Class 12 Physics Chapter 11 NCERT Solutions - Dual Nature of Radiation and Matter: Questions with Answers PDF

Q.11.1 Find the

(a) maximum frequency, and

(b) minimum wavelength of X-rays produced by 30 kV electrons.

 Ans.11.1 Potential of the electrons, V = 30 kV = 3 $\times {10}^4$  V

Energy of the electron, E = e $\times V$  where e = charge of an electron = 1.6 $\times {10}^{-19}C$

 

• Maximum frequency produced by the X-ray = $\nu$

The energy of the electron is given by the relation, E = h $\nu$ ,

where h = Planck’s constant = 6.626 $\times {10}^{-34}$  Js

$\nu =\frac{E}{h}$ = $\frac{3\mathrm{}\times {10}^4\times 1.6\times {10}^{-19}}{6.626\times {10}^{-34}}$  = 7.244 $\times {10}^{18}$  Hz

$\mathrm{H}\mathrm{e}\mathrm{n}\mathrm{c}\mathrm{e}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{m}\mathrm{a}\mathrm{x}\mathrm{i}\mathrm{u}\mathrm{m}\mathrm{}\mathrm{f}\mathrm{r}\mathrm{e}\mathrm{q}\mathrm{u}\mathrm{e}\mathrm{n}\mathrm{c}\mathrm{y}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{X}-\mathrm{r}\mathrm{a}\mathrm{y}\mathrm{}\mathrm{p}\mathrm{r}\mathrm{o}\mathrm{d}\mathrm{u}\mathrm{c}\mathrm{e}\mathrm{d}\mathrm{}\mathrm{i}\mathrm{s}\mathrm{}$ 7.244 $\times {10}^{18}$  Hz

 

• The minimum wavelength produced is given as

  $\lambda =\frac{c}{\nu }$ where c = Speed of light in air, c = 3 $\times {10}^{18}$  m/s

  $\lambda =\frac{3\mathrm{}\times {10}^8}{7.244\times {10}^{18}}$ = 4.14 $\times {10}^{-11}$  m = 0.0414 nm

Hence, the minimum wavelength produced is 0.414 nm.

 

Q.11.2 The work function of cesium metal is 2.14 eV. When light of frequency 6  $\times {10}^{14}$ Hz is incident on the metal surface, photoemission of electrons occurs. What is the

(a) maximum kinetic energy of the emitted electrons,

(b) Stopping potential, and

(c) maximum speed of the emitted photoelectrons?

Ans.11.2 Work function of cesium metal,  ${\varnothing }_0$ = 2.14 eV

Frequency of light,  $\nu$ = 6  $\times {10}^{14}$ Hz

The maximum kinetic energy is given by the photoelectric effect,  $K$ =  $h\nu -{\varnothing }_0$ ,

Where  $h$ = Planck’s constant = 6.626  $\times {10}^{-34}$ Js, 1 eV = 1.602  $\times {10}^{-19}$ J

$K$ =  $\frac{6.626\times {10}^{-34}\times 6\times {10}^{14}}{1.602\times {10}^{-19}}-2.14$ = 0.345 eV

$\mathrm{H}\mathrm{e}\mathrm{n}\mathrm{c}\mathrm{e}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{m}\mathrm{a}\mathrm{x}\mathrm{i}\mathrm{m}\mathrm{u}\mathrm{m}\mathrm{}\mathrm{k}\mathrm{i}\mathrm{n}\mathrm{e}\mathrm{t}\mathrm{i}\mathrm{c}\mathrm{}\mathrm{e}\mathrm{n}\mathrm{e}\mathrm{r}\mathrm{g}\mathrm{y}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{e}\mathrm{m}\mathrm{i}\mathrm{t}\mathrm{t}\mathrm{e}\mathrm{d}\mathrm{}\mathrm{e}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{t}\mathrm{r}\mathrm{o}\mathrm{n}\mathrm{}\mathrm{i}\mathrm{s}\mathrm{}0.3416\mathrm{}\mathrm{e}\mathrm{V}$

For stopping potential  $V_0$ , we can write the equation for kinetic energy as:

$K$ =  $V_0$ , where e = charge of an electron =  $1.6\times {10}^{-19}$

or  $V_0=\frac{K}{e}$ =  $\frac{0.345\times 1.602\times {10}^{-19}}{1.6\times {10}^{-19}}$ = 0.345 V

Hence, the stopping potential is 0.345 V

Maximum speed of the emitted photoelectrons = v

Kinetic energy K =  $\frac{1}{2}$ m  $v^2$ , where m = mass of the electron = 9.1  $\times {10}^{-31}$ kg

Hence, v =  $\sqrt{\frac{2K}{m}}$ =  $\sqrt{\frac{2\times 0.345\times 1.602\times {10}^{-19}}{9.1\mathrm{}\times {10}^{-31}}}$ = 348.5  $\times {10}^3$ m/s = 346.8 km/s

Therefore, the maximum speed of the emitted photoelectrons is 346.8 km/s

 

Q.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.11.3 Photoelectric cut-off voltage,  $V_0$ = 1.5 V

Maximum kinetic energy of the photoelectrons emitted is given as

$K=e{V}_0$ , where e = charge of an electron = 1.6  $\times {10}^{-19}$ C

K = 1.6  $\times {10}^{-19}\times 1.5$ = 2.4  $\times {10}^{-19}$ J

Hence, maximum kinetic energy of the photoelectrons emitted is 2.4  $\times {10}^{-19}$ J

 

Q.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?

Ans.11.4 Wavelength of the monochromatic light,  $\lambda =632.8nm$ = 632.8  $\times {10}^{-9}$ m

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

Planck’s constant, h = 6.626  $\times {10}^{-34}$ Js

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

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

The energy of each photon is given as,  $E$ =  $\frac{hc}{\lambda }$ =  $\frac{6.626\mathrm{}\times {10}^{-34}\times 3\mathrm{}\times {10}^8}{632.8\times {10}^{-9}}$ J = 3.141  $\times {10}^{-19}$ J

The momentum of each photon is given by  $p$ =  $\frac{h}{\lambda }$ =  $\frac{6.626\mathrm{}\times {10}^{-34}}{632.8\times {10}^{-9}}$ = 1.047  $\times {10}^{-27}$ kg-m/s

Number of photons arriving per second at a target irradiated by the beam = n

Assume that the beam has a uniform cross-section that is less than the target area.

Hence, the equation for power can be written as,  $P=nE$

n =  $\frac{P}{E}$ =  $\frac{9.42\times {10}^{-3}}{3.141\times {10}^{-19}}$ = 3.0  $\times {10}^{16}$

Momentum of the hydrogen atom = momentum of the proton = 1.047  $\times {10}^{-27}$ kg-m/s

The momentum of the hydrogen atom ,  $p=mv$ , where m = mass of hydrogen atom and v = velocity

Hence, v =  $\frac{p}{m}$ =  $\frac{1.047\times {10}^{-27}}{1.66\mathrm{}\times {10}^{-27}}$ = 0.631 m/s

Q.11.5 The energy flux of sunlight reaching the surface of the earth is 1.388×10³ W/m2. How many photons (nearly) per square meter are incident on the Earth per second? Assume that the photons in the sunlight have an average wavelength of 550 nm.

Ans.11.5 The energy flux of sunlight,  $\varphi$ = 1.388  $\times {10}^3$ W/  $m^2$

Hence power of the sunlight per square meter, P = 1.388  $\times {10}^{3}W$

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

Planck’s constant, h = 6.626  $\times {10}^{-34}$ Js

Average wavelength of photon,  $\lambda$ = 550 nm = 550  $\times {10}^{-9}$ m

If n is the number of photon per square meter, incident on earth per second, the equation of power can be written as

$P=nE$ or  $n$ =  $\frac{P}{E}$

We know  $E$ =  $\frac{hc}{\lambda }$

Hence,  $n$ =  $\frac{P\lambda }{hc}$ =  $\frac{1.388\times {10}^3\times 550\mathrm{}\times {10}^{-9}}{6.626\mathrm{}\times {10}^{-34}\times 3\mathrm{}\times {10}^8}$ = 3.84  $\times {10}^{21}$ photons  $m^2$ /s

Therefore, every second 3.84  $\times {10}^{21}$ photons are incident per square meter on earth.

 

Q.11.6 In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be 4.12  $\times \mathit{}{10}^{-15}$ Vs. Calculate the value of Planck’s constant.

Ans.11.6 The slope of the cut-off voltage (V) versus frequency (  $\nu )$ of an incident light is given as:

$\frac{\mathrm{V}}{\nu }=$ 4.12  $\times {10}^{-15}$ Vs

The relationship of V and  $\nu$ is given as  $h\nu$ =  $eV$ or  $\frac{\mathrm{V}}{\nu }=$  $\frac{h}{e}$

where e = Charge of an electron = 1.6  $\times {10}^{-19}$ C and h = Plank’s constant

Therefore, h =  $e\times \frac{\mathrm{V}}{\nu }$ = 1.6  $\times {10}^{-19}$  $\times$ 4.12  $\times {10}^{-15}$ = 6.592  $\times {10}^{-34}$ Js

Plank’s constant = 6.592  $\times {10}^{-34}$ Js

 

Q.11.7 A 100W 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?

Ans.11.7 Power of the sodium lamp, P = 100 W

Wavelength of the emitted sodium light,  $\lambda$ = 589 nm = 589  $\times {10}^{-9}$ m

Planck’s constant, h = 6.626  $\times {10}^{-34}$ Js

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

Energy per photon associated with the sodium light is given as:

$E$ =  $\frac{hc}{\lambda }$ =  $\frac{6.626\mathrm{}\times {10}^{-34}\times 3\mathrm{}\times {10}^8}{589\mathrm{}\times {10}^{-9}}$ = 3.37  $\times {10}^{-19}$ J =  $\frac{3.37\mathrm{}\times {10}^{-19}}{1.6\times {10}^{-19}}$ eV = 2.11 eV

Let the number of photon delivered to the sphere = n

The equation of power can be written as  $P=nE$

$n$ =  $\frac{P}{E}$ =  $\frac{100}{3.37\mathrm{}\times {10}^{-19}}$ photons/sec = 2.97  $\times {10}^{20}$ photons/s

Therefore, every second, 2.97  $\times {10}^{20}$ are delivered to the sphere.

 

Q.11.8 The threshold frequency for a certain metal is 3.3 × 10¹⁴ Hz. If light of frequency 8.2 × 10¹⁴ Hz is incident on the metal, predict the cutoff voltage for the photoelectric emission.

Ans.11.8 Threshold frequency of the metal,  ${\nu }_0=3.3\times {10}^{14}$ Hz

Frequency of the light incident on metal,  $\nu =8.2\times {10}^{14}$ Hz

Charge of an electron, e = 1.6  $\times {10}^{-19}$ C

Planck’s constant, h = 6.626  $\times {10}^{-34}$ Js

Let the cut-off voltage for the photoelectric emission from the metal be  $V_0$

The equation of the cut-off energy is given as:

$e{V}_0$ =  $h(\nu -{\nu }_0)$ or

$V_0=\frac{h(\nu -{\nu }_0)}{e}$ =  $\frac{6.626\mathrm{}\times {10}^{-34}\times (8.2\times {10}^{14}-3.3\times {10}^{14})}{1.6\mathrm{}\times {10}^{-19}}$ V = 2.0292 V

Therefore, the cut-off voltage for the photoelectric emission is 2.0292V

 

Q.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?

Ans.11.9 Work function of the metal,  ${\varnothing }_0$ =  $4.2eV$

Charge of an electron, e = 1.6  $\times {10}^{-19}$ C

Planck’s constant, h = 6.626  $\times {10}^{-34}$ Js

Wavelength of the incident radiation,  $\lambda$ = 330 nm= 330  $\times {10}^{-9}$ m

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

The energy of the incident photon is given as:

$E$ =  $\frac{hc}{\lambda }$ =  $\frac{6.626\mathrm{}\times {10}^{-34}\times 3\mathrm{}\times {10}^8}{330\mathrm{}\times {10}^{-9}}$ = 6.02  $\times {10}^{-19}$ J =  $\frac{6.02\mathrm{}\times {10}^{-19}}{1.6\times {10}^{-19}}$ eV = 3.76 eV

Since the energy of the incident photon (3.76  $eV$ ) is less than the work function of the metal (4.2  $eV$ ), there will be no photoelectric emission.

 

Q.11.10 Light of frequency 7.21 × 10¹⁴ Hz is incident on a metal surface. Electrons with a maximum speed of 6.0 × 10⁵ m/s are ejected from the surface. What is the threshold frequency for photoemission of electrons?

Ans.11.10 Frequency of the incident photon,  $\nu =7.21\times {10}^{14}$ Hz

Maximum speed of electron, v = 6  $\times {10}^5$ m/s

Planck’s constant, h = 6.626  $\times {10}^{-34}$ Js

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

Let the threshold frequency =  ${\nu }_0$

From the relation of threshold frequency and kinetic energy, we can write

$\frac{1}{2}$ m  $v^2$ =  $h$ (  $\nu -{\nu }_0)$

${\nu }_0=$  $\nu -\frac{m{v}^2}{2h}$ =  $7.21\times {10}^{14}-\frac{9.1\mathrm{}\times {10}^{-31}\times {(6\mathrm{}\times {10}^5)}^2}{2\times 6.626\mathrm{}\times {10}^{-34}}$ = 4.738  $\times {10}^{14}$ Hz

Therefore, the threshold frequency is 4.738  $\times {10}^{14}$ Hz

 

Q.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.11.11 Wavelength of the light produced by Argon laser,  $\lambda$ = 488 nm = 488  $\times {10}^{-9}$ m

Stopping potential of the photoelectrons,  $V_0$ = 0.38 V =  $\frac{0.38}{1.6\times {10}^{-19}}$  $eV$

Planck’s constant, h = 6.626  $\times {10}^{-34}$ Js

Charge of an electron, e = 1.6  $\times {10}^{-19}$ C

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

From Einstein’s photoelectric effect, we have the relation involving the work function  ${\varnothing }_0$ of the material of the emitter as:

$e{V}_0=\frac{hc}{\lambda }-{\varnothing }_0$

${\varnothing }_0=\frac{hc}{\lambda }-e{V}_0$ =  $\frac{6.626\mathrm{}\times {10}^{-34}\times 3\mathrm{}\times {10}^8}{488\mathrm{}\times {10}^{-9}\times 1.6\times {10}^{-19}}$  $-$ 1.6  $\times {10}^{-19}\times \frac{0.38}{1.6\times {10}^{-19}}$ = 2.54 – 0.38 = 2.165 eV

Therefore,, the material with which the emitter is made has the work function of 2.165 eV.

 

Q.11.12 Calculate the

(a) momentum, and

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

Ans.11.12 Potential difference, V = 56V

Planck’s constant, h = 6.626  $\times {10}^{-34}$ Js

Charge of an electron, e = 1.6  $\times {10}^{-19}$ C

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

At equilibrium, the kinetic energy of each electron is equal to the acceleration potential. If v is the velocity of each electron, we can write

$\frac{1}{2}m{v}^2$ =  $eV$

$v=\sqrt{\frac{2eV}{m}}$ =  $\sqrt{\frac{2\times 1.6\times {10}^{-19}\times 56}{9.1\mathrm{}\times {10}^{-31}}}$ = 4.44  $\times {10}^6$ m/s

The momentum of each electron = mv = 9.1  $\times {10}^{-31}\times$ 4.44  $\times {10}^6$ kgm/s

= 4.04  $\times {10}^{-24}$ kgm/s

De Broglie wavelength of an electron accelerating through a potential V, is given by the relation:

$\lambda =\frac{12.27}{\sqrt{V}}$ Å =  $\frac{12.27}{\sqrt{56}}$ Å = 1.64 Å = 1.64  $\times {10}^{-10}$ m = 0.164 nm

 

Q.11.13 What is the

(a) momentum,

(b) speed, and

(c) de Broglie wavelength of an electron with kinetic energy of 120 eV.

Ans.11.13 Kinetic energy of electron,  $E_k=$ 120 eV = 120  $\times 1.6\times {10}^{-19}$ J

Planck’s constant, h = 6.626  $\times {10}^{-34}$ Js

Charge of an electron, e = 1.6  $\times {10}^{-19}$ C

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

The kinetic energy of electron can be written as  $E_k=\frac{1}{2}$ m  $v^2$

$v$ =  $\sqrt{\frac{2E_k}{m}}$ =  $\sqrt{\frac{2\times 120\mathrm{}\times 1.6\times {10}^{-19}}{9.1\mathrm{}\times {10}^{-31}}}$ = 6.496  $\times {10}^6$ m/s

Momentum of the electron, p = mv = 9.1  $\times {10}^{-31}\times$ 6.496  $\times {10}^6$ = 5.911  $\times {10}^{-24}$ kgm/s

Speed of the electron = 6.496  $\times {10}^6$ m/s

De Broglie wavelength of an electron with momentum p is given as

$\lambda =\frac{h}{p}$ =  $\frac{6.626\mathrm{}\times {10}^{-34}}{5.911\mathrm{}\times {10}^{-24}}$ = 1.121  $\times {10}^{-10}$ m = 0.1121 nm

 

Q.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.

Ans.11.14 Wavelength of light of a sodium line,  $\lambda$ = 589 nm = 589  $\times {10}^{-9}$ m

Mass of electron,  $m_e$ = 9.1  $\times {10}^{-31}$ kg

Mass of a neutron,  $m_n$ = 1.66  $\times {10}^{-27}$ kg

Planck’s constant, h = 6.626  $\times {10}^{-34}$ Js

The kinetic energy of the electron  $E_k=\frac{1}{2}{m}_e{v}^2$ ….(1)

The equation for De Broglie wavelength  $\lambda =\frac{h}{m_e\times v}$ or  ${\lambda }^2=\frac{h^2}{{m_e}^2\times v^2}$ ………(2)

$v^2=\frac{h^2}{{m_e}^2\times {\lambda }^2}$

Combining equation (1) and (2), we get

$E_k=\frac{1}{2}{m}_e\frac{h^2}{{m_e}^2\times {\lambda }^2}$ =  $\frac{h^2}{2{\lambda }^2{m}_e}$ =  $\frac{{(6.626\mathrm{}\times {10}^{-34})}^2}{2{(589\mathrm{}\times {10}^{-9})}^2\times 9.1\mathrm{}\times {10}^{-31}}$ = 6.95  $\times {10}^{-25}$ J =  $\frac{6.95\times {10}^{-25}}{1.6\times {10}^{-19}}$ eV

= 4.345  $\times {10}^{-6}$ eV = 4.345  $\mu eV$

(b)The kinetic energy of the neutron  $E_n=\frac{1}{2}{m}_n{v}^2$ ….(1)

The equation for De Broglie wavelength  $\lambda =\frac{h}{m_n\times v}$ or  ${\lambda }^2=\frac{h^2}{{m_n}^2\times v^2}$ ………(2)

$v^2=\frac{h^2}{{m_n}^2\times {\lambda }^2}$

Combining equation (1) and (2), we get

$E_k=\frac{1}{2}{m}_n\frac{h^2}{{m_n}^2\times {\lambda }^2}$ =  $\frac{h^2}{2{\lambda }^2{m}_n}$ =  $\frac{{(6.626\mathrm{}\times {10}^{-34})}^2}{2{(589\mathrm{}\times {10}^{-9})}^2\times 1.66\mathrm{}\times {10}^{-27}}$ = 3.81  $\times {10}^{-28}$ J =  $\frac{3.81\times {10}^{-28}}{1.6\times {10}^{-19}}$ eV

= 2.382  $\times {10}^{-9}$ eV = 2.382  $neV$

 

Q.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.11.15 Mass of the bullet, m = 0.04 kg

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

Planck’s constant, h = 6.626  $\times {10}^{-34}$ Js

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

$\lambda =\frac{h}{m\times v}$ =  $\frac{6.626\mathrm{}\times {10}^{-34}}{0.04\times 1000}$ = 1.65  $\times {10}^{-35}$ m

Mass of the ball, m = 0.06 kg

Speed of the ball, v = 1.0 m/s

Planck’s constant, h = 6.626  $\times {10}^{-34}$ Js

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

$\lambda =\frac{h}{m\times v}$ =  $\frac{6.626\mathrm{}\times {10}^{-34}}{0.06\times 1}$ = 1.10  $\times {10}^{-32}$ m

Mass of the dust particle, m = 1.0  $\times {10}^{-9}$ kg

Speed of the dust particle, v = 2.2 m/s

Planck’s constant, h = 6.626  $\times {10}^{-34}$ Js

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

$\lambda =\frac{h}{m\times v}$ =  $\frac{6.626\mathrm{}\times {10}^{-34}}{1.0\mathrm{}\times {10}^{-9}\times 2.2}$ = 3.01  $\times {10}^{-25}$ m

 

Q.11.16 An electron and a photon each have a wavelength of 1.00 nm. Find

(a) their momentum,

(b) the energy of the photon, and

(c) the kinetic energy of electron.

Ans.11.16 Wavelength of electron,  ${\lambda }_e$ = Wavelength of proton,  ${\lambda }_p$ = 1.0 nm = 1  $\times {10}^{-9}$ m

Planck’s constant, h = 6.626  $\times {10}^{-34}$ Js

From De Broglie wavelength relation,  $\lambda =\frac{h}{p},$ where p = momentum

$p$ =  $\frac{h}{\lambda }$ =  $\frac{6.626\mathrm{}\times {10}^{-34}}{1\mathrm{}\times {10}^{-9}}$ = 6.626  $\times {10}^{-25}$ kg.m/s. Since  ${\lambda }_e={\lambda }_p$ , their momentum will be also equal.

The energy of photon is given by the relation:  $E$ =  $\frac{hc}{\lambda },$

where c = speed of light = 3  $\times {10}^8$ m/s

$E$ =  $\frac{6.626\mathrm{}\times {10}^{-34}\times 3\times {10}^8}{1\mathrm{}\times {10}^{-9}}$ Js =  $\frac{6.626\mathrm{}\times {10}^{-34}\times 3\times {10}^8}{1\mathrm{}\times {10}^{-9}\times 1.6\times {10}^{-19}}$ eV = 1242.38 eV = 1.242 keV

Kinetic energy of electron, having momentum p is given by the relation

$E_k$ =  $\frac{1}{2}\frac{p^2}{m}$ where m = mass of electron = 9.1  $\times {10}^{-31}$ kg.

Hence,  $E_k$ =  $\frac{1}{2}\frac{{(6.626\mathrm{}\times {10}^{-25})}^2}{9.1\mathrm{}\times {10}^{-31}}$ J = 2.412  $\times {10}^{-19}$ J =  $\frac{2.412\times {10}^{-19}}{1.6\times {10}^{-19}}$ eV = 1.51 eV

 

Q.11.17 (a) For what kinetic energy of a neutron will the associated de Broglie wavelength be 1.40 × 10^–10 m?

(b) Also find the de Broglie wavelength of a neutron, in thermal equilibrium with matter, having an average kinetic energy of (3/2) k T at 300 K.

Ans.11.17 De Broglie wavelength of the neutron,  $\lambda$ = 1.40  $\times {10}^{-10}$ m

Mass of neutron, m = 1.66  $\times {10}^{-27}$ kg

Planck’s constant, h = 6.626  $\times {10}^{-34}$ Js

Kinetic energy,  $E_k$ =  $\frac{1}{2}m{v}^2$ …………..(1)

De Broglie wavelength and velocity (v) are related as

$\lambda =\frac{h}{mv}$ ……………..(2)

Combining equation (1) and (2), we get

$E_k$ =  $\frac{1}{2}m({\frac{h}{m\lambda })}^2$ =  $\frac{h^2}{2m{\lambda }^2}$ =  $\frac{{(6.626\mathrm{}\times {10}^{-34})}^2}{2\times 1.66\mathrm{}\times {10}^{-27}\times ({1.40\mathrm{}\times {10}^{-10})}^2}$ = 6.75  $\times {10}^{-21}$ J =  $\frac{6.75\times {10}^{-21}}{1.6\times {10}^{-19}}$ eV

= 42.17  $\times {10}^{-3}eV$

Temperature of the neutron, T = 300 K

Average kinetic energy of the neutron,  $E_{k-avg}$ =  $\frac{3}{2}kT$ ,

where k = Boltzmann constant = 1.38  $\times {10}^{-23}$ kg  $m^2{s}^{-2}{K}^{-1}$

$E_{k-avg}$ =  $\frac{3}{2}kT$ =  $\frac{3}{2}\times 1.38\times {10}^{-23}\times 300=6.21\times {10}^{-21}$ J

The relationship of De Broglie wavelength is given as:

$\lambda =\frac{h}{\sqrt{2\times E_{k-avg}\times m}}$ =  $\frac{6.626\mathrm{}\times {10}^{-34}}{\sqrt{2\times 6.21\times {10}^{-21}\times 1.66\mathrm{}\times {10}^{-27}}}$ = 1.46  $\times {10}^{-10}$ m = 0.146 nm

 

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

Ans.11.18 The momentum of a photon having energy (  $h\nu )$ is given as  $p$ =  $\frac{h\nu }{c}$ =  $\frac{h}{\lambda }$

So,  $\lambda =\frac{h}{p}$ …………….(1)

Where,  $\lambda =$ Wavelength of the electromagnetic radiation

h = Planck’s constant

c = speed of light

De Broglie wavelength of the photon is given as  $\lambda =\frac{h}{mv}$

But momentum,  $=mv$ , where  $m$ = mass of the photon,  $v$ = velocity of the photon

Hence  $\lambda =\frac{h}{p}$ ………..(2)

Hence, it can be inferred from equation (1) and (2) that wavelength of the electromagnetic radiation is equal to the De Broglie wavelength of the photon.

 

Q.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.11.19 Temperature of the Nitrogen molecule, T = 300 K

Atomic mass of nitrogen = 14.0076 u

Hence, mass of Nitrogen molecule, m = 2  $\times$ 14.0076 u = 28.0152 u

We know, 1 u = 1.66  $\times {10}^{-27}$ kg

So, m = 28.0152  $\times$ 1.66  $\times {10}^{-27}$ kg = 4.65  $\times {10}^{-26}$ kg

Planck’s constant, h = 6.626  $\times {10}^{-34}$ Js

Boltzmann constant, k = 1.38  $\times {10}^{-23}$ kg  $m^2{s}^{-2}{K}^{-1}$

We have the expression that relates to mean kinetic energy (  $\frac{3}{2}kT)$ of the nitrogen molecule with root mean square speed (  $v_{rms}$ ) as:

$\frac{1}{2}m{v_{rms}}^2$ =  $\frac{3}{2}kT$

$v_{rms}=\sqrt{\frac{3kT}{m}}$ =  $\sqrt{\frac{3\times 1.38\times {10}^{-23}\times 300}{4.65\times {10}^{-26}}}$ = 516.814 m/s

De Broglie wavelength of the nitrogen molecule

$\lambda =\frac{h}{m{v}_{rms}}$ =  $\frac{6.626\mathrm{}\times {10}^{-34}}{4.65\times {10}^{-26}\times 516.814}$ = 2.76  $\times {10}^{-11}$ m = 0.0276 nm

Therefore, De Broglie wavelength is 0.0276 nm.

 

Q.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 small 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?

Ans.11.20 Potential difference across evacuated tube, V = 500 V

Specific charge of electron, e/m = 1.76  $\times {10}^{11}$ C/kg

The speed of each emitted electron is given by the relation of kinetic energy as

$E_{kinetic}$ =  $\frac{1}{2}m{v}^2$ =  $eV$

$v=\sqrt{\frac{2eV}{m}}$ =  $\sqrt{2V\frac{e}{m}}$ =  $\sqrt{2\times 500\times 1.76\times {10}^{11}}$ = 13.27  $\times {10}^6$ m/s

Therefore, the speed of each emitted electron is 13.27  $\times {10}^6$ m/s

Collector potential, V = 10 MV = 10  $\times {10}^6$ V

The speed is given by  $v=\sqrt{\frac{2eV}{m}}$  $=\sqrt{2V\frac{e}{m}}$ =  $\sqrt{2\times 10\mathrm{}\times {10}^6\mathrm{}\times 1.76\times {10}^{11}}$

= 1.876  $\times {10}^9$ m/s

This is not possible nothing can move faster than the light. In the above formula

$E_{kinetic}$ =  $\frac{1}{2}m{v}^2$ can only be used in the non-relativistic limit, i.e. v << c

For very high speed problems, relativistic equations must be considered for solving them. In the relativistic limit, the total energy is given as :

$E=m{c}^2$ , where m = relativistic mass

$E=m_0\sqrt{(1-\frac{v^2}{c^2}})$ where  $m_0$ = Mass of the particle at rest.

Hence, kinetic energy is given as

$E_{kinetic}$ = m  $c^2$ -  $m_0{c}^2$

 

Q.11.21 (a) A monoenergetic electron beam with electron speed of 5.20 × 10⁶ m s ^–1 is subject to a magnetic field of 1.30 × 10^–4 T normal to the beam velocity. What is the radius of the circle traced by the beam, given e/m for electron equals 1.76 × 10¹¹C kg^–1.

(b) Is the formula you employ in (a) valid for calculating radius of the path of a 20 MeV electron beam? If not, in what way is it modified?

[Note: Exercises 11.20(b) and 11.21(b) take you to relativistic mechanics which is beyond the scope of this book. They have been inserted here simply to emphasize the point that the formulas you use in part (a) of the exercises are not valid at very high speeds or energies. See answers at the end to know what ‘very high speed or energy’ means.]

Ans.11.21 Speed of the electron, v = 5.20  $\times {10}^{6}m/s$

Magnetic field experienced by the electron, B = 1.30  $\times {10}^{-4}$ T

Specific charge of electron, e/m = 1.76  $\times {10}^{11}$ C/kg

Charge of an electron e = 1.60  $\times {10}^{-19}$ C

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

The force exerted on the electron is given as

F =  $e\left|\vec{v}+\vec{B}\right|$

=  $evB\mathit{sin}\theta$ , where  $\theta$ = angle between the magnetic field and the beam velocity

The magnetic field is normal to the direction of beam, hence  $\theta =90°$

Therefore,  $F=evB$ ………………(1)

The beam traces a circular path of radius r. The magnetic field due to its bending nature provides a centrifugal force (F =  $\frac{m{v}^2}{r}$ ) for the beam.

Therefore,  $evB$ =  $\frac{m{v}^2}{r}$

r =  $\frac{m{v}^2}{evB}$ =  $\frac{mv}{eB}$

=  $\frac{9.1\times {10}^{-31}\times 5.20\times {10}^6}{1.60\times {10}^{-19}\times 1.30\times {10}^{-4}}$ = 0.2275 m = 22.75 cm

Energy of the electron beam, E = 20 MeV = 20  $\times {10}^6$ eV = 20  $\times {10}^6\times 1.6\times {10}^{-19}$ J

E = 3.2  $\times {10}^{-12}$ J

The energy of the electron beam is given as:

E =  $\frac{1}{2}$ m  $v^2$

v =  $\sqrt{\frac{2E}{m}}$ =  $\sqrt{\frac{2\times 3.2\times {10}^{-12}}{9.1\times {10}^{-31}}}$ = 2.652  $\times {10}^9$ m/s

This is not possible nothing can move faster than the light. In the above formula

$E_{kinetic}$ =  $\frac{1}{2}m{v}^2$ can only be used in the non-relativistic limit, i.e. v << c

For very high speed problems, relativistic equations must be considered for solving them. In the relativistic limit, the mass is given as :

$m=m_0\sqrt{(1-\frac{v^2}{c^2}})$ where  $m_0$ = Mass of the particle at rest.

Hence, the radius of the circular path is given as

r =  $\frac{mv}{eB}$ =  $\frac{v}{eB}\times m_0\sqrt{(1-\frac{v^2}{c^2}})$

 

Q.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 ( 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.

Ans.11.22 Potential, V = 100 V

Magnetic field experienced by electron, B = 2.83  $\times {10}^{-4}$ T

Radius of the circular orbit, r = 12.0 cm = 12  $\times {10}^{-2}$ m

Mass of each electron = m

Charge on each electron = e

Velocity of each electron= v

The energy of each electron is equal to its kinetic energy, i.e.

$\frac{1}{2}$ m  $v^2$ =  $eV$

$v^2=\frac{2eV}{m}$ ……..(1)

Since centripetal force (  $\frac{m{v}^2}{r})$ = Magnetic force (evB), we can write

$\frac{m{v}^2}{r}=evB$

$v$ =  $\frac{eBr}{m}$ ………………(2)

Equating equations (1) and (2) we get

$\frac{2eV}{m}=\frac{e^2{B}^2{r}^2}{m^2}$

$\frac{e}{m}$ =  $\frac{2V}{B^2{r}^2}$ =  $\frac{2\times 100}{({2.83\mathrm{}\times {10}^{-4})}^2\times ({12\times {10}^{-2})}^2}$ = 1.734  $\times {10}^{11}$ C/kg

Therefore, the specific charge ratio (e/m) is 1.734  $\times {10}^{11}$ C/kg

 

Q.11.23

(a) An X-ray tube produces a continuous spectrum of radiation with its short wavelength end at 0.45 Å. What is the maximum energy of a photon in the radiation?

(b) From your answer to (a), guess what order of accelerating voltage (for electrons) is required in such a tube?

Ans.11.23 Wavelength produced by X-ray,  $\lambda =$ 0.45 Å = 0.45  $\times {10}^{-10}$ m

Planck’s constant, h = 6.626  $\times {10}^{-34}$ Js

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

The maximum energy of a photon is given as:

$E$ =  $\frac{hc}{\lambda }$ =  $\frac{6.626\times {10}^{-34}\times 3\times {10}^8}{0.45\mathrm{}\times {10}^{-10}}$ = 4.417  $\times {10}^{-15}$ J =  $\frac{4.417\times {10}^{-15}}{1.6\times {10}^{19}}$ eV = 27.6  $\times {10}^3$ eV = 27.6 keV

Therefore, the maximum energy of an X-ray photon is 27.6 keV

To get an X-ray of 27.6 keV, the incident electrons must possess at least 27.6 keV of kinetic energy. Hence an accelerating voltage in the order of 30 keV will be required.

 

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

Ans.11.24 The total energy of two X-rays = 10.2 BeV = 10.2  $\times {10}^{9}eV$ = 10.2  $\times {10}^9\times 1.6\times {10}^{-19}$ J

Hence energy of each X-ray  $E$ =  $\frac{10.2\mathrm{}\times {10}^9\times 1.6\times {10}^{-19}}{2}$ = 8.16  $\times {10}^{-10}$ J

Planck’s constant, h = 6.626  $\times {10}^{-34}$ Js

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

From the relation of energy and wavelength, we get

$E$ =  $\frac{hc}{\lambda }$ or

$\lambda =\frac{hc}{E}$ =  $\frac{6.626\times {10}^{-34}\times 3\times {10}^8}{8.16\times {10}^{-10}}$ = 2.436  $\times {10}^{-16}$ m

Therefore the wavelength associated with each X-ray is 2.436  $\times {10}^{-16}$ m

 

Q.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.

Ans.11.25 The power of the medium wave transmitter, P = 10 kW = 10  $\times {10}^3$ W =  ${10}^4$ J/s

Hence energy emitted by the transmitter per second, E =  ${10}^4$ J

Wavelength of the radio wave,  $\lambda$ = 500 m

Planck’s constant, h = 6.626  $\times {10}^{-34}$ Js

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

Energy of the wave is given as :

$E_w$ =  $\frac{hc}{\lambda }$ =  $\frac{6.626\times {10}^{-34}\times 3\times {10}^8}{500}$ = 3.98  $\times {10}^{-28}$ J

Let n be number of photons emitted by the transmitter. Hence, total energy transmitted is given by:

n  $E_w$ = E

n =  $\frac{E}{E_w}$ =  $\frac{{10}^4}{3.98\times {10}^{-28}}$ = 2.52  $\times {10}^{31}$

Intensity of light perceived by the human eye, I =  ${10}^{-10}$ W  $m^{-2}$

Area of the pupil, A = 0.4  ${cm}^2$ = 0.4  $\times {10}^{-4}{m}^2$

Frequency of white light,  $\nu$ = 6  $\times {10}^{14}$ Hz

Planck’s constant, h = 6.626  $\times {10}^{-34}$ Js

Energy emitted by photon is given as:

$E=h\nu$ = 6.626  $\times {10}^{-34}\times 6\times {10}^{14}$ = 3.9756  $\times {10}^{-19}$ J

If n is total number of photons falling per second, per unit area of the pupil,

Total energy per unit area of n photons is = n  $\times$ 3.9756  $\times {10}^{-19}$ J = Intensity of light

n  $\times$ 3.9756  $\times {10}^{-19}$ =  ${10}^{-10}$

n = 2.52  $\times {10}^8$

Total number of photons entering the pupil per second is given as:

$n_A$ = n  $\times A$ = 2.52  $\times {10}^8\times$ 0.4  $\times {10}^{-4}$ = 10.06  $\times {10}^3$ /s

The number is not as large as the one found in problem (a), but it is large enough for the human eye to never see the individual photons.

 

Q.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 ?

Ans.11.26 Wavelength of ultraviolet light,  $\lambda$ = 2271Å = 2271  $\times {10}^{-10}$ m

Stopping potential of the metal,  $V_0$ = 1.3 V

Planck’s constant, h = 6.626  $\times {10}^{-34}$ Js

Charge of an electron, e = 1.6  $\times {10}^{-19}$ C

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

Work function of the metal,  ${\varnothing }_0$

Frequency of light =  $\nu$

We have the photo-energy relation from the photoelectric effect as:

${\varnothing }_0=h\nu -e{V}_0$

=  $\frac{hc}{\lambda }-e{V}_0$

=  $\frac{6.626\times {10}^{-34}\times 3\times {10}^8}{2271\times {10}^{-10}}-1.6\times {10}^{-19}\times 1.3$

= 8.75  $\times {10}^{-19}$ - 2.08  $\times {10}^{-19}$

= 6.67  $\times {10}^{-19}$ J

=  $\frac{6.67\times {10}^{-19}}{1.6\times {10}^{-19}}$ eV

= 4.17 eV

Let  ${\nu }_0$ be the threshold frequency of the metal.

Therefore,  ${\varnothing }_0$ = h  ${\nu }_0$

${\nu }_0=$  $\frac{{\varnothing }_0}{h}$ =  $\frac{6.67\times {10}^{-19}}{6.626\times {10}^{-34}}$ Hz = 1.007  $\times {10}^{15}$ Hz

Wavelength of red light,  ${\lambda }_r$ = 6328 Å = 6328  $\times {10}^{-10}$ m

Frequency of red light,  ${\nu }_r$ =  $\frac{c}{{\lambda }_r}$ =  $\frac{3\times {10}^8}{6328\times {10}^{-10}}$ = 4.74  $\times {10}^{14}$ Hz

Since  ${\nu }_0>{\nu }_r,\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{p}\mathrm{h}\mathrm{o}\mathrm{t}\mathrm{o}\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l}\mathrm{}\mathrm{w}\mathrm{i}\mathrm{l}\mathrm{l}\mathrm{}\mathrm{n}\mathrm{o}\mathrm{t}\mathrm{}\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{p}\mathrm{o}\mathrm{n}\mathrm{d}\mathrm{}\mathrm{t}\mathrm{o}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{r}\mathrm{e}\mathrm{d}\mathrm{l}\mathrm{i}\mathrm{g}\mathrm{h}\mathrm{t}\mathrm{}\mathrm{p}\mathrm{r}\mathrm{o}\mathrm{d}\mathrm{u}\mathrm{c}\mathrm{e}\mathrm{d}\mathrm{}\mathrm{b}\mathrm{y}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{l}\mathrm{a}\mathrm{s}\mathrm{e}\mathrm{r}.$

 

Q.11.27 Monochromatic radiation of wavelength 640.2 nm (1nm = 10^–9 m) from a neon lamp irradiates photosensitive material made of cesium 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.

Ans.11.27 Wavelength of the monochromatic light,  $\lambda$ = 640.2 nm = 640.2  ${\times 10}^{-9}m$

Stopping potential of neon lamp,  $V_0$ = 0.54 V

Charge of an electron, e = 1.6  $\times {10}^{-19}C$

Planck’s constant, h = 6.626  $\times {10}^{-34}$ Js

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

Let  ${\varnothing }_0$ be the work function and  $\nu$ frequency of emitted light

We have the photo-energy relation from the photoelectric effect as:

$e{V}_0$ =  $h\nu -{\varnothing }_0$ = h  $\frac{c}{\lambda }$ -  ${\varnothing }_0$

${\varnothing }_0=\frac{hc}{\lambda }$ -  $e{V}_0$

=  $\frac{6.626\times {10}^{-34}\times 3\times {10}^8}{640.2{\times 10}^{-9}}$  $-$ 1.6  $\times {10}^{-19}\times 0.54$

= 3.105  $\times {10}^{-19}$ - 0.864  $\times {10}^{-19}$

= 2.241  $\times {10}^{-19}$ J

=  $\frac{2.241\times {10}^{-19}}{1.6\times {10}^{-19}}$ eV

=1.40 eV

The wavelength of the radiation emitted from an iron source,  $\lambda \text{'}$ = 427.2 nm = 427.2  $\times {10}^{-9}$ m

Let the new stopping potential be  $V_0\text{'}$

From the relation  ${\varnothing }_0=\frac{hc}{\lambda \text{'}}$ -  $e{V}_0\text{'}$ , we get

$e{V}_0^{\text{'}}=\frac{hc}{{\lambda }^{\text{'}}}-{\varnothing }_0$

=  $\frac{6.626\times {10}^{-34}\times 3\times {10}^8}{427.2\times {10}^{-9}}$ - 2.241  $\times {10}^{-19}$

= 4.653  $\times {10}^{-19}$ - 2.241  $\times {10}^{-19}$

= 2.412  $\times {10}^{-19}$ J

=  $\frac{2.412\times {10}^{-19}}{1.6\times {10}^{-19}}$ eV

= 1.51 eV

The new stopping potential is 1.51 eV.

 

Q.11.28 A 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 photo-cell, the following lines from a mercury source were used:

${\mathit{\lambda }}_1$ = 3650 Å,  ${\mathit{\lambda }}_2$ = 4047 Å,  ${\mathit{\lambda }}_3$ = 4358 Å,  ${\mathit{\lambda }}_4$ = 5461 Å,  ${\mathit{\lambda }}_5$ = 6907 Å,

The stopping voltages, respectively, were measured to be:

${\mathit{V}}_{01}$ = 1.28 V,  ${\mathit{V}}_{02}$ = 0.95 V,  ${\mathit{V}}_{03}$ = 0.74 V,  ${\mathit{V}}_{04}$ = 0.16 V,  ${\mathit{V}}_{05}$ = 0 V

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

[Note: You will notice that to get h from the data, you will need to know e (which you can take to be 1.6  $\times {10}^{-19}$ C). Experiments of this kind on Na, Li, K, etc. were performed by Millikan, who, using his own value of e (from the oil-drop experiment) confirmed Einstein’s photoelectric equation and at the same time gave an independent estimate of the value of h.]

Ans.11.28 Einstein’s photoelectric equation is given as:

$e{V}_0$ =  $h\nu -{\varnothing }_0$

$V_0=\frac{h}{e}\nu -\frac{{\varnothing }_0}{e}$ ………..(1)

where

$V_0=$ Stopping potential

h = Planck’s constant

e = Charge of an electron

$\nu =\mathrm{F}\mathrm{r}\mathrm{e}\mathrm{q}\mathrm{u}\mathrm{e}\mathrm{n}\mathrm{c}\mathrm{y}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{r}\mathrm{a}\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}$

${\varnothing }_0=\mathrm{W}\mathrm{o}\mathrm{r}\mathrm{k}\mathrm{}\mathrm{f}\mathrm{u}\mathrm{n}\mathrm{c}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{a}\mathrm{}\mathrm{m}\mathrm{a}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{l}$

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

It can be concluded from equation (1) that  $V_0$ is directly proportional to frequency  $\nu$

Now frequency  $\nu$ can be expressed as

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

Then,

  ${\nu }_1$ =  $\frac{c}{{\lambda }_1}$ =  $\frac{3\times {10}^8}{3650\mathrm{}\mathrm{\AA }}$ Hz = 8.219 $\times {10}^{14}$ Hz

  ${\nu }_2$ =  $\frac{c}{{\lambda }_2}$ =  $\frac{3\times {10}^8}{4047\mathrm{}\mathrm{\AA }}$  =   $\frac{3\times {10}^8}{3650\times {10}^{-10}\mathrm{}}$Hz = 7.413 $\times {10}^{14}$ $\frac{\mathrm{<br>}}{\mathrm{}}$ Hz

${\nu }_3$ =  $\frac{c}{{\lambda }_3}$ =  $\frac{3\times {10}^8}{4358\mathrm{}\mathrm{\AA }}$ = $\frac{3\times {10}^8}{4047\times {10}^{-10}}$Hz = 6.883 $\times {10}^{14}$  Hz

${\nu }_4$ =  $\frac{c}{{\lambda }_4}$ =  $\frac{3\times {10}^8}{5461\mathrm{}\mathrm{\AA }}$ = $\frac{3\times {10}^8}{4358\times {10}^{-10}}$Hz = 5.493 $\times {10}^{14}$  $\frac{\mathrm{}}{}$ Hz

${\nu }_5$ =  $\frac{c}{{\lambda }_5}$ =  $\frac{3\times {10}^8}{6907\mathrm{}\mathrm{\AA }}$ = $\frac{3\times {10}^8}{5461\times {10}^{-10}}$Hz = 4.343 $\times {10}^{14}$Hz

From the given data of stopping potential, we get the following table

Frequency $\times {10}^{14}$  Hz	8.219	7.413	6.883	5.493	4.343
Stopping potential $V_0$V	1.28	0.95	0.74	0.16	0

It can be observed that the obtained curve is a straight line. It intersects the frequency axis at 5  $\times {10}^{14}$ Hz, which is the threshold frequency ( ${\nu }_0$ ${\mathrm{<br>}}^{}$ ) of the material. Point D corresponds to a frequency less than threshold frequency. Hence, there is no photoelectric emission for the ${\lambda }_5$ line and therefore, no stopping voltage is required to stop the current.

Slope of the straight line = $\frac{AB}{BC}$ =  $\frac{1.28-0.16}{(8.219-5.493)\times {10}^{14}}$

From equation (1), the slope $\frac{h}{e}$ ${\mathrm{<br>}}_{}$ can be written as

  $\frac{h}{e}=\frac{1.28-0.16}{(8.219-5.493)\times {10}^{14}}$ =  $\frac{1.12}{2.726\times {10}^{14}}$ Js

  $h$ =  $\frac{1.12}{2.726\times {10}^{14}}\times 1.6\times {10}^{-19}$ = 6.573  $\times {10}^{-34}$ Js

The work function of the material is given by, ${\varnothing }_0$  = $h{\nu }_0$  = 6.573 $\times {10}^{-34}\times$ 5 $\times {10}^{14}$

= 3.286 $\times {10}^{-19}$  J = $\frac{3.286\times {10}^{-19}}{1.6\times {10}^{-19}}$ eV = 2.054eV

.

Q.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 a 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 ?

Ans.11.29 Wavelength of the radiation, $\lambda$ = 3300 Å = 3300 $\times {10}^{-10}$ m

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

Planck’s constant, h = 6.626 $\times {10}^{-34}$ Js

The energy of the incident radiation is given as:

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

= $\frac{6.626\times {10}^{-34}\times 3\times {10}^8\mathrm{}}{3300\times {10}^{-10}}$ = 6.024 $\times {10}^{-19}$ J = $\frac{6.024\times {10}^{-19}}{1.6\times {10}^{-19}}$ eV = 3.765 eV

It can be observed that the energy of incident radiation is greater than the work function of Na and K only. Mo and Ni having more work function will not show the photoelectric emission.

If the source of light is brought near the photocells and placed 50 cm away from them, then the intensity of radiation will increase, but it will not affect the energy of the radiation. Hence, the result will be the same as before. However, the photoelectrons emitted from Na and K will increase in proportion to intensity.

 

Q.11.30 Light of intensity 10^–5 W m^–2 falls on a sodium photo-cell of surface area 2 cm². Assuming that the top 5 layers of sodium absorb the incident energy, estimate 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?

Ans.11.30 Intensity of the incident light, I = ${10}^{-5}$  W $m^{-2}$

Surface area of the sodium photocell, A = 2 ${cm}^2$  = 2 $\times {10}^{-4}{m}^2$

The incident power of the light, P = I $\times A$  = ${10}^{-5}$ $\times {10}^{-4}{m}^2$ W = 2 $\times {10}^{-9}$  W

Work function of the metal, ${\varnothing }_0$ = 2 eV = 2 $\times 1.6\times {10}^{-19}$  J = 3.2 $\times {10}^{-19}$  J

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

The effective atomic area of a sodium atom, $A_e$  = ${10}^{-20}$ $m^2$

 

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

n’ = n $\times \frac{A}{A_e\mathrm{}}$  = 5 $\times \frac{2\times {10}^{-4}}{{10}^{-20}}$  = 1 $\times {10}^{17}$

Since the incident power is uniformly absorbed by all the electrons continuously, the amount of energy absorbed per second per electron is

$E=\frac{P}{n\text{'}}$  = $\frac{2\times {10}^{-9}}{1\times {10}^{17}}$  = 2 $\times {10}^{-26}$  J/s

Time required for photoelectric emission,

t = $\frac{{\varnothing }_0}{E}$  = $\frac{3.2\times {10}^{-19}}{2\times {10}^{-26}}$ s = 16 $\times {10}^6$  seconds = 0.507 years

 

The time required for the photoelectric emission is nearly half a year, which is not practical. Hence, the wave picture is in disagreement with the given experiment.

 

Q.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 inter-atomic spacing in the lattice) ( = 9.11 × 10^–31 kg).

Ans.11.31 Wavelength of the light emitted from probe, $\lambda$ = 1 Å = 1 $\times {10}^{-10}$ m

Mass of the electron,  ${\mathit{m}}_{\mathit{e}}$ = 9.11 $\times {10}^{-31}$ kg

Planck’s constant, h = 6.626 $\times {10}^{-34}$ Js

Charge of an electron, e = 1.6 $\times {10}^{-19}$ C

Speed of light, c = 3  $\times {10}^{8\mathrm{<br>}}$ m/s

The kinetic energy of the electron is given as :

$E_k$ = $\frac{1}{2}{m}_e{v}^2$ ……………(1)

The de Broglie wavelength is given by

$\lambda =\frac{h}{p}$ ,where p = momentum = $m_{e}v$

$\lambda =\frac{h}{m_{e}v}$

$v$ =  $\frac{h}{m_e\lambda }$

Substituting the value of v in equation (1), we get

$E_k$ = $\frac{1}{2}{m}_e{\left(\frac{h}{m_e\lambda }\right)}^2$ = $\frac{m_e{h}^2}{2{m_e}^2{\lambda }^2}$ = $\frac{h^2}{2{m_e\lambda }^2}$

= $\frac{{(6.626\times {10}^{-34})}^2}{2{\times 9.11\times {10}^{-31}\times (1\times {10}^{-10})}^2}$

= $\frac{2.41\times {10}^{-17}}{1.6\times {10}^{-19}}$ ev

= 150.6 eV

Energy of a photon, E’ =  $\frac{hc}{\lambda }$ = $\frac{6.626\times {10}^{-34}\times 3\times {10}^8}{1\times {10}^{-10}\times 1.6\times {10}^{-19}}$ = 12.42 $\times {10}^3$ eV= 12.42 keV

Hence, a photon has a greater energy than an electron for the same wavelength.

 

Q.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. (  = 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

Ans.11.32 Kinetic energy of neutron, $E_{kn}$ == 150 eV = 150 $\times 1.6\times {10}^{-19}$ J = 2.4 $\times {10}^{-27}$ J

Mass of a neutron, $m_n$ = 1.675 $\times {10}^{-27}$

kg

Planck’s constant, h = 6.626 $\times {10}^{-34}$ Js

Kinetic energy of a neutron is given by the relation:

 =  $E_{kn}=\frac{1}{2}{m}_n{v}_n^2$………..(1)

Where, $v_n$ is the velocity of neutron

The de Broglie wavelength is given by

$\lambda =\frac{h}{p}$ , where p = momentum = $m_n{v}_n$

$\lambda =\frac{h}{m_n{v}_n}$

$v_n$ = $\frac{h}{m_n\lambda }$

Substituting the value of  $v_n$ in equation (1), we get

$E_{kn}=\frac{1}{2}{m}_n{\left(\frac{h}{m_n\lambda }\right)}^2$ = $\frac{h^2}{2{m_n\lambda }^2}$

   ${\lambda }^2$ =  $\frac{h^2}{2m_n{E}_{kn}}$

$\lambda =$ $\sqrt{\frac{h^2}{2m_n{E}_{kn}}}$ = $\frac{h}{\sqrt{2m_n{E}_{kn}}}$ = $\frac{6.626\times {10}^{-34}}{\sqrt{2\times 1.675\times {10}^{-27}\times 2.4\times {10}^{-17}}}$  = 2.337 $\times {10}^{-12}$ m

 It is given in the previous problem that the inter-atomic spacing of a crystal is about 1 Å, i.e. ${10}^{-10}$ m. Hence, the inter-atomic spacing is about ${10}^2$  times greater. Hence, a neutron beam of energy 150 eV is not suitable for diffraction experiments.

• Room temperature, T = 27 $℃$ = 27 + 273 = 300 K

The average kinetic energy is given by the equation, $E_{kn}$ = $\frac{3}{2}$ kT

Where, k = Boltzmann constant = 1.38 $\times {10}^{-23}$ J ${mol}^{-1}{K}^{-1}$

 

The wavelength of neutron is given by

$\lambda =$ $\sqrt{\frac{h^2}{2m_n{E}_{kn}}}$ = $\frac{h}{\sqrt{2m_n{E}_{kn}}}$ = $\frac{h}{\sqrt{2m_n\times \frac{3}{2}\mathrm{k}\mathrm{T}}}$ = $\frac{h}{\sqrt{3m_{n}kT}}$

= $\frac{6.626\times {10}^{-34}}{{(3\times 1.675\times {10}^{-27}\times 1.38\times {10}^{-23}\times 300)}^{\frac{1}{2}}}$  = 1.45 $\times {10}^{-10}$ m

This wavelength is comparable to the inter-atomic spacing of a crystal ( ${10}^{-10}$ m). Hence, the high energy neutron beam should first be thermalised, before using for diffraction.

 

Q.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?

Ans.11.33 The accelerating voltage of the electrons, V = 50 kV = 50 $\times {10}^3$  V

Mass of the electron, $m_e$ = 9.11 $\times {10}^{-31}$  kg

Planck’s constant, h = 6.626 $\times {10}^{-34}$  Js

Charge of an electron, e = 1.6 $\times {10}^{-19}$  C

Wavelength of yellow light, $\lambda$ = 5.9 $\times {10}^{-7}$ m

The kinetic energy of the electron is given as, $E_k$  = e $\times V$  = 1.6 $\times {10}^{-19}\times$ 50 $\times {10}^3$  J

= 8 $\times {10}^{-15}$ J

De Broglie wavelength is given by the relation,

$\lambda =\frac{h}{\sqrt{2\times m_e\times E_k\mathrm{}}}$ = $\frac{6.626\times {10}^{-34}}{\sqrt{2\times 9.11\times {10}^{-31}\times 8\times {10}^{-15}}}$  = 5.5 $\times {10}^{-12}$  m

This wavelength is nearly ${10}^5$ times less than the wavelength of a yellow light.

The resolving power of a microscope is inversely proportional to the wavelength of light used. Thus, the resolving power of an electron microscope is nearly ${10}^5$ times that of an optical microscope.

 

Q.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 early 1970’s 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.)

Ans.11.34 Wavelength of a proton or a neutron, $\lambda \approx {10}^{-15}$ m

Rest mass energy of an electron: $m_0{c}^2$ = 0.511 MeV = 0.511 $\times {10}^6$  eV

= 0.511 $\times {10}^6\times 1.6\times {10}^{-19}$  J

$m_{0}c$ = 8.176 $\times {10}^{-14}$ J

Planck’s constant, h = 6.626 $\times {10}^{-34}$ Js

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

 

The momentum of a proton or a neutron is given as:

$p$ = $\frac{h}{\lambda }$  = $\frac{6.626\times {10}^{-34}}{{10}^{-15}}$  = 6.626 $\times {10}^{-19}$  kgm/s

The relativistic relation for energy (e) is given as:

$E^2$ = $p^2{c}^2$ + $m_0^2{c}^2$

= ${(6.626\times {10}^{-19})}^2\times {(3\times {10}^8)}^2$ + ${(8.176\times {10}^{-14})}^2$

= 4.390 $\times {10}^{-37}\times 9\times {10}^{16}$  + 6.684 $\times {10}^{-27}$

=3.951 $\times {10}^{-20}$

E = 1.988 $\times {10}^{-10}$  J = $\frac{1.988\times {10}^{-10}}{1.6\times {10}^{-19}}$  eV = 1.242 $\times {10}^9$  eV = 1.242 BeV

 

Q.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.

Ans.11.35 Room temperature, T = 27 $℃$  = 300 K

Atmospheric pressure, P = 1 atm = 1.01 $\times {10}^5$  Pa

Atomic weight of helium atom = 4

Avogadro’s number, $N_A$  = 6.023 $\times {10}^{23}$

Boltzmann’s constant, k = 1.38 $\times {10}^{-23}$  J ${mol}^{-1}{K}^{-1}$

Planck’s constant, h = 6.626 $\times {10}^{-34}$  Js

Average energy of a gas at temperature T is given as:

E = $\frac{3}{2}$ kT = $\frac{3}{2}\times$ 1.38 $\times {10}^{-23}\times 300$  = 6.21 $\times {10}^{-21}$ J

De Broglie wavelength is given as

$\lambda =$ $\frac{h}{\sqrt{2mE}}$  , where m = mass of He atom = $\frac{Atomicweight}{Avogadr{o}^{\text{'}}snumber}$  = $\frac{4}{6.023\times {10}^{23}}$

m = 6.641 $\times {10}^{-24}$  gm = 6.641 $\times {10}^{-27}$  kg

 

  $\lambda =$ $\frac{6.626\times {10}^{-34}}{\sqrt{2\times 6.641\times {10}^{-27}\times 6.21\times {10}^{-21}}}$ = 7.29 $\times {10}^{-11}$ m

We have ideal gas formula:

PV = RT

PV = kNT

     $\frac{V}{N}=\frac{kT}{P}$          

Where, V = volume of the gas

N = number of moles of the gas

Mean separation between two atoms of the gas is given by the relation:

r = $({\frac{V}{N})}^{1/3}$   = ${\left(\frac{kT}{P}\right)}^{1/3}$

=( ${\frac{1.38\times {10}^{-23}\times 300}{1.01\times {10}^5})}^{1/3}$

= 3.45 $\times {10}^{-9}$  m

 

Hence, the mean separation between the atoms is much greater than the De Broglie wavelength.

 

Q.11.36 Compute the typical de Broglie wavelength of an electron in a 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 indistinguishibility has many fundamental implications which you will explore in more advanced Physics courses.]

 Ans.11.36 Temperature, T = 27 $℃$  = 300 K

Mean separation between two electrons, r = 2 $\times {10}^{-10}$  m

De Broglie wavelength of an electron is given as:

  $\lambda =$ $\frac{h}{\sqrt{3mkT}}$ , where

Planck’s constant, h = 6.626 $\times {10}^{-34}$  Js

m = mass of an electron = 9.11 $\times {10}^{-31}$  kg

k = Boltzmann constant = 1.38 $\times {10}^{-23}$  J ${mol}^{-1}{K}^{-1}$

  $\lambda =$ $\frac{6.626\times {10}^{-34}}{\sqrt{3\times 9.11\times {10}^{-31}\times 1.38\times {10}^{-23}\times 300}}$ = 6.23 $\times {10}^{-9}$  m

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

 

Q.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 $\mathit{\nu }$ , p = $\frac{\mathit{h}}{\mathit{\lambda }}$  ,

But while the value of  is physically significant, the value of  (and therefore, the value of the phase speed  ) has no physical significance. Why?

Ans.11.37 Quarks inside protons and neutrons carry fractional charges. This is because nuclear force increases extremely if they are pulled apart. Therefore, fractional charges may exist in nature; observable charges are still the integral multiple of an electrical charge.

The basic relations for electric field and magnetic field are

$\left\{eV=\frac{1}{2}m{v}^2\right\}$ and $\left\{eBV=\frac{m{v}^2}{r}\right\}$  respectively

These relations include e (electric charge), v (velocity), m (mass), V (potential), r (radius) and B (magnetic field. These relations give the value of the velocity of an electron as

$\left(v=\sqrt{2V\left(\frac{e}{m}\right)}\right)and\left(v=Br\left(\frac{e}{m}\right)\right)$ respectively

It can be observed from these relations that the dynamics of an electron is determined not by e and m separately, but by the ratio of  . $\frac{e}{m}$

(c) At atmospheric pressure, the ions of gases have no chance of reaching their respective electrons because of collision and recombination with other gas molecules. Hence, gases are insulators at atmospheric pressure. A low pressures, ions have a chance of reaching their respective electrodes and constitute a current. Hence, they conduct electricity at these pressures.

The work function of a metal is the minimum energy required for a conduction electron to get out of the metal surface. All the electrons in an atom do not have the same energy level. When a ray having some photon energy is incident on a metal surface, the electrons come out from different levels with different energies. Hence, these emitted electrons show different energy distributions

The absolute value of energy of a particle is arbitrary within the additive constant. Hence, wavelength( $\lambda )$ ( is significant, but the frequency( $\nu )$ (  associated with an electron has no direct physical significance.

Therefore, the product $\nu \lambda$  (phase speed) has no physical significance.

Group speed is given as:

$v_G$ = $\frac{dv}{dk}$ = $\frac{dv}{d\left(\frac{1}{\lambda }\right)}$ = dE/dp = (d(p^2/2m))/dp = p/m

This quantity has a physical meaning.