .Verify that the RHS has dimensions of length.

Solution:

Dimensions of Єo = [A2T2]

Dimensions of π = [L2MLT-2]

Dimensions of e = [AT]

Dimensions of h = [ML2T-1]

Bohr’s radius = [M2L4T-2]/[M2L3T-2]

Or a0 = L, which is in the dimension of length.

Question 2: Find the wavelength of the radiation emitted by hydrogen in the transitions (a) n = 3 to n = 2, (b) n=5 to n=4 and (c) n=10 to n=9

Solution:

We know the formula to find the wavelength:

HC Verma Class 12 Ch 21 Solution 2

Where n = quantum number of final state and m = quantum number of state

Here Z = 1 [As, atom is hydrogen]

(a) n=2 and m=3

1/λ = 1.1 x 107[1/22 – 1/32]

or λ = 6.54 x 10-7 = 654 nm

(b) n=4, m=5

1/λ = 1.1 x 107[1/42 – 1/52]

or λ = 400/[1.1 x 107x9] = 4050 nm

(c) n=9, m=10

1/λ = 1.1 x 107[1/92 – 1/102]

or λ = [81×100]/[1.1 x 107x19] = 38849 nm

Question 3: Calculate the smallest wavelength of radiation that may be emitted by (a) hydrogen (b) He+ and (c) Li++

Solution:

For smallest wavelength, energy should be maximum. So, electron will jump from ground state to the state where energy is maximum.

Let us say: ground state is n=1 and the final stage is m = ∞

we know, 1/λ = RZ2 (1/n2 – 1/m2)

(a) Here Z = 1, for Hydrogen atom

=> 1/λ = 1.1 x 107[1/12 – 1/∞]

Or λ = 91 nm

(b) Here Z = 2, for He atom

=> 1/λ = 1.1 x 107 x 22[1/12 – 1/∞]

Or λ = 23 nm

(b) Here Z = 3, for Li atom

=> 1/λ = 1.1 x 107 x 32[1/12 – 1/∞]

Or λ = 10 nm

Question 4: Evaluate Rydberg constant by putting the values of the fundamental constants in its expression.

Solution:

Rydberg constant = R = me4/8Єo2h3c

Where,

Charge of electron =e = 1.6 x 10-19 C

Mass of electron = m = 9.1 x 10-31 kg

Velocity of light = c = 3 x 108 m/s

Planks constant = h = 6.63 x 10-34 J-s

Permittivity = Єo = 8.85 x 10-12

Substituting all the above values, we have

R = [9.1 x 10-31x(1.6 x 10-19)4]/[8x(6.63 x 10-34)3x3 x 108x(8.85 x 10-12)2]

= 1.097 x 107

Question 5: Find the binding energy of a hydrogen atom in the state n=2.

Solution:

Energy of hydrogen atom = E = -13.6/n2 eV

Where 13.6 is the binding energy of the hydrogen atom

In this case, initial state of the atom will be m=2

If we consider transition state from infinity to n = 2

E = -13.6/n2 – (-13.6/m2)

or = 13.6(1/∞2 – 1/22)

or E = -13.6 x 1/4 = -3.4 eV

Question 6: Find the radius and energy of He+ ion in the states

(a) n=1 (b) n=4 (c) n=10

Solution:

We know, radius of an atomic orbit = r = n2ao/Z

Here ao = 0.053 nm

or r = (0.053xn2)/Z

The energy of the atomic orbit = E = (-13.6xZ2)/n2

Here Z = 2, As atomic number of He is 2

(a) n = 1

r = 0.053×12/2 = 0.265 angstrom

E = -13.6×22/12 = -54.4 eV

(b) n = 4

r = 0.053×42/2 = 4.24 angstrom

E = -13.6×22/42 = -3.4 eV

(c) n = 10

r = 0.053×102/2 = 26.5 angstrom

E = -13.6×22/102 = -0.544 eV

Question 7: A hydrogen atom emits ultraviolet radiation of wavelength 102.5 nm. What are the quantum numbers of the states involved in the transition?

Solution:

we know, 1/λ = RZ2(1/n2 – 1/m2)

Given, λ = 102.5nm , n = 1 and Z=1 (atomic number of hydrogen is 1)

=>1/[102.5×10-9] = 1.1 x 107(1/12 – 1/m2)

=>109/[102.5×1.1 x 109] = 1 – 1/m2

=>m = 2.97 = 3 (approx.)

Question 8: (a) Find the first excitation potential of He+ ion.

(b) Find the ionisation potential of Li++ ion.

Solution:

(a) Using relation, E = -13.6xz2/n2

As He+ makes first transition, so n=1 and m=2.

Energy at stage 1, E1 = -13.6×22/12 = -54.4 eV

Energy at stage 2, E2 = -13.6×22/22 = -13.6 eV

The difference of the two energies will give us the excitation potential of He+. So

E = E2 – E1 = -13.6 – (-54.4) = 40.8 V

(b) For Li++, atomic number, Z = 3.

Ionisation potential = 13.6 x Z2 = 13.6×9 = 122.4

So, ionisation potential of Li+ is 122.4 V

Question 9: A group of hydrogen atoms are prepared in n=4 states. List the wavelengths that are emitted as the atoms makes the transitions and return to n=2 states.

Solution: There will be 3 wavelengths:

we know, 1/λ = RZ2(1/n2 – 1/m2)

As atomic number of hydrogen = 1, so Z = 1.

Here R = 1.1 x 107 m-1

For the first case, n=3 and m=4

1/ λ = 1.1 x 107 x (1/9 – 1/16)

Or λ = 1875 nm

For the second case, n=2 and m=3

1/ λ = 1.1 x 107 x (1/4 – 1/9)

Or λ = 655 nm

For the third case, n=2 and m=4

1/ λ = 1.1 x 107 x (1/4 – 1/16)

Or λ = 487 nm

Question 10: A positive ion having just one electron ejects it if a photon of wavelength 228Å or less is absorbed by it. Identify the ion.

Solution:

Since the electron is ejected, so it makes a transition from its ground state n=1 to the immediate excited state m=2.

we know, 1/λ = RZ2(1/n2 – 1/m2) and E = hc/ λ

From both the equations, we have

-13.6xZ2 (1/22 – 1/12) = 0.0872 x 10-16 eV

Or -13.6xZ2(-3/4) = 0.0872 x 10-16 x 1.6 x 10-19 V

Or Z2 = 5.3

Or Z = 2.3 (approx.)

The ion may be He+.

Question 11: Find the maximum Coulomb force that can act on the electron due to the nucleus in a hydrogen atom.

Solution:

Using Coulomb force, F, is

F = q1q2/4πЄor2

Here r = distance between nucleus and the electron = Bohr’s radius = 0.053 angstrom

Also value of 1/4πЄo = 9 x 109

q1 = q2 = q = 1.6 x 10-19 C

So, F = 9 x 109 x (1.6 x 10-1)2/(0.053×10-10)2 = 8.22 x 10-8 N

Question 12: A hydrogen atom in a state having a binding energy of 0.85eV makes transition to a state with excitation energy 10.2eV.

(a) Identify the quantum number n of the upper and the lower energy states involved in the transition.

(b) Find the wavelength of the emitted radiation.

Solution:

We know, binding energy = E = 13.6Z2/n2 = 13.6/n12

Given: binding energy in the first case 0.85eV and second case 10.2eV

0.85 = 13.6/n12 => n1 = 4

Again 10.2 = 13.6/n22 => n2 = 2

Thus, the states are 4 and 2.

(b) using part (a) results

1/λ = 1.1 x 107[1/22 – 1/42]

Atomic number of Hydrogen = Z = 1

=> λ = 487 nm

Question 13: Whenever a photon is emitted by hydrogen in Balmer series, it is followed by another photon in Lyman series. What wavelength does this latter photon correspond to?

Solution:

Given: photon is emitted in Balmer series and then next in Lyman series.

Let n = 1 and m=2, then we obtain the wavelength as,

1/λ = 1.1 x 107[1/12 – 1/22]

Atomic number of Hydrogen = Z = 1

=> λ = 122 nm

Question 14: A hydrogen atom in state n=6 makes two successive transitions and reaches the ground state. In the first transition a photon of 1.13eV is emitted.

(a) Find the energy of the photon emitted in the second transition.

(b) What is the value of n in the intermediate state?

Solution:

(a) Energy at n = 6

E = -13.6xz2/n2 = -13.6×12/62 = -0.37777

Energy for hydrogen in the ground state = E = -13.6

Energy in the second transition = [Energy for hydrogen in the ground state] – [Energy at sixth state (n = 6)] + [first transition energy]

= -13.6 + (-0.37777) + 1.13 = -12.1

-ve sign here indicates that energy is being given for making the transitions from 6th state to the ground state.

=>E = 12.1 eV

(b)

Energy in the intermediate state =1.13+0.3777777 = 1.507 eV

We know, E = 13.6xz2/n2 = 13.6×12/n2

=>1.507 = 13.6×12/n2

=> n = 3 (approx)

Question 15: What is the energy of a hydrogen atom in the first excited state if the potential energy is taken to be zero in the ground state?

Solution:

We know, hydrogen requires 10.2 eV of energy to get excited from the ground state.

As the energy of hydrogen atom in the ground state is 13.6 eV. So, the energy needed to make the hydrogen atom reach the first excited state is,

E = 13.6 + 10.2 = 23.8 eV

Question 16: A hot gas emits radiation of wavelengths 46 nm,82.8 nm and 103.5 nm only. Assume that the atoms have only two excited states and the difference between consecutive energy levels decreases as energy is increased. Taking the energy of the highest energy state to be zero, find the energies of the ground state and the first excited state.

Solution:

Energy of the ground state will be the energy acquired in the transition of 2 excitation state to ground state.

As per given statement, higher energy state to be zero, so n=2 is zero in terms of energy, as atom makes transition to n=2 from ground state.

Energy of the ground state = E = hc/ λ1 = [6.634 x 1034 x 3 x 108]/[ 46 x 10-9] = 27 eV

Energy of the first excited state = E = hc/ λ2 = [6.634 x 1034 x 3 x 108]/[103.5 x 10-9] = 12 eV

Question 17: A gas of hydrogen like ions is prepared in a particular excited state A. It emits photons having wavelength equal to the wavelength of the first line of the Lyman series together with photons of five other wavelengths. Identify the gas and find the principal quantum number of the state A.

Solution:

We know, (n(n-1))/2 = 6 …(1)

where n =principal quantum number

As gas emits total of 6 wavelengths.

=> Solving (1), we have n = 4

And gas is in the 4th excited state. Gas is He+.

Question 18: Find the maximum angular speed of the electron of a hydrogen atom in a stationary orbit.

Solution:

Let m be the mass of the electron, v be the linear speed, ω angular speed and r be the radius of hydrogen’s first stationary orbit.

We know, mvr = nh/2π

Where n = any integer

and V = rω

So, from both the equations, we have

mr2ω = nh/2π

=> ω = nh/2πmr2 = [1x(6.63 x 1034)] / [2×3.14x(9.1093 x 10-31)x(0.53 x 10-10)2]

=> ω = 0.413 x 1017

= 4.13 x 10^16 rad s-1.

Question 19: A spectroscopic instrument can resolve two nearby wavelengths λ and λ+ Δλ if λ/Δλ is smaller than 8000.This is used to study the spectral lines of the Balmer series of hydrogen. Approximately how many lines will be resolved by the instrument?

Solution:

The range of Balmer series is from 656.3nm to 365 nm. (known)

Now,

Number of wavelengths in the given range= [656.3-365 ]/8000 = 36

Two lines will be extra for the 1st and last wavelength, so total number of lines = 36+2 = 38

Question 20: Suppose, in certain conditions only those transitions are allowed to hydrogen atoms in which the principal quantum number n changes by 2.

(a) Find the smallest wavelength emitted by hydrogen.

(b) List the wavelengths emitted by hydrogen in the visible range (380 nm to 780 nm).

Solution:

Let n = 1 and m = 3

n = for minimum wavelength and m = for given condition

Using Einstein-Planck equation:

E = hc/λ and

Also, we know, E = 13.6 (1/n2 – 1/m2)

=> 13.6 (1/n2 – 1/m2) = hc/λ

=> 13.6 (1/12 – 1/32) = [6.64×10-34x3x108]/λ

=> 13.6 (1/1 – 1/9) = [4.14×10-15x3x108]/λ

=> λ = 1.027 x 10-7 m = 103 nm

(b) Wavelength obtained in above result does not fall in visible range so let us consider another transition possible that is from n = 2 to m = 4

=> 13.6 (1/22 – 1/42) = [6.64×10-34x3x108]/λ

=> λ = 487 nm

Above wavelength falls in visible range. This is the wavelength emitted by hydrogen in the visible range.

Question 21: According to Maxwell’s theory of electrodynamics, an electron going in a circle should emit radiation of frequency equal to its frequency of revolution. What should be the wavelength of the radiation emitted by a hydrogen atom in ground state if this rule is followed?

Solution:

Let r be the radius of the ground state and v be the velocity of the electron moving in ground state.

Frequency of the revolution of electron in the circle, f = v/2πr

Also, f = c/λ

Where f = frequency of radiation emitted by hydrogen atom in ground state

=> λ = 2πcr/v = [2x3x108x3.14×0.53×10-10]/[2187×103] = 45.686 nm

Question 22: The average kinetic energy of molecules in a gas at temperature T is 1.5 kT. Find the temperature at which the average kinetic energy of the molecules of hydrogen equals the binding energy of its atoms. Will hydrogen remain in molecular from at this temperature? Take k = 8.62 × 10–5 eV K–1.

Solution:

As, Binding energy is equal to the average kinetic energy of the molecules.

(3/2)KT = 13.6 eV

=> T = 13.6/[1.5×8.62×10-5] = 1.05 x 105 K

No, it is not possible for hydrogen molecules to remain in molecular state at such a high temp.

Question 23: Find the temperature at which the average thermal kinetic energy is equal to the energy needed to take a hydrogen atom from its ground state to n = 3 state. Hydrogen can now emit red light of wavelength 653.1 nm. Because of Maximillian distribution of speeds, a hydrogen sample emits red light at temperatures much lower than that obtained from this problem. Assume that hydrogen molecules dissociate into atoms.

Solution:

Energy needed to take hydrogen atom from ground to second exited state is

E = 13.6(1/12 – 1/32) = 12.08 eV

The average thermal kinetic energy = 3/2 KT

As average thermal kinetic energy is equal to the energy.

So, 3/2 KT = 12.08 e V

1.5 x 8.62 x 10-5 x T = 12.08

=> T = 12.08/[1.5 x 8.62 x 10-5] = 9.4 x 104 kelvin

Question 24: Average lifetime of a hydrogen atom excited to n = 2 state is 10-8 s. Find the number of revolutions made by the electron on the average before it jumps to the ground state.

Solution:

From Bohr’s model, frequency is

f = me4/4ε2n3h3

Where n = 2, h = 6.63 x 10-34 and m = 9.1 x 10-31 kg

Time period (T) = 1/f = 4ε2n3h3/me4

= [4×8.852x23x6.633]/[9.1×1.64] x 10-19

= 1.224 x 10-15 seconds

Now,

Number of revolution = [Average lifetime]/[Time period]

Given: Average lifetime = 10–8 s

so, Number of revolution = 10-8/[1.224×10-15] = 8.16 x 106 revolutions

Question 25: Calculate the magnetic dipole moment corresponding to the motion of the electron in the ground state of a hydrogen atom.

Solution:

Since hydrogen atom is in ground state, this implies n = 1

Dipole moment of the electron = μ = niA = qfA

= e x me4/[4Єo2h3n3] x πr2 n2

= [me5 πr2 n2]/[4Єo2h3n3]

HC Verma Class 12 Ch 21 Solution 25