Dual Nature of Radiation and Matter NCERT Solutions

SimplyAcad has provided the best NCERT Solutions Class 12 Dual Nature of Radiation and Matter to ensure students’ foundations are built strongly through comprehensive answers. The solutions are solved in a detailed way and elaborately explains each step. This allows no space for any doubts and confusions in students’ minds. 

The NCERT Solutions Class 12 Dual Nature of Radiation and Matter offers great support to students. It prepares them for the approaching board examinations as well as many entrance exams where radiation forms an essential segment of the physics paper. Scroll below to find the solutions of the entire exercise prescribed in the syllabus. 

NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter

 

Q.11.1:Question

Solution

Let’s solve the problem step by step.

Part A: Maximum Frequency of X-rays

Given:

• Potential of the electrons,
  $V = 30 \, \text{kV} = 3 \times 10^4 \, \text{V}$ 

  V=30kV=3×104V

• Planck’s constant,
  $h = 6.626 \times 10^{-34} \, \text{J} \cdot \text{s}$ 

  h=6.626×10−34J⋅s

• Energy of 1 electron volt,
  $1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}$ 

  1eV=1.6×10−19J

Step 1: Calculate the energy of the electrons.

The energy

$E$

E of the electrons can be calculated using the relation:

 

$$E = V \times \text{eV}$$

E=V×eV

Substituting the values:

 

$$E = 3 \times 10^4 \times 1.6 \times 10^{-19} \, \text{J}$$

E=3×104×1.6×10−19J

$$E = 4.8 \times 10^{-15} \, \text{J}$$

E=4.8×10−15J

Step 2: Calculate the maximum frequency

$v_{\text{max}}$

vmax​ of X-rays.

The maximum frequency of X-rays is given by:

 

$$v_{\text{max}} = \frac{E}{h}$$

vmax​=hE​

Substituting the values:

 

$$v_{\text{max}} = \frac{4.8 \times 10^{-15}}{6.626 \times 10^{-34}} \, \text{Hz}$$

vmax​=6.626×10−344.8×10−15​Hz

$$v_{\text{max}} = 7.24 \times 10^{18} \, \text{Hz}$$

vmax​=7.24×1018Hz

Final Answer for Part A:

 

$$v_{\text{max}} = 7.24 \times 10^{18} \, \text{Hz}$$

vmax​=7.24×1018Hz

Part B: Minimum Wavelength of X-rays

Step 3: Calculate the minimum wavelength

$\lambda_{\text{min}}$

λmin​.

The minimum wavelength

$\lambda_{\text{min}}$

λmin​ can be calculated using the relation:

 

$$\lambda_{\text{min}} = \frac{c}{v_{\text{max}}}$$

λmin​=vmax​c​

Where:

• 
  $c = 3 \times 10^8 \, \text{m/s}$ 

  c=3×108m/s (speed of light)

Substituting the values:

 

$$\lambda_{\text{min}} = \frac{3 \times 10^8}{7.24 \times 10^{18}} \, \text{m}$$

λmin​=7.24×10183×108​m

$$\lambda_{\text{min}} = 4.14 \times 10^{-11} \, \text{m}$$

λmin​=4.14×10−11m

Convert the wavelength into nanometers (nm):

 

$$\lambda_{\text{min}} = 0.0414 \, \text{nm}$$

λmin​=0.0414nm

Final Answer for Part B:

 

$$\lambda_{\text{min}} = 0.0414 \, \text{nm}$$

λmin​=0.0414nm

Q.11.2: The work function of caesium metal is 2.14 eV. When the light of frequency 6 × 1014 Hz is incident on the metal surface, photoemission of electrons occurs. Find the following:

(a) The maximum kinetic energy of the emitted electrons

(b) Stopping potential

(c) The maximum speed of the emitted photoelectrons

 Ans:

Work function of caesium,

Φ𝑜

= 2.14eV

Frequency of light, v = 6.0 x 1014 Hz

(a) The maximum kinetic energy of the emitted electrons:

K = hν –

Φ𝑜

Where,

h = Planck’s constant = 6.626 x 10-34 Js

Therefore,

K =

= 2.485 – 2.140 = 0.345 eV

Hence, 0.345 eV is the maximum kinetic energy of the emitted electrons.

(b) For stopping potential Vo, we can write the equation for kinetic energy as:

K = eVo

Therefore, Vo =

Hence, 0.345 V is the stopping potential of the material.

(c) Maximum speed of photoelectrons emitted = ν

Following is the kinetic energy relation:

K =

Where,

m = mass of electron = 9.1 x 10-31 Kg

Therefore, ν = 3.323 x 105 m/s = 332.3 km/s

Hence, 332.3 km/s is the maximum speed of the emitted photoelectrons.

Question 11.3: The photoelectric cut-off voltage in a certain experiment is 1.5 V.
What is the maximum kinetic energy of photoelectrons emitted?

 Ans:

Photoelectric cut-off voltage, Vo = 1.5 V

For emitted photoelectrons, the maximum kinetic energy is:

Ke = eVo

Where,

e = charge on an electron = 1.6 x 10-19 C

Therefore, Ke = 1.6 x 10-19 x 1.5 = 2.4 x 10-19 J

Therefore, 2.4 x 10-19 J is the maximum kinetic energy emitted by the photoelectrons.

Question 11.4: Monochromatic light of wavelength 632.8 nm is produced by a helium-neon laser. The power emitted is 9.42 mW. (a) Find the energy and momentum of each photon in the light beam, (b) How many photons per second, on the average, arrive at a target irradiated by this beam? (Assume the beam to have uniform cross-section which is less than the target area), and (c) How fast does a hydrogen atom have to travel in order to have the same momentum as that of the photon?

Solution

Given:

• Wavelength of monochromatic light,
  $\lambda = 632.8 \, \text{nm} = 632.8 \times 10^{-9} \, \text{m}$ 

  λ=632.8nm=632.8×10−9m

• Power emitted by the laser,
  $P = 9.42 \, \text{mW} = 9.42 \times 10^{-3} \, \text{W}$ 

  P=9.42mW=9.42×10−3W

• Mass of a hydrogen atom,
  $m = 1.66 \times 10^{-27} \, \text{kg}$ 

  m=1.66×10−27kg

• Planck’s constant,
  $h = 6.626 \times 10^{-34} \, \text{J} \cdot \text{s}$ 

  h=6.626×10−34J⋅s

• Speed of light,
  $c = 3 \times 10^8 \, \text{m/s}$ 

  c=3×108m/s

Part A: Energy and Momentum of the Photon

Step 1: Calculate the energy of each photon.

The energy

$E$

E of a photon is given by:

 

$$E = \frac{hc}{\lambda}$$

E=λhc​

Substituting the values:

 

$$E = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{632.8 \times 10^{-9}} \, \text{J}$$

E=632.8×10−96.626×10−34×3×108​J

$$E = 3.14 \times 10^{-19} \, \text{J}$$

E=3.14×10−19J

Step 2: Calculate the momentum of the photon.

The momentum

$p$

p of a photon is given by:

 

$$p = \frac{h}{\lambda}$$

p=λh​

Substituting the values:

 

$$p = \frac{6.626 \times 10^{-34}}{632.8 \times 10^{-9}} \, \text{kg} \cdot \text{m/s}$$

p=632.8×10−96.626×10−34​kg⋅m/s

$$p = 1.05 \times 10^{-27} \, \text{kg} \cdot \text{m/s}$$

p=1.05×10−27kg⋅m/s

Final Answers for Part A:

• Energy of each photon:
  $E = 3.14 \times 10^{-19} \, \text{J}$ 

  E=3.14×10−19J

• Momentum of each photon:
  $p = 1.05 \times 10^{-27} \, \text{kg} \cdot \text{m/s}$ 

  p=1.05×10−27kg⋅m/s

Part B: Number of Photons per Second Arriving on the Target

Step 3: Calculate the number of photons per second.

The number of photons

$n$

n arriving per second is given by:

 

$$n = \frac{P}{E}$$

n=EP​

Substituting the values:

 

$$n = \frac{9.42 \times 10^{-3}}{3.14 \times 10^{-19}} \, \text{photons/s}$$

n=3.14×10−199.42×10−3​photons/s

$$n \approx 3 \times 10^{16} \, \text{photons/s}$$

n≈3×1016photons/s

Final Answer for Part B:

• Number of photons per second:
  $n \approx 3 \times 10^{16} \, \text{photons/s}$ 

  n≈3×1016photons/s

Part C: Speed of the Hydrogen Atom

Step 4: Calculate the speed of the hydrogen atom.

Since the momentum of the hydrogen atom is the same as the momentum of the photon, the speed

$v$

v of the hydrogen atom is given by:

 

$$v = \frac{p}{m}$$

v=mp​

Substituting the values:

 

$$v = \frac{1.05 \times 10^{-27}}{1.66 \times 10^{-27}} \, \text{m/s}$$

v=1.66×10−271.05×10−27​m/s

$$v \approx 0.63 \, \text{m/s}$$

v≈0.63m/s

Final Answer for Part C:

• Speed of the hydrogen atom:
  $v \approx 0.63 \, \text{m/s}$ 

  $v\approx 0.63\text{m/s}$

Question 11.5: The energy flux of sunlight reaching the surface of the earth is 1.388 × 103 W/m2. How many photons are incident on the earth per second/square metre? Assume an average wavelength of 550 nm.

 Ans:

Sunlight reaching the surface of the earth has an energy flux of

𝜙

= 1.388 × 103 W/m2

Hence, the power of sunlight per square metre, P = 1.388 × 103 W

Speed of light, c = 3 × 108 m/s

Planck’s constant, h = 6.626 × 10-34 Js

𝜆

= 550 nm = 550 x 10-9m is the average wavelength of the photons from the sunlight

Number of photons per square metre incident on earth per second = n

Hence, the equation for power be written as:

P = nE

Therefore, n=

= 3.84 x 1021 photons/m2/s

Therefore, 3.84 x 1021 photons are incident on the earth per square metre.

Question 11.6: In an experiment on the photoelectric effect, the slope of the cut-off
voltage versus frequency of incident light is found to be 4.12 × 10-15 V s. Calculate the value of Planck’s constant.

 Ans:

Given that the slope of cut-off voltage (V) versus frequency (v) is:

V and frequency are related by the equation:

Hν = eV

Where,

e = Charge on an electron = 1.6 x 10-19 C

h = Planck’s constant

Therefore, h = e x

= 1.6 x 10-19 x 4.12 x 10-15 = 6.592 x 10-34 Js

Therefore, 6.592 x 10-34 Js is the Planck’s constant that is determined from the above equation.

Question 11.7:A 100 W sodium lamp radiates energy uniformly in all directions. The lamp is located at the centre of a large sphere that absorbs all the sodium light which is incident on it. The wavelength of the sodium light is 589 nm. (a) What is the energy per photon associated with the sodium light? (b) At what rate are the photons delivered to the sphere?

Question 11.8: The threshold frequency for a certain metal is 3.3 x 1014 Hz. If the light of frequency 8.2 x 1014 Hz is incident on the metal, predict the cut-off voltage for the photoelectric emission.

Ans:

Threshold frequency of the metal, vo= 3.3 × 1014 Hz

Frequency of light incident on the metal, v= 8.2 × 1014 Hz

Charge on an electron, e = 1.6 × 10-19 C

Planck’s constant, h = 6.626 × 10-34 Js

Cut-off voltage for the photoelectric emission from the metal = Vo

The equation for the cut–off energy is given as:

eVo = h(ν – νo)

Vo =

Therefore, the cut-off voltage for the photoelectric emission is 2.0291 V.

Question 11.9:The work function for a certain metal is 4.2 eV. Will this metal give photoelectric emission for incident radiation of wavelength 330 nm?

Question 11.10: Light of frequency $7.21\times {10}^{14}\text{Hz}$ is incident on a metal surface. Electrons with a maximum speed of $6.0\times {10}^{5}m/s$ are ejected from the surface. What is the threshold frequency for photoemission of electrons?

Solution

Definition: Threshold Frequency

• Threshold Frequency (
  $v_0$ 

  v0​): It is the minimum frequency of incident light required to cause photoelectric emission from a metal surface.

Given:

• Frequency of incident light,
  $v = 7.21 \times 10^{14} \, \text{Hz}$ 

  v=7.21×1014Hz

• Speed of the ejected electron,
  $v = 6.0 \times 10^5 \, \text{m/s}$ 

  v=6.0×105m/s

• Planck’s constant,
  $h = 6.626 \times 10^{-34} \, \text{J} \cdot \text{s}$ 

  h=6.626×10−34J⋅s

• Mass of an electron,
  $m = 9.1 \times 10^{-31} \, \text{kg}$ 

  m=9.1×10−31kg

Formula Used:

The energy equation for the photoelectric effect is given by:

 

$$E = h(v – v_0) = \frac{1}{2} m v^2$$

E=h(v−v0​)=21​mv2

Rearranging for threshold frequency (

$v_0$

v0​):

 

$$v_0 = v – \frac{m v^2}{2h}$$

v0​=v−2hmv2​

Step 1: Calculate the threshold frequency.

Substituting the given values into the equation:

 

$$v_0 = 7.21 \times 10^{14} – \frac{9.1 \times 10^{-31} \times (6 \times 10^5)^2}{2 \times 6.626 \times 10^{-34}}$$

v0​=7.21×1014−2×6.626×10−349.1×10−31×(6×105)2​

Calculating the term involving the speed of the ejected electron:

 

$$v_0 = 7.21 \times 10^{14} – \frac{9.1 \times 10^{-31} \times 36 \times 10^{10}}{2 \times 6.626 \times 10^{-34}}$$

v0​=7.21×1014−2×6.626×10−349.1×10−31×36×1010​

$$v_0 = 7.21 \times 10^{14} – \frac{327.6 \times 10^{-21}}{13.252 \times 10^{-34}}$$

v0​=7.21×1014−13.252×10−34327.6×10−21​

$$v_0 = 7.21 \times 10^{14} – 2.47 \times 10^{14} \, \text{Hz}$$

v0​=7.21×1014−2.47×1014Hz

$$v_0 = 4.73 \times 10^{14} \, \text{Hz}$$

v0​=4.73×1014Hz

Final Answer:

• The threshold frequency for the photoemission of electrons is
  $4.73 \times 10^{14} \, \text{Hz}$ 

  4.73×1014Hz.

Question 11.11: Light of wavelength 488 nm is produced by an argon laser which is used in the photoelectric effect. When light from this spectral line is incident on the emitter, the stopping (cut-off) potential of photoelectrons is 0.38 V. Find the work function of the material from which the emitter is made.

Ans:

The wavelength of light produced by the argon laser,

𝜆

= 488 nm = 488 x 10-9 m

Stopping potential of the photoelectrons, Vo = 0.38 V

1eV = 1.6 x 10-19 J

Therefore, Vo =

Planck’s constant, h = 6.6 x 10-34 Js

Charge on an electron, e = 1.6 x 10-19 C

Speed of light, c = 3 x 108 m/s

Using Einstein’s photoelectric effect, the following is the relation for the work function:

Φ𝑜

of the material of the emitter as:

eVo =

Therefore, 2.16 eV is the work function of the material with which the emitter is made.

Question 11.12: Calculate the following:

(a) Momentum

(b) The de Broglie wavelength of the electrons accelerated through a potential difference of 56 V.

Ans:

The potential difference, V = 56V

Planck’s constant, h = 6.6 x 10-34 Js

Mass of an electron, m = 9.1x 10-31 Kg

Charge on an electron, e = 1.6 x 10-19C

(a) At equilibrium, the kinetic energy of each electron is equal to the accelerating potential, i.e. we can write the relation of velocity (v) of each electron as:

The momentum of each accelerated electron is given as:

p = mv

= 9.1 x 10-31 x 4.44 x 106 = 4.04 x 10-24 Kg m/s

Therefore, 4.04 x 10-24 Kg m/s is the momentum of each electron.

(b) de Broglie wavelength of an electron accelerating through a potential V is given by the relation:

𝜆

=

Therefore, 0.1639 nm is the de Broglie wavelength of each electron.

Question 11.13: What is the (a) momentum, (b) speed, and (c) de Broglie wavelength of an electron with kinetic energy of 120 eV.

Solution

Given:

• Kinetic energy of the electron,
  $E_k = 120 \, \text{eV}$ 

  Ek​=120eV

a) Momentum of the Electron

The momentum of an electron (

$p$

p) is calculated using the relation:

 

$$p = \sqrt{2m E_k}$$

p=2mEk​​

Where:

• 
  $m$ 

  $m$ is the mass of the electron (

  $9.1 \times 10^{-31} \, \text{kg}$ 

  9.1×10−31kg)

• 
  $E_k$ 

  Ek​ is the kinetic energy in joules (1 eV =

  $1.6 \times 10^{-19} \, \text{J}$ 

  1.6×10−19J)

Substituting the given values:

 

$$p = \sqrt{2 \times 9.1 \times 10^{-31} \times (120 \times 1.6 \times 10^{-19})}$$

p=2×9.1×10−31×(120×1.6×10−19)​

$$p = \sqrt{2 \times 9.1 \times 10^{-31} \times 1.92 \times 10^{-17}}$$

p=2×9.1×10−31×1.92×10−17​

$$p = \sqrt{34.944 \times 10^{-48}}$$

p=34.944×10−48​

$$p = 5.91 \times 10^{-24} \, \text{kg} \cdot \text{m/s}$$

p=5.91×10−24kg⋅m/s

Thus, the momentum of the electron is

$5.91 \times 10^{-24} \, \text{kg} \cdot \text{m/s}$

5.91×10−24kg⋅m/s.

b) Speed of the Electron

The speed (

$v$

v) of the electron is given by:

 

$$v = \frac{p}{m}$$

v=mp​

Substituting the values:

 

$$v = \frac{5.91 \times 10^{-24}}{9.1 \times 10^{-31}}$$

v=9.1×10−315.91×10−24​

$$v = 6.49 \times 10^6 \, \text{m/s}$$

v=6.49×106m/s

Thus, the speed of the electron is

$6.49 \times 10^6 \, \text{m/s}$

6.49×106m/s.

c) de Broglie Wavelength of the Electron

The de Broglie wavelength (

$\lambda$

λ) is given by:

 

$$\lambda = \frac{h}{p}$$

λ=ph​

Where:

• 
  $h$ 

  $h$ is Planck’s constant (

  $6.626 \times 10^{-34} \, \text{J} \cdot \text{s}$ 

  6.626×10−34J⋅s)

Substituting the values:

 

$$\lambda = \frac{6.626 \times 10^{-34}}{5.91 \times 10^{-24}}$$

λ=5.91×10−246.626×10−34​

$$\lambda = 1.121 \times 10^{-10} \, \text{m}$$

λ=1.121×10−10m

$$\lambda = 0.112 \, \text{nm}$$

λ=0.112nm

Thus, the de Broglie wavelength of the electron is

$0.112 \, \text{nm}$

0.112nm

Question 11.14:

The wavelength of light from the spectral emission line of sodium is 589 nm. Find the kinetic energy at which

(a) an electron, and

(b) a neutron, would have the same de Broglie wavelength.

Question 11.15: What is the de Broglie wavelength of:

(a) a bullet of mass 0.040 kg travelling at the speed of 1.0 km/s,

(b) a ball of mass 0.060 kg moving at a speed of 1.0 m/s, and

(c) a dust particle of mass 1.0 × 10-9 kg drifting with a speed of 2.2 m/s?

Ans:

(a) Mass of the bullet, m = 0.040 Kg

The speed of the bullet, v = 1.0 km/s = 1000 m/s

Planck’s constant, h = 6.6 x 10-34 Js

de Broglie wavelength of the bullet is given by the relation:

(b) Mass of the ball, m = 0.060 Kg

The speed of the ball, v = 1.0 m/s

de Broglie wavelength of the ball is given by the relation:

=

(c) Mass of the dust particle, m = 1 x 10-9 Kg

The speed of the dust particle, v = 2.2 m/s

de Broglie wavelength of the dust particle is given by the relation:

=

 

Question 11.16: An electron and a photon each have a wavelength of 1.00 nm. Find the following:

(a) Momenta

(b) The energy of the photon

(c) The kinetic energy of the electron

Ans:

Wavelength of an electron

𝜆𝑒

and a photon

𝜆𝑝

,

𝜆𝑒

=

𝜆𝑝

=

𝜆

= 1 nm = 1 x 10-9 m

Planck’s constant, h = 6.63 x 10-34 Js

(a) The momentum of an elementary particle is given by de Broglie relation:

It is clear that momentum depends only on the wavelength of the particle. Since the wavelengths of an electron and a photon are equal, both have an equal momentum.

Therefore, p =

Therefore, both the electron and photon have an equal momentum of 6.63 x 10-25 Kg m/s.

(b) The energy of a photon is given by the relation:

E =

Where,

Speed of light, c = 3 x 108 m/s

Therefore, E =

Therefore, the energy of the photon is 1.243 keV.

(c) The kinetic energy (K) of an electron having momentum p, is given by the relation:

K =

Where, m = Mass of the electron = 9.1 x 10-31 Kg

p = 6.63 x 10-25 Kg m/s

Therefore, K =

Therefore, 1.51eV is the kinetic energy of the electron.

Question 11.17:

(a)A) For what kinetic energy of a neutron will the associated de Broglie wavelength be $1.40\times {10}^{-10}m?$

B) Find the de Broglie wavelength of a neutron, in thermal equilibrium with matter, having an average kinetic energy of $\frac{3}{2}kT$ at $300K$.

Solution

A) Energy of the Neutron

Given:

• Wavelength of the neutron,
  $\lambda = 1.4 \times 10^{-10} \, \text{m}$ 

  λ=1.4×10−10m

• Planck’s constant,
  $h = 6.63 \times 10^{-34} \, \text{J} \cdot \text{s}$ 

  h=6.63×10−34J⋅s

• Mass of the neutron,
  $m = 1.67 \times 10^{-27} \, \text{kg}$ 

  m=1.67×10−27kg

The energy (

$E$

E) of the neutron can be found using the relationship between energy and momentum:

 

$$E = \frac{p^2}{2m}$$

E=2mp2​

Where

$p$

p (momentum) is given by:

 

$$p = \frac{h}{\lambda}$$

p=λh​

Substituting this into the energy formula:

 

$$E = \left(\frac{h}{\lambda}\right)^2 \times \frac{1}{2m}$$

E=(λh​)2×2m1​

Substituting the given values:

 

$$E = \left(\frac{6.63 \times 10^{-34}}{1.4 \times 10^{-10}}\right)^2 \times \frac{1}{2 \times 1.67 \times 10^{-27}}$$

E=(1.4×10−106.63×10−34​)2×2×1.67×10−271​

$$E = \left(\frac{6.63 \times 10^{-34}}{1.4 \times 10^{-10}}\right)^2 \times \frac{1}{3.34 \times 10^{-27}}$$

E=(1.4×10−106.63×10−34​)2×3.34×10−271​

$$E = \frac{(4.7357 \times 10^{-24})^2}{3.34 \times 10^{-27}}$$

E=3.34×10−27(4.7357×10−24)2​

$$E = 6.71 \times 10^{-21} \, \text{J}$$

E=6.71×10−21J

Converting the energy to electron volts (eV):

 

$$E = \frac{6.71 \times 10^{-21}}{1.6 \times 10^{-19}} \, \text{eV}$$

E=1.6×10−196.71×10−21​eV

$$E = 4.196 \times 10^{-2} \, \text{eV}$$

E=4.196×10−2eV

Final Answer:
The energy of the neutron is

$6.71 \times 10^{-21} \, \text{J}$

6.71×10−21J or

$4.196 \times 10^{-2} \, \text{eV}$

4.196×10−2eV.

B) De-Broglie Wavelength of the Neutron at 300 K

Given:

• Temperature of the neutron,
  $T = 300 \, \text{K}$ 

  $T=300\text{K}$

• Boltzmann constant,
  $k = 1.38 \times 10^{-23} \, \text{J/K}$ 

  k=1.38×10−23J/K

• Mass of the neutron,
  $m = 1.67 \times 10^{-27} \, \text{kg}$ 

  m=1.67×10−27kg

The average kinetic energy (

$E$

E) of the neutron is:

 

$$E = \frac{3}{2} k T$$

E=23​kT

The De-Broglie wavelength (

$\lambda$

λ) is given by:

 

$$\lambda = \frac{h}{\sqrt{2mE}} = \frac{h}{\sqrt{2m \times \frac{3}{2} k T}} = \frac{h}{\sqrt{3mkT}}$$

λ=2mE​h​=2m×23​kT​h​=3mkT​h​

Substituting the given values:

 

$$\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{3 \times 1.67 \times 10^{-27} \times 1.38 \times 10^{-23} \times 300}}$$

λ=3×1.67×10−27×1.38×10−23×300​6.63×10−34​

$$\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{2.057 \times 10^{-26}}}$$

λ=2.057×10−26​6.63×10−34​

$$\lambda = \frac{6.63 \times 10^{-34}}{4.535 \times 10^{-13}}$$

λ=4.535×10−136.63×10−34​

$$\lambda = 1.46 \times 10^{-10} \, \text{m} = 0.146 \, \text{nm}$$

λ=1.46×10−10m=0.146nm

Final Answer:
The De-Broglie wavelength of the neutron is

$0.146 \, \text{nm}$

0.146nm

Question 11.18: Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon).

Ans:

The momentum of a photon having energy (hv) is given as:

p =

Where,

𝜆

= wavelength of the electromagnetic radiation

c = speed of light

h = Planck’s constant

De Broglie wavelength of the photon is given as:

Where, m = mass of the photon

v = velocity of the photon

From equations (i) and (ii), it can be concluded that the wavelength of the electromagnetic radiation and the de Broglie wavelength of the photon is equal.

Question 11.19: What is the de Broglie wavelength of a nitrogen molecule in air at 300 K? Assume that the molecule is moving with the root-mean-square speed of molecules at this temperature. (Atomic mass of nitrogen = 14.0076 u)

 Ans:

The temperature of the nitrogen molecule, T = 300 K

The atomic mass of nitrogen = 14.0076 u

Hence, the mass of the nitrogen molecule, m = 2 × 14.0076 = 28.0152 u

But, 1 u = 1.66 × 10-27 kg

Therefore, m = 28.0152 ×1.66 × 10-27 kg

Planck’s constant, h = 6.63 × 10-34 Js

Boltzmann constant, k = 1.38 × 10-23 J/K

We have the expression that relates mean kinetic energy

of the nitrogen molecule with the root mean square speed (Vrms ) as:

For nitrogen molecule, the de Broglie wavelength is given as:

Therefore, the de Broglie wavelength of the nitrogen molecule is 0.028 nm.

Question 11.20: (a) Estimate the speed with which electrons emitted from a heated emitter of an evacuated tube impinge on the collector maintained at a potential difference of 500 V with respect to the emitter. Ignore the low initial speeds of the electrons. The specific charge of the electron, i.e. its e/m, is given to be 1.76 × 1011 C kg–1.
(b) Use the same formula you employ in (a) to obtain electron speed for a collector potential of 10 MV. Do you see what is wrong? In what way is the formula to be modified?

Answer:

(a) Potential difference of the evacuated tube = 500 V

Specific charge of the electron, e/m = 1.76 × 1011 C kg–1

Kinetic energy = (1/2) mv2 = eV

Speed of the emitted electron, v = (2Ve/m)1/2

= (2 x 500 x 1.76 x 1011)1/2  

= 1.32 x 107 m/s.

(b) Collector potential, V = 10 MV = 10 x 106 V.

Speed of electron = v = (2Ve/m)1/2

= (2 x 107 x 1.76 x 1011)1/2

= 1.88 x 109 m/s.

This answer is not correct. Since the value is greater than the speed of light (c), the expression (1/2) mv2 for energy should be used in the non -relativistic limit, i.e., v << c.

In the relativistic limits, the total energy is given as

E = mc2

Here,

m is the relativistic mass

m = m0 (1- v2/c2)1/2

m0 = mass of the particle at rest

Kinetic energy is given as

K = mc2 – m0c2.

Question 11.21:

A) A monoenergetic electron beam with electron speed of

$5.20 \times 10^6 \, \text{m/s}$

5.20×106m/s is subject to a magnetic field of

$1.30 \times 10^{-4} \, \text{T}$

1.30×10−4T normal to the beam velocity. What is the radius of the circle traced by the beam, given that

$\frac{e}{m}$

me​ for the electron equals

$1.76 \times 10^{11} \, \text{C/kg}$

1.76×1011C/kg?

B) A monoenergetic electron beam with electron speed of

$5.20 \times 10^6 \, \text{m/s}$

5.20×106m/s is subject to a magnetic field of

$1.30 \times 10^{-4} \, \text{T}$

1.30×10−4T normal to the beam velocity. The radius of the circle traced by the beam is calculated, given that

$\frac{e}{m}$

me​ for the electron equals

$1.76 \times 10^{11} \, \text{C/kg}$

1.76×1011C/kg. Is the formula you employed above valid for calculating the radius of the path of a

$20 \, \text{MeV}$

20MeV electron beam? If not, in what way is it modified?

Solution

A) Radius of the Circle Traced by the Electron Beam

Given:

• Speed of the electron,
  $v = 5.20 \times 10^6 \, \text{m/s}$ 

  v=5.20×106m/s

• Magnetic field,
  $B = 1.30 \times 10^{-4} \, \text{T}$ 

  B=1.30×10−4T

• Charge-to-mass ratio for electron,
  $\frac{e}{m} = 1.76 \times 10^{11} \, \text{C/kg}$ 

  me​=1.76×1011C/kg

The force on an electron moving in a magnetic field is given by:

 

$$F = q(v \times B) = qvB \sin\theta$$

F=q(v×B)=qvBsinθ

Since the magnetic field is normal to the velocity of the electron (

$\theta = 90^\circ$

θ=90∘), the force becomes:

 

$$F = qvB$$

F=qvB

This force provides the centripetal force necessary for circular motion:

 

$$\frac{mv^2}{R} = qvB$$

Rmv2​=qvB

Rearranging to find the radius

$R$

R:

 

$$R = \frac{mv}{qB} = \frac{v}{\frac{e}{m}B}$$

R=qBmv​=me​Bv​

Substituting the given values:

 

$$R = \frac{5.20 \times 10^6 \, \text{m/s}}{1.76 \times 10^{11} \, \text{C/kg} \times 1.30 \times 10^{-4} \, \text{T}}$$

R=1.76×1011C/kg×1.30×10−4T5.20×106m/s​

$$R = \frac{5.20 \times 10^6}{2.288 \times 10^7} \, \text{m}$$

R=2.288×1075.20×106​m

$$R = 0.227 \, \text{m} = 22.7 \, \text{cm}$$

R=0.227m=22.7cm

Final Answer:
The radius of the circle traced by the electron beam is

$22.7 \, \text{cm}$

22.7cm.

B) Validity of the Formula for a 20 MeV Electron Beam

For a non-relativistic electron, the formula used is:

 

$$R = \frac{mv}{qB}$$

R=qBmv​

However, for a 20 MeV electron, the electron moves at relativistic speeds. The rest mass energy of an electron is approximately

$0.51 \, \text{MeV}$

0.51MeV, so a 20 MeV electron’s speed is close to the speed of light (

$c$

c). In such a case, the non-relativistic formula is not valid.

Relativistic Correction:

The relativistic mass

$m$

m of the electron is given by:

 

$$m = m_0 \left(1 – \frac{v^2}{c^2}\right)^{-\frac{1}{2}}$$

m=m0​(1−c2v2​)−21​

Where:

• 
  $m_0$ 

  m0​ is the rest mass of the electron.

• 
  $v$ 

  $v$ is the velocity of the electron.

• 
  $c$ 

  $c$ is the speed of light.

The relativistic formula for the radius becomes:

 

$$R = \frac{m v}{q B} = \frac{m_0 \left(1 – \frac{v^2}{c^2}\right)^{-\frac{1}{2}} v}{q B}$$

R=qBmv​=qBm0​(1−c2v2​)−21​v​

Thus, for relativistic speeds, the radius of the path is larger than predicted by the non-relativistic formula due to the increase in the effective mass of the electron.

Final Answer:
The non-relativistic formula

$R = \frac{mv}{qB}$

R=qBmv​ is not valid for a 20 MeV electron beam. The formula needs to be modified to include relativistic effects, using:

 

$$R = \frac{m_0 \left(1 – \frac{v^2}{c^2}\right)^{-\frac{1}{2}} v}{q B}$$

R=qBm0​(1−c2v2​)−21​v​.

Question 11.22: An electron gun with its collector at a potential of 100 V fires out electrons in a spherical bulb containing hydrogen gas at low pressure (∼10–2 mm of Hg). A magnetic field of 2.83 × 10–4 T curves the path of the electrons in a circular orbit of radius 12.0 cm. (The path can be viewed because the gas ions in the path focus the beam by attracting electrons, and emitting light by electron capture; this method is known as the ‘fine beam tube’ method.) Determine e/m from the data.

Answer:

Potential of the collector, V = 100 V

Magnetic field experienced by the electron, B = 2.83 × 10–4 T

Radius of the circular orbit, r = 12 cm = 12.0 x 10-2 m

Kinetic energy, (1/2)mv2 = eV

v2 = 2eV/m ——–(1)

The magnetic field that curves the path of the electron provides the centripetal force

evB = mv2/r

eB = mv/r

v = eBr/m ——–(2)

Substituting (2) in (1)

Therefore, the specific charge ratio e/m is 1.73 x 1011 Ckg-1.

Question 11. 23: (a) An X-ray tube produces a continuous spectrum of radiation with its short wavelength ending at 0.45 Å. What is the maximum energy of a photon in radiation?
(b) From your answer to (a), guess what order of accelerating voltage (for electrons) is required in such a tube?

Answer:

(a) Wavelength produced by the X-ray tube, λ = 0.45 Å = 0.45 x 10-10 m

Speed of light, c = 3 x 108 m/s

Planck’s constant, h = 6.626 x 10-34 Js

The maximum energy of a photon is given as

Emax = hc/λmin

= (6.626 x 10-34)(3 x 108 m/s)/(0.45 x 10-10 m x 1.6 x 10-19)

= 19.878 x 10-26/0.72 x 10-29

= 27.60 x 103 eV = 27.6 keV.

(b) To incident electron should have an energy of 27.6 keV to get an X-ray of 27.6 keV. Therefore, the accelerating voltage of the order of 30 keV is required for producing X-rays.

Question 11.24: In an accelerator experiment on high-energy collisions of electrons with positrons, a certain event is interpreted as the annihilation of an electron-positron pair of total energy 10.2 BeV into two γ-rays of equal energy. What is the wavelength associated with each γ-ray? (1BeV = 109 eV)

Answer:

Total energy of the electron-positron pair, E = 10.2 BeV = 10.2 x 109 eV = 10.2 x 109 x 1.6 x 10-19 J

Hence, the energy of each γ-ray , E’ = E/2 = (10.2 x 109 x 1.6 x 10-19 )/2 = 8.16 x 10-10 J

Energy and wavelength relation is given as,

E’ = hc/λ

Therefore, λ = hc/E’

Here, h = 6.626 x 10-34 Js

c = 3 x 108 m/s

Therefore, the wavelength associated with each γ-ray is 2.436 x 10-16 m.

Question 11.25: Estimating the following two numbers should be interesting. The first number will tell you why radio engineers do not need to worry much about photons. The second number tells you why our eye can never ‘count photons’, even in barely detectable light.
(a) The number of photons emitted per second by a medium wave transmitter of 10 kW power, emitting radio waves of wavelength 500 m.
(b) The number of photons entering the pupil of our eye per second corresponding to the minimum intensity of white light that we humans can perceive (∼10–10 W m–2). Take the area of the pupil
to be about 0.4 cm2, and the average frequency of white light to be about 6 × 1014 Hz.

Answer:

(a) Power of the medium wave transmitter, P = 10 kW = 104 W

The energy emitted by the transmitter per second, E = 104

The wavelength of the radio waves, λ = 500 m

The energy of the wave is given as, E’ = hc/λ

E’ = (6.6 x 10-34 x 3 x 108)/500

= 3.96 x 10-28 J

Let n be the number of photons emitted by the transmitter

nE’ = E

n = E/E’

= 104/(3.96 x 10-28 )

= 0.2525 x 1032

The energy E’ of the radio photon is very less, but the number of photons emitted is large. The total energy of the radio ways can be considered as continuous, and the existence of the minimum quantum energy can be ignored.

(b) Intensity of the light perceived by the human eye, I =  10–10 W m–2

Area of the pupil, A = 0.4 cm2 = 0.4 x 10-4 m2

Frequency of the white light, ν = 6 x 1014 Hz

h = Planck’s constant = 6.6 x 10-34 Js

The energy of the emitted photon, E = hν

= 6.6 x 10-34  x6 x 1014

= 3.96 x 10-19 J

Let n be the total number of photons falling per unit area per unit time. The total energy per unit for n photons is

E = n x 3.96 x 10-19 J/s/m2

The total energy per unit for n photons is equal to the intensity of the light.

E = I

I = n x 3.96 x 10-19 J/s/m2

n = I/3.96 x 10-19

= 10-10/3.96 x 10-19

= 2.52 x 108 m2/s

The total number of photons entering the pupil is given as,

nA = 2.52 x 108 x 0.4 x 10-4

= 1.008 x 104 s-1

The number is large, so the human eye can never count the number of individual photons.

Question 11.26: Ultraviolet light of wavelength 2271 Å from a 100 W mercury source irradiates a photo-cell made of molybdenum metal. If the stopping potential is −1.3 V, estimate the work function of the metal. How would the photo-cell respond to a high intensity (∼10⁵ W m⁻²) red light of wavelength 6328 Å produced by a He-Ne laser?

Question 11.27: Monochromatic radiation of wavelength 640.2 nm (1nm = 10–9 m) from a neon lamp irradiates photosensitive material made of caesium on tungsten. The stopping voltage is measured to be 0.54 V. The source is replaced by an iron source, and its 427.2 nm line irradiates the same photo-cell. Predict the new stopping voltage.

Answer:

Wavelength of the monochromatic radiation, λ = 640.2 nm = 640.2 × 10−9 m

The stopping potential of the neon lamp, V0 = 0.54 V

Charge on an electron, e = 1.6 × 10−19 C

Planck’s constant, h = 6.6 × 10−34 Js

From the photoelectric effect, we have the photon-electron relation as eV0 = hν – Φ0

The work function of the metal, Φ0 = hν – eV0

= (hc/λ) – eV0

= 3.093 x 10-19 – 0.864 x 10-19

= 2.229 x 10-19 J

The wavelength of the radiation emitted by the iron source, λ’ = 427.2 nm = 427.2 x 10-9 m

Let V0‘ be the new stopping potential

Therefore, eV0‘ = (hc/λ’) – Φ0

= 2.401 x 10-19 J

V0‘ = 2.401 x 10-19 J/ 1.6 x 10-19 J

= 1.5 eV

Therefore, the new stopping potential = 1.50 eV.

Question 11.28A mercury lamp is a convenient source for studying frequency dependence of photoelectric emission, since it gives a number of spectral lines ranging from the UV to the red end of the visible spectrum. In our experiment with rubidium photocell, the following lines from a mercury source were used: ${\lambda }_1=3650\stackrel{\circ }{A}$,
${\lambda }_2=4047\stackrel{\circ }{A},{\lambda }_3=4358\stackrel{\circ }{A}$
${\lambda }_4=5461\stackrel{\circ }{A},{\lambda }_5=6907\stackrel{\circ }{A}$

The stopping voltages, respectively, were measured to be:

V_01= 1.28 V,

V_02= 0.95 V,

V_03= 0.74 V,

V_04= 0.16 V,

V_05= 0 V

Determine the value of Planck’s constant h, the threshold frequency and work function for the material

Solution

Step 1: Find the frequencies of the various lines.

The formula used is:

$v = \frac{c}{\lambda}$

v=λc​

According to Einstein’s photoelectric equation:

$eV_0 = h v – W_0$

eV0​=hv−W0​

where:

• 
  $V_0$ 

  V0​ is the stopping potential,

• 
  $h$ 

  $h$ is Planck’s constant,

• 
  $e$ 

  $e$ is the charge on an electron,

• 
  $v$ 

  $v$ is the frequency of radiation,

• 
  $W_0$ 

  W0​ is the work function of the material.

The frequency

$v$

v can be calculated using:

$v = \frac{c}{\lambda}$

v=λc​

Calculating frequencies for the given wavelengths:

$v_1 = \frac{c}{\lambda_1} = \frac{3 \times 10^8}{3650 \times 10^{-10}} = 8.219 \times 10^{14} \, \text{Hz}$

v1​=λ1​c​=3650×10−103×108​=8.219×1014Hz

$v_2 = \frac{c}{\lambda_2} = \frac{3 \times 10^8}{4047 \times 10^{-10}} = 7.412 \times 10^{14} \, \text{Hz}$

v2​=λ2​c​=4047×10−103×108​=7.412×1014Hz

$v_3 = \frac{c}{\lambda_3} = \frac{3 \times 10^8}{4358 \times 10^{-10}} = 6.884 \times 10^{14} \, \text{Hz}$

v3​=λ3​c​=4358×10−103×108​=6.884×1014Hz

$v_4 = \frac{c}{\lambda_4} = \frac{3 \times 10^8}{5461 \times 10^{-10}} = 5.493 \times 10^{14} \, \text{Hz}$

v4​=λ4​c​=5461×10−103×108​=5.493×1014Hz

$v_5 = \frac{c}{\lambda_5} = \frac{3 \times 10^8}{6907 \times 10^{-10}} = 4.341 \times 10^{14} \, \text{Hz}$

v5​=λ5​c​=6907×10−103×108​=4.341×1014Hz

Tabulated data:

 

$$\begin{array}{|c|c|} \hline \text{Frequency} \, (v \times 10^{14} \text{Hz}) & \text{Stopping Potential} \, (V_0) \\ \hline 8.219 & 1.28 \\ 7.412 & 0.95 \\ 6.884 & 0.74 \\ 5.493 & 0.16 \\ 4.341 & 0 \\ \hline \end{array}$$

Frequency(v×1014Hz)8.2197.4126.8845.4934.341​Stopping Potential(V0​)1.280.950.740.160​​

Step 2: Draw the graph of stopping potential vs frequency from the calculated data.

Plotting

$V_0$

V0​ against

$v$

v gives a straight line. The intersection of this line with the

$v$

v-axis indicates the threshold frequency

$v_0$

v0​ of the material.

Step 3: Find the value of Planck’s constant

$h$

h.

The slope of the graph is given by:

$\text{slope} = \frac{h}{e}$

slope=eh​

From the relation between frequency

$v$

v and stopping potential

$V_0$

V0​:

$eV_0 = h v – W_0$

eV0​=hv−W0​

Using the slope:

$\text{slope} = \frac{V_4 – V_1}{v_4 – v_1}$

slope=v4​−v1​V4​−V1​​

Substituting the values:

$\text{slope} = \frac{1.28 – 0.16}{8.219 \times 10^{14} – 5.493 \times 10^{14}}$

slope=8.219×1014−5.493×10141.28−0.16​

$\text{slope} = 4.15 \times 10^{-15} \, \text{V s}$

slope=4.15×10−15V s

$\frac{h}{e} = 4.15 \times 10^{-15}$

eh​=4.15×10−15

$h = \left[ \frac{1.28 – 0.16}{8.219 – 5.493} \times 1.6 \right]$

h=[8.219−5.4931.28−0.16​×1.6]

$h = 6.64 \times 10^{-34} \, \text{J s}$

h=6.64×10−34J s

Step 4: Find the value of the work function

$\phi_0$

ϕ0​.

The work function of the metal is given by:

$\phi_0 = h v_0$

ϕ0​=hv0​

Substituting:

$\phi_0 = 6.64 \times 10^{-34} \times 5 \times 10^{14}$

ϕ0​=6.64×10−34×5×1014

$\phi_0 = 3.32 \times 10^{-19} \, \text{J}$

ϕ0​=3.32×10−19J

Since

$1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}$

1eV=1.6×10−19J:

$\phi_0 = 2.1 \, \text{eV}$

ϕ0​=2.1eV

Final Answer:

• The Planck’s constant is
  $h = 6.64 \times 10^{-34} \, \text{J s}$ 

  h=6.64×10−34J s.

• The work function is
  $\phi_0 = 2.1 \, \text{eV}$ 

  ϕ0​=2.1eV.

Question 11.29: The work function for the following metals is given:  Na: 2.75 eV, K: 2.30 eV, Mo: 4.17 eV, Ni: 5.15 eV. Which of these metals will not give photoelectric emission for radiation of wavelength 3300 Å from a He-Cd laser placed 1 m away from the photocell? What happens if the laser is brought nearer and placed 50 cm away?

Answer:

Wavelength λ = 3300 Å

Speed of light = 3 x 108 m/s

Planck’s constant = 6.63 x 10-34 Js

The energy of the photon of the incident light

E = hc/λ = (6.63 x 10-34 x 3 x 108)/3300x 10-10

⇒ 6.018 x 10-19 J

⇒  (6.018 x 10-19 J)/1.6 x 10-19

= 3.7 eV

The energy of the incident radiation is greater than the work function of Na and K. It is lesser for Mo and Ni. Therefore, Mo and Ni will not show a photoelectric effect.

If the laser is brought nearer and placed 50 cm away, then the intensity of the radiation will increase. The energy of the radiation will not be affected. Therefore, the result will be the same. However, the photoelectrons from Na and K will increase in proportion to intensity.

Question 11.30: Light of intensity 10–5 W m–2 falls on a sodium photo-cell of a surface area of 2 cm2. Assuming that the top 5 layers of sodium absorb the incident energy, estimate the time required for photoelectric emission in the wave picture of radiation. The work function for the metal is given to be about 2 eV. What is the implication of your answer?

Answer:

The intensity of the light = 10–5 W m–2

The surface area of the sodium photocell, A = 2 cm2

Incident power of the light, P = I x A

= 10-5 x 2 x 10-4

= 2 x 10-9 W

Work function of the metal, Φ0 = 2 eV

= 2 x 1.6 x 10-19 

= 3.2 x 10-19 J

The number of layers of sodium that absorbs the incident energy, n = 5

The atomic area of the sodium atom, Ae, is 10-20 m2

Hence, the number of conduction electrons in n layers is given as:

n’ = n x (A/Ae)

= 5 x [(2 x 10-4)/10-20] = 1017

The incident power is absorbed by all the electrons continuously. The amount of energy absorbed per electron per second is

E = P/n’

= (2 x 10-9)/1017

= 2 x 10-26 J/s

The time for photoelectric emission

t = Φ0/E

= (3.2 x 10-19)/(2 x 10-26) = 1.6 x 107 s ≈ 0.507 years

The time required for the photoelectric emission is almost half a year. This is not practical. Therefore, the wave function is in disagreement with the given experiment.

Question 11.31: Crystal diffraction experiments can be performed using X-rays or electrons accelerated through appropriate voltage. Which probe has greater energy? (For quantitative comparison, take the wavelength of the probe equal to 1 Å, which is of the order of interatomic spacing in the lattice) (me =9.11 × 10–31 kg).

Answer:

For electrons, kinetic energy, K.E= (1/2) mev2

= (mev)2/2m

K.E = p2/2me

⇒

For photon of X-rays, energy, E = hc/λe

= (6.63 x 10-34 x 3 x 108)/ (10-10 x 1.6 x 10-19)

= 12.375 x 103 eV

= 12.375 keV

Hence, the energy of the photons of X-rays is more than the energy of the electron.

Question 11.32: (a) Obtain the de Broglie wavelength of a neutron of kinetic energy 150 eV. As you have seen in Exercise 11.31, an electron beam of this energy is suitable for crystal diffraction experiments. Would a neutron beam of the same energy be equally suitable? Explain. (mn = 1.675 × 10–27 kg) (b) Obtain the de Broglie wavelength associated with thermal neutrons at room temperature (27 °C). Hence explain why a fast neutron beam needs to be thermalised with the environment before it can be used for neutron diffraction experiments.

Solution

(a) Determining the de Broglie Wavelength of a Neutron

Given:

• Kinetic energy of the neutron,
  $K = 150 \, \text{eV}$ 

  $K=150\text{eV}$

• Mass of the neutron,
  $m_n = 1.675 \times 10^{-27} \, \text{kg}$ 

  mn​=1.675×10−27kg

• 
  $1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}$ 

  1eV=1.6×10−19J

Step 1: Convert the Kinetic Energy to Joules

 

$K = 150 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 2.4 \times 10^{-17} \, \text{J}$

K=150eV×1.6×10−19J/eV=2.4×10−17J

Step 2: Calculate the Velocity of the Neutron

The kinetic energy is given by:

 

$K = \frac{1}{2} m_n v^2$

K=21​mn​v2

Rearrange to solve for

$v$

v:

 

$v = \sqrt{\frac{2K}{m_n}}$

v=mn​2K​​

Substitute the values:

 

$v = \sqrt{\frac{2 \times 2.4 \times 10^{-17}}{1.675 \times 10^{-27}}} \approx 3.46 \times 10^5 \, \text{m/s}$

v=1.675×10−272×2.4×10−17​​≈3.46×105m/s

Step 3: Determine the de Broglie Wavelength

The de Broglie wavelength

$\lambda$

λ is given by:

 

$\lambda = \frac{h}{p}$

λ=ph​

where

$p$

p is the momentum,

$p = m_n v$

p=mn​v.

Thus:

 

$\lambda = \frac{h}{m_n v} = \frac{h}{\sqrt{2m_n K}}$

λ=mn​vh​=2mn​K​h​

Substitute the values:

 

$\lambda = \frac{6.626 \times 10^{-34}}{\sqrt{2 \times 1.675 \times 10^{-27} \times 2.4 \times 10^{-17}}} \approx 2.337 \times 10^{-12} \, \text{m}$

λ=2×1.675×10−27×2.4×10−17​6.626×10−34​≈2.337×10−12m

Conclusion:

The de Broglie wavelength of the neutron is

$2.337 \times 10^{-12} \, \text{m}$

2.337×10−12m. This wavelength is significantly smaller than the inter-atomic spacing (about

$10^{-10} \, \text{m}$

10−10m), indicating that a neutron beam with this energy is not suitable for diffraction experiments because the wavelength must be on the order of the inter-atomic spacing for effective diffraction.

(b) De Broglie Wavelength of a Thermal Neutron

Given:

• Temperature of the neutron beam, T = 27 \, ^\circ \text{C} = 300 \, \text{K}

Step 1: Calculate the Average Kinetic Energy

The average kinetic energy of a neutron is given by:

 

$K = \frac{3}{2} k T$

K=23​kT

where

$k$

k is Boltzmann’s constant,

$k = 1.38 \times 10^{-23} \, \text{J/K}$

k=1.38×10−23J/K.

Substitute the values:

 

$K = \frac{3}{2} \times 1.38 \times 10^{-23} \times 300 \approx 1.242 \times 10^{-21} \, \text{J}$

K=23​×1.38×10−23×300≈1.242×10−21J

Step 2: Determine the de Broglie Wavelength

Using the de Broglie wavelength formula:

 

$\lambda = \frac{h}{\sqrt{2 m_n K}}$

λ=2mn​K​h​

Substitute the values:

 

$\lambda = \frac{6.626 \times 10^{-34}}{\sqrt{2 \times 1.675 \times 10^{-27} \times 1.242 \times 10^{-21}}} \approx 1.45 \times 10^{-10} \, \text{m}$

λ=2×1.675×10−27×1.242×10−21​6.626×10−34​≈1.45×10−10m

Question 11.33: An electron microscope uses electrons accelerated by a voltage of 50 kV. Determine the de Broglie wavelength associated with the electrons. If other factors (such as numerical aperture, etc.) are taken to be roughly the same, how does the resolving power of an electron microscope compare with that of an optical microscope which uses yellow light?

Answer:

Electrons are accelerated by a voltage = 50 kV

Charge on an electron, e = 1.6 x 10-19 C

Mass of the electron, me = 9.11 x 10-31 kg

Wavelength of the yellow light = 5.9 x 10-7 m

The kinetic energy of the electron, E = eV

= (1.6 x 10-19) x (50 x 103)

= 8 x 10-15 J

De Broglie wavelength of an electron is given as

The wavelength is 105 times lesser than the wavelength of yellow light. The resolving power of the microscope and the wavelength of the light used is inversely proportional. Therefore, the resolving power of the electron microscope is 105 times greater than the optical microscope.

Question 11.34: The wavelength of a probe is roughly a measure of the size of a structure that it can probe in some detail. The quark structure of protons and neutrons appears at the minute length scale of 10–15 m or less. This structure was first probed in the early 1970s using high-energy electron beams produced by a linear accelerator at Stanford, USA. Guess what might have been the order of energy of these electron beams. (Rest mass energy of electron = 0.511 MeV.)

Answer:

The wavelength of the proton or neutron, λ ≈ 10-15 m

Rest mass-energy of an electron:

m0c2 = 0.511 MeV

= 0.511 x 106 x 1.6 x 10-19

= 0.8176 x 10-13 J

Planck’s constant, h = 6.6 x 10-34 Js

Speed of light, c = 3 x 108 m/s

The momentum  of the proton or a neutron is given as

p = h/λ

=  6.6 x 10-34 /10-15

= 6.6 x 10-19 kg m/s

The relativistic relation for energy (E) is given as

E2 = p2c2 + m20C4

= (6.6 x 10-19 x 3 x 108)2 + (0.8176 x 10-13)2

= 392.04 x 10-22 + 0.6685 x 10-26

≈ 392.04 x 10-22

⇒ E = 19.8 x 10-11

= 19.8 x 10-11/1.6 x 10-19

= 12.375 x 108 eV

Thus, the order of energy of these electron beams is 12.375 x 108 eV.

Question 11.35: Find the typical de Broglie wavelength associated with a He atom in helium gas at room temperature (27 °C) and 1 atm pressure, and compare it with the mean separation between two atoms under these conditions.

Answer:

Room temperature, T = 270 C = 27 + 273 = 300 K

Atmospheric pressure, P = 1 atm = 1.01 x 105 Pa

The atomic weight of the He atom = 4

Avogadro’s number, NA = 6.023 x 1023

Boltzmann’s constant, k = 1.38 x 10-23 J/mol/K

The average energy of a gas at temperature T,

De Broglie wavelength is given as

E = (3/2) kT

m = mass of the He atom

= Atomic weight/NA

= 4/(6.023 x 1023)

= 6.64 x 10-24 g

m = 6.64 x 10-27 kg

= 0.7268 x 10-10 m

We have the ideal gas formula

PV = RT

PV = kNT

V/N = kT/P

Here,

V is the volume of the gas

N is the number of moles of the gas

The mean separation between the two atoms of the gas is given as

= 3.35 x 10-9 m

Therefore, the mean separation between the atoms is greater than the de Broglie wavelength.

Question 11.36: Compute the typical de Broglie wavelength of an electron in metal at 27 °C and compare it with the mean separation between two electrons in a metal which is given to be about 2 × 10–10 m. [Note: Exercises 11.35 and 11.36 reveal that while the wave-packets associated with gaseous molecules under ordinary conditions are non-overlapping, the electron wave-packets in a metal strongly overlap with one another. This suggests that whereas molecules in an ordinary gas can be distinguished apart, electrons in a metal cannot be distinguished apart from one another. This
indistinguishability has many fundamental implications, which you will explore in more advanced Physics courses.]

Answer:

Temperature, T = 27°C = 27 + 273 = 300 K

Mean separation between two electrons, r = 2 × 10−10 m

De Broglie wavelength of an electron is

h = Planck’s constant = 6.6 × 10−34 Js

m = Mass of an electron = 9.11 × 10−31 kg

k = Boltzmann constant = 1.38 × 10−23 J/mol/K.

≈ 6.2 x 10-9 m

Hence, the de Broglie wavelength is much greater than the given inter-electron separation.

Question 11.37: Answer the following questions:
(a) Quarks inside protons and neutrons are thought to carry fractional charges [(+2/3)e; (–1/3)e]. Why do they not show up in Millikan’s oil-drop experiment?
(b) What is so special about the combination e/m? Why do we not simply talk of e and m separately?
(c) Why should gases be insulators at ordinary pressures and start conducting at very low pressures?
(d) Every metal has a definite work function. Why do all photoelectrons not come out with the same energy if incident radiation is monochromatic? Why is there an energy distribution of photoelectrons?
(e) The energy and momentum of an electron are related to the frequency and wavelength of the associated matter wave by the relations:
E = h ν, p = h/λ
But while the value of λ is physically significant, the value of ν (and therefore, the value of the phase speed ν λ) has no physical significance. Why?

Answer:

(a) Quarks inside protons and neutrons are thought to carry fractional charges [(+2/3)e ; (–1/3)e]. This is because nuclear forces grow stronger if they are pulled apart. Therefore, it seems that fractional charges may exist in nature. The observable charges are an integral multiple of electrical charges (e).

(b) The relation between the electric field and the magnetic field,

eV = (1/2) mv2  and eBv = mv2/r

Here,

e = electric charge

v = velocity

V = potential

r = Radius

B = magnetic field

From these equations, it can be understood that the dynamics of an electron can be determined only by the ratio e/m and not by e and m separately.

(c)  At the atmospheric pressure, the ions in the gas do not have a chance of reaching their respective electrons due to collision and recombination with other molecules in the gas. At low pressures, ions have a chance to reach their respective electrons, which results in the flow of current.

(d)  The minimum energy required for an electron in the conduction band to get out of the metal is called the work function. These electrons occupy different energy levels, because of which, for the same incident radiation, electrons come out with different energies.

(e) The absolute value of the energy of a particle is arbitrary within the addictive constant. Therefore, the wavelength(λ) is significant, but the frequency (ν) of the electron does not have direct physical significance. Therefore, product νλ has no physical significance.