# HC Verma ii Solutions for Class 12 Science Physics Chapter 20 - Photoelectric Effect And Wave–particle Duality

HC Verma ii Solutions for Class 12 Science Physics Chapter 20 Photoelectric Effect And Wave–particle Duality are provided here with simple step-by-step explanations. These solutions for Photoelectric Effect And Wave–particle Duality are extremely popular among class 12 Science students for Physics Photoelectric Effect And Wave–particle Duality Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the HC Verma ii Book of class 12 Science Physics Chapter 20 are provided here for you for free. You will also love the ad-free experience on Meritnation’s HC Verma ii Solutions. All HC Verma ii Solutions for class 12 Science Physics are prepared by experts and are 100% accurate.

#### Page No 363:

#### Question 1:

Planck's constant has the same dimensions as

(a) force × time

(b) force × distance

(c) force × speed

(d) force × distance time

#### Answer:

(d) force × distance time

Planck's constant,

*h* = $\frac{E}{v}=\frac{\mathrm{Force}\times \mathrm{distance}}{\mathrm{frequency}}$

$\Rightarrow $ *h* = force $\times $ distance time

#### Page No 363:

#### Question 2:

Two photons of

(a) equal wavelength have equal linear momenta

(b) equal energies have equal linear momenta

(c) equal frequencies have equal linear momenta

(d) equal linear momenta have equal wavelengths

#### Answer:

Two photons having equal linear momenta have equal wavelengths is correct. As in the rest of the options magnitude of momentum or energy can be same because energy and momentum are inversely proportional to wavelength. But the direction of propagation of the photons can be different.

Hence the correct option is D.

#### Page No 363:

#### Question 3:

Let *p* and *E* denote the linear momentum and energy of a photon. If the wavelength is decreased,

(a) both *p* and *E* increase

(b) *p* increases and *E* decreases

(c) *p* decreases and *E* increases

(d) both *p* and *E* decrease

#### Answer:

(a) both *p* and *E* increase

From the de-Broglie relation, wavelength,

$\lambda =\frac{h}{p}...\left(1\right)$

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

Here,* h* = Planck's constant

* p* = momentum of electron

It is clear from the above equation that $p\propto \frac{1}{\lambda}$.

Thus, if the wavelength $\left(\lambda \right)$ is decreased, then momentum $\left(p\right)$ will be increase.

Relation between momentum and energy:

$p=\sqrt{2mE}$

Here, *E *= energy of electron

*m* = mass of electron

Substituting the value of *p* in equation (1), we get:

$\lambda =\frac{h}{\sqrt{2mE}}\phantom{\rule{0ex}{0ex}}\Rightarrow \sqrt{E}=\frac{h}{\lambda \sqrt{2m}}\phantom{\rule{0ex}{0ex}}\Rightarrow E=\frac{{h}^{2}}{2m{\lambda}^{2}}\phantom{\rule{0ex}{0ex}}$

Thus, on decreasing $\lambda $, the energy will increase.

#### Page No 363:

#### Question 4:

Let *n _{r}* and

*n*be the number of photons emitted by a red bulb and a blue bulb, respectively, of equal power in a given time.

_{b}(a)

*n*=

_{r}*n*

_{b}(b)

*n*<

_{r}*n*

_{b}(c)

*n*>

_{r}*n*

_{b}(d) The information is insufficient to derive a relation between

*n*and

_{r}*n*.

_{b}#### Answer:

(c) *n _{r}* >

*n*

_{b}The two bulbs are of equal power. It means that they consume equal amount of energy per unit time.

Now, as the frequency of blue light $\left({f}_{b}\right)$ is higher than the frequency of red light$\left({f}_{r}\right)$, $h{f}_{b}>h{f}_{r}$.

Hence, the energy of a photon of blue light is more than the energy of a photon of red light.

Thus, a photon of blue light requires more energy than a photon of red light to be emitted.

For the same energy given to the bulbs in a certain time, the number of photons of blue light will be less than that of red light.

$\therefore $

*n*>

_{r}*n*(As the amount of energy emitted from the two bulb is same)

_{b}#### Page No 363:

#### Question 5:

The equation *E* = *pc* is valid

(a) for an electron as well as for a photon

(b) for an electron but not for a photon

(c) for a photon but not for an electron

(d) neither for an electron nor for a photon

#### Answer:

(c) for a photon but not for an electron

The equation *E* = *pc* is valid for a particle with zero rest mass. The rest mass of a photon is zero, but the rest mass of an electron is not zero. So, the equation will be valid for photon, and not electron.

#### Page No 363:

#### Question 6:

The work function of a metal is *hv*_{0}. Light of frequency *v* falls on this metal. Photoelectric effect will take place only if

(a) *v* ≥ *v*_{0}

(b) *v* > 2*v*_{0}

(c) *v* < *v*_{0}

(d) *v* < *v*_{0}/2

#### Answer:

(a) *v* ≥ *v*_{0}

As the work function of the metal is *hv*_{0}, the threshold frequency of the metal is *v*_{0}.

For photoelectric effect to occur, the frequency of the incident light should be greater than or equal to the threshlod frequency of the metal on which light is incident.

#### Page No 363:

#### Question 7:

Light of wavelength λ falls on a metal with work-function *hc/**λ*_{0}. Photoelectric effect will take place only if

(a) λ ≥ λ_{0}

(b) λ ≥ 2λ_{0}

(c) λ ≤ λ_{0}

(d) λ < λ_{0}/2

#### Answer:

(c)* λ* ≤ *λ*_{0}

As the work-function of the metal is *hc/**λ*_{0}, its threshold wavelength is *λ*_{0}.

For photoelectric effect, the wavelength of the incident light should be less than or equal to the threshold wavelength of the metal on which light is incident.

#### Page No 363:

#### Question 8:

It is found that yellow light does not eject photoelectrons from a metal. Is it advisable to try with orange light or with green light?

#### Answer:

Photoelectrons are emitted from a metal's surface if the frequency of incident radiation is more than the threshold frequency of the given metal surface. As yellow light does not eject photoelectrons from a metal it means that the threshold frequency of the metal is more than the frequency of yellow light. Since the frequency of orange light is less than the frequency of yellow light, therefore it will not be able to eject photoelectrons from the metal's surface. The frequency of green light is more than the frequency of yellow light. Hence, when it is incident on the metal surface, it will eject electrons from the metal.

#### Page No 363:

#### Question 9:

It is found that photosynthesis starts in certain plants when exposed to sunlight, but it does not start if the plants are exposed only to infrared light. Explain.

#### Answer:

Photosynthesis starts when a plant is exposed to visible light. The visible light's photons possess just enough energy to excite the electrons of molecules of the plant without causing damage to its cells. Infrared rays have less frequency than visible light. Due to this, the energy of the photons of infrared rays are not sufficient to initiate photosynthesis. Therefore, photosynthesis does not start if plants are exposed only to infrared light.

#### Page No 363:

#### Question 10:

The threshold wavelength of a metal is *λ*_{0}. Light of wavelength slightly less than λ_{0} is incident on an insulated plate made of this metal. It is found that photoelectrons are emitted for some time and after that the emission stops. Explain.

#### Answer:

When light of wavelength less than *λ*_{0} is incident on the metal surface, the free electrons of the metal will gain energy and come out of the metal surface. As the metal plate is insulated (it is not connected with the battery), the free electrons of the metal will not be replaced by the other electrons. Hence, photoelectron emission will stop after some time.

#### Page No 363:

#### Question 11:

Is *p* − *E*/c valid for electrons?

#### Answer:

From Einstein's mass- energy equation,

*E* =* **mc*^{2}

$\Rightarrow E=\frac{{m}_{0}{c}^{2}}{\sqrt{1-{\displaystyle \frac{{v}^{2}}{{c}^{2}}}}}$

Relativistic momentum,

$p=mv\phantom{\rule{0ex}{0ex}}\Rightarrow p=\frac{{m}_{0}{c}^{2}}{\sqrt{1-{\displaystyle \frac{{v}^{2}}{{c}^{2}}}}}$

Combining the above equations, we get:

${E}^{2}={{m}_{0}}^{2}{c}^{4}+{p}^{2}{c}^{2}$

From the above equation, it is clear that for *p* = *E*/c to be valid, the rest mass of the body should be zero. As electrons do not have zero rest mass, this equation is not valid for electrons.

#### Page No 363:

#### Question 12:

Consider the de-Broglie wavelength of an electron and a proton. Which wavelength is smaller if the two particles have (a) the same speed (b) the same momentum (c) the same energy?

#### Answer:

de-Broglie wavelength,

$\lambda =\frac{h}{mv}$,

where *h* = Planck's constant

$m$ = mass of the particle

*v* = velocity of the particle

(a) It is given that the speed of an electron and proton are equal.

It is clear from the above equation that

$\lambda \propto \frac{1}{m}$

As mass of a proton, *m*_{p} > mass of an electron, *m*_{e}, the proton will have a smaller wavelength compared to the electron.

(b) $\lambda =\frac{h}{p}$ $\left(\because p=mv\right)$

So, when the proton and the electron have same momentum, they will have the same wavelength.

(c) de-Broglie wavelength $\left(\lambda \right)$ is also given by

$\lambda =\frac{h}{\sqrt{2mE}}$,

where *E* = energy of the particle.

Let the energy of the proton and electron be *E.*

Wavelength of the proton,

${\lambda}_{p}=\frac{h}{\sqrt{2{m}_{\mathrm{p}}E}}...\left(1\right)$

Wavelength of the electron,

${\lambda}_{e}=\frac{h}{\sqrt{2{m}_{\mathrm{e}}E}}...\left(2\right)$

Dividing (2) by (1), we get:

$\frac{{\lambda}_{e}}{{\lambda}_{p}}=\frac{\sqrt{{m}_{e}}}{\sqrt{{m}_{p}}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{\lambda}_{e}}{{\lambda}_{p}}<1\phantom{\rule{0ex}{0ex}}\Rightarrow {\lambda}_{e}<{\lambda}_{p}$

It is clear that the proton will have smaller wavelength compared to the electron.

#### Page No 363:

#### Question 13:

If an electron has a wavelength, does it also have a colour?

#### Answer:

Colour is a characteristic of electromagnetic waves. Electrons behave as a de-Broglie wave because of their velocity. A de-Broglie wave is not an electromagnetic wave and is one dimensional. Hence, no colour is shown by an electron.

#### Page No 364:

#### Question 8:

When stopping potential is applied in an experiment on photoelectric effect, no photoelectric is observed. This means that

(a) the emission of photoelectrons is stopped

(b) the photoelectrons are emitted but are re-absorbed

(c) the photoelectrons are accumulated near the collector plate

(d) the photoelectrons are dispersed from the sides of the apparatus

#### Answer:

(b) the photoelectrons are emitted but are re-absorbed by the emitter metal

In an experiment on photoelectric effect, the photons incident at the metal plate cause photoelectrons to be emitted. The metal plate is termed as "emitter". The electrons ejected are collected at the other metal plate called "collector". When the potential of the collector is made negative with respect to the emitter (or the stopping potential is applied), the electrons emitted from the emitter are repelled by the collector. As a result, some electrons go back to the cathode and the current decreases.

#### Page No 364:

#### Question 9:

If the frequency of light in a photoelectric experiment is doubled, the stopping potential will

(a) be doubled

(b) be halved

(c) become more than double

(d) become less than double

#### Answer:

(c) become more than double

According to Einstein's equation of photoelectric effect,

$e{V}_{0}=hv-\phi \phantom{\rule{0ex}{0ex}}\Rightarrow {V}_{0}=\frac{hv-\phi}{e}...\left(1\right)$

Here,* **V*_{0} = stopping potential

* v* = frequency of light

$\phi $ = work function

Let the new frequency of light be 2*ν* and the corresponding stopping potential be *V*_{0}'.

Therefore,

$e{V}_{0}\text{'}=2hv-\phi \phantom{\rule{0ex}{0ex}}{V}_{0}\text{'}=\frac{2hv-\phi}{e}...\left(2\right)$

Multiplying both sides of equation (1) by 2, we get:

$2{V}_{0}=\frac{2hv-2\phi}{e}...\left(3\right)$

Now if we compare (2) and (3), it can be observed that:

$\frac{2hv-\phi}{e}>\frac{2hv-2\phi}{e}\phantom{\rule{0ex}{0ex}}\Rightarrow {V}_{0}\text{'}2{V}_{0}$

It is clear from the above equation that if the frequency of light in a photoelectric experiment is doubled, the stopping potential will be more than doubled.

#### Page No 364:

#### Question 10:

The frequency and intensity of a light source are doubled. Consider the following statements.

(A) The saturation photocurrent remains almost the same.

(B) The maximum kinetic energy of the photoelectrons is doubled.

(a) A and B are true.

(b) A is true but B is false.

(c) A is false but B is true.

(d) A and B are false.

#### Answer:

(b) A is true but B is false.

Saturated current varies directly with the intensity of light. As the intensity of light is increased, a large number of photons fall on the metal surface. As a result, a large number of electrons interact with the photons. As a result, the number of emitted electrons increases and, hence, the current also increases.

At the same time, the frequency of the light source also increases.Also, with the increase in frequency of light, the stopping potential increases as well. This will reduce the current. The combined effect of these two is that the current will remain the same

Hence, A is true.

From the Einstein's photoelectric equation.

${K}_{max}=hv-\phi $

Where *K*_{max} = kinetic energy of electron

* v* = frequency of light

$\phi $ = work function of metal

It is clear from the above equation. As the frequency of light source is doubled, kinetic energy of electron increases. But, it becomes more than the double.

Hence, B is false.

#### Page No 364:

#### Question 11:

A point source of light is used in a photoelectric effect. If the source is removed farther from the emitting metal, the stopping potential

(a) will increase

(b) will decrease

(c) will remain constant

(d) will either increase or decrease

#### Answer:

(c) will remain constant

As the source is removed farther from the emitting metal, the intensity of light will decrease. As the stopping potential does not depend on the intensity of light, it will remain constant.

#### Page No 364:

#### Question 12:

A point source causes photoelectric effect from a small metal plate. Which of the following curves may represent the saturation photocurrent as a function of the distance between the source and the metal?

Figure

#### Answer:

From the given curves ,curve (d) is correct.

As the relation between intensity (*I*) of light and distance (*r*) is

$I\propto \frac{1}{{r}^{2}}$

As the distance between the source and the metal is increased, it will result in decrease in the intensity of light. As the saturation current is directly proportional to the intensity of light ($i\propto I$), it can be concluded that current varies as $i\propto \frac{1}{{r}^{2}}$. Thus, curve d is correct.

#### Page No 364:

#### Question 13:

A non-monochromatic light is used in an experiment on photoelectric effect. The stopping potential

(a) is related to the mean wavelength

(b) is related to the longest wavelength

(c) is related to the shortest wavelength

(d) is not related to the wavelength

#### Answer:

(c) is related to the shortest wavelength

For photoelectric effect to be observed, wavelength of the incident light $\left(\lambda \right)$ should be less than the threshold wavelength $\left({\lambda}_{0}\right)$ of the metal.

Einstein's photoelectric equation:

$e{V}_{0}=\frac{hc}{{\lambda}_{0}}-\phi $

Here, *V*_{0} = stopping potential

${\lambda}_{0}$ = threshold wavelength

*h* = Planck's constant

$\phi $ = work-function of metal

It is clear from the above equation that stopping potential is related to the shortest wavelength (threshold wavelength).

#### Page No 364:

#### Question 14:

A proton and an electron are accelerated by the same potential difference. Let λ* _{e}* and λ

_{p}denote the de Broglie wavelengths of the electron and the proton, respectively.

(a) λ

*= λ*

_{e}

_{p}(b) λ

*< λ*

_{e}

_{p}(c) λ

*> λ*

_{e}

_{p}(d) The relation between λ

*and λ*

_{e}*depends on the accelerating potential difference.*

_{p}#### Answer:

(c) *λ*_{e} > *λ*_{p}

Let *m*_{e} and *m*_{p} be the masses of electron and proton, respectively.

Let the applied potential difference be *V*.

Thus, the de-Broglie wavelength of the electron,

${\lambda}_{e}=\frac{h}{\sqrt{2{m}_{\mathrm{e}}eV}}...\left(1\right)$

And de-Broglie wavelength of the proton,

${\lambda}_{p}=\frac{h}{\sqrt{2{m}_{\mathrm{p}}eV}}...\left(2\right)$

Dividing equation (2) by equation (1), we get:

$\frac{{\lambda}_{\mathrm{p}}}{{\lambda}_{\mathrm{e}}}=\frac{\sqrt{{m}_{\mathrm{e}}}}{\sqrt{{m}_{\mathrm{p}}}}$

*m*_{e} <* **m*_{p}

$\therefore $ $\frac{{\lambda}_{p}}{{\lambda}_{e}}<1$

$\Rightarrow {\lambda}_{p}<{\lambda}_{e}$

#### Page No 364:

#### Question 1:

When the intensity of a light source in increased,

(a) the number of photons emitted by the source in unit time increases

(b) the total energy of the photons emitted per unit time increases

(c) more energetic photons are emitted

(d) faster photons are emitted

#### Answer:

(a) the number of photons emitted by the source in unit time increases

(b) the total energy of the photons emitted per unit time increases

When the intensity of a light source in increased, a large number of photons are emitted from the light source. Hence, option (a) is correct.

Due to increase in the number of photons, total energy of the photons emitted per unit time also increases. Hence, option (b) is also correct.

Increase in the intensity of light increases only the number of photons, not the energy of photons, Hence, option (c) is incorrect.

The speed of photons is not affected by the intensity of light, Hence, option (d) is incorrect.

#### Page No 364:

#### Question 2:

Photoelectric effect supports quantum nature of light because

(a) there is a minimum frequency below which no photoelectrons are emitted

(b) the maximum kinetic energy of photoelectrons depends only on the frequency of light and not on its intensity

(c) even when the metal surface is faintly illuminated the photoelectrons leave the surface immediately

(d) electric charge of the photoelectrons is quantised

#### Answer:

(a) there is a minimum frequency below which no photoelectrons are emitted

(b) the maximum kinetic energy of photoelectrons depends only on the frequency of light and not on its intensity

(c) even when the metal surface is faintly illuminated the photoelectrons leave the surface immediately

Photoelectric effect can be explained on the basis of quantum nature of light. According to the quantum nature of light, energy in light is not uniformly spread. It is contained in packets or quanta known as photons.

Energy of a photon, *E* = *hv*, where *h* is Planck's constant and *v *is the frequency of light.

Above a particular frequency, called threshold frequency, energy of a photon is sufficient to emit an electron from the metal surface and below which, no photoelectron is emitted, as the energy of the photon is low. Hence, option (a) supports the quantum nature of light.

Now, kinetic energy of an electron,

$K=h{v}_{0}-\phi $

Thus, kinetic energy of a photoelectron depends only on the frequency of light (or energy). This shows that if the intensity of light is increased, it only increases the number of photons and not the energy of photons. Kinetic energy of photons can be increased by increasing the frequency of light or by increasing the energy of photon, which supports* E* = *hv* and, hence, the quantum nature of light. Hence, option (b) also supports the quantum nature of light.

Photoelectrons are emitted from a metal surface even if the metal surface is faintly illuminated; it means that less photons will interact with the electrons. However, few electrons absorb energy from the incident photons and come out from the metal. This shows the quantum nature of light. Hence, (c) also supports the quantum nature of light.

Electric charge of the photoelectrons is quantised; but this statement does not support the quantum nature of light.

#### Page No 364:

#### Question 3:

A photon of energy *hv* is absorbed by a free electron of a metal with work-function *hv* − φ.

(a) The electron is sure to come out.

(b) The electron is sure to come out with kinetic energy *hv* − φ.

(c) Either the electron does not come out or it comes our with kinetic energy *hv* − φ.

(d) It may come out with kinetic energy less than *hv* − φ.

#### Answer:

(d) It may come out with kinetic energy less than *hv* − φ.

When light is incident on the metal surface, the photons of light collide with the free electrons. In some cases, a photon can give all the energy to the free electron. If this energy is more then the work-function of the metal,then there are two possibilities. The electron can come out of the metal with kinetic energy *hv* − φ or it may lose energy on collision with the atoms of the metal and come out with kinetic energy less than *hv* − φ. Thus, it may come out with kinetic energy less than *hv* − φ.

#### Page No 364:

#### Question 4:

If the wavelength of light in an experiment on photoelectric effect is doubled,

(a) photoelectric emission will not take place

(b) photoelectric emission may or may not take place

(c) the stopping potential will increase

(d) the stopping potential will decrease

#### Answer:

(b) photoelectric emission may or may not take place

(d) the stopping potential will decrease

For photoelectric effect to be observed, wavelength of incident light should not be more than the largest wavelength called threshold wavelength $\left({\lambda}_{0}\right)$. If the wavelength of light in an experiment on photoelectric effect is doubled and if it is equal to or less than the threshold wavelength, then photoelectric emission will take place. If it is greater than the threshold wavelength, photoelectric emission will not take place. The photoelectric emission may or may not take place.Photoelectric emission depends on the wavelength of incident light.

Hence, option (b) is correct and (a) is incorrect.

From Einstein's photoelectric equation,

$e{V}_{0}=\frac{hc}{{\lambda}_{0}}-\phi $,

where *V*_{0} = stopping potential

${\lambda}_{0}$ = threshold wavelength

*h* = Planck's constant

$\phi $ = work-function of metal

It is clear that

${V}_{0}\propto \frac{1}{{\lambda}_{0}}$

Thus, if the wavelength of light in an experiment on photoelectric effect is doubled, its stopping potential will become half.

#### Page No 364:

#### Question 5:

The photo current in an experiment on photoelectric effect increases if

(a) the intensity of the source is increased

(b) the exposure time is increased

(c) the intensity of the source is decreased

(d) the exposure time is decreased

#### Answer:

(a) the intensity of the source is increased

When the intensity of the source is increased, the number of photons emitted from the source increases. As a result, a large number of electrons of the metal interact with these photons and hence, the number of electrons emitted from the metal increases. Thus, the photocurrent in an experiment of photoelectric effect increases. The photocurrent does not depend on the exposure time. Hence, option (a) is correct.

#### Page No 364:

#### Question 6:

The collector plate in an experiment on photoelectric effect is kept vertically above the emitter plate. A light source is put on and a saturation photocurrent is recorded. An electric field is switched on that has a vertically downward direction.

(a) The photocurrent will increase.

(b) The kinetic energy of the electrons will increase.

(c) The stopping potential will decrease.

(d) The threshold wavelength will increase.

#### Answer:

(b) The kinetic energy of the electrons will increase.

As there is no effect of electric field on the number of photons emitted, the photoelectric current will remain same. Hence, option (a) is incorrect.

When an electric field is applied, then electric force will act on the electron moving opposite the direction of electric field, which will increase the kinetic energy of the electron. Hence, option (b) is correct.

As the kinetic energy of the electron is increasing, its stopping potential will increase. Hence, option (c) is incorrect.

Threshold wavelength is the characteristic property of the metal and will not change. Hence, (d) is incorrect.

#### Page No 364:

#### Question 7:

In which of the following situations, the heavier of the two particles has smaller de Broglie wavelength? The two particles

(a) move with the same speed

(b) move with the same linear momentum

(c) move with the same kinetic energy

(d) have fallen through the same height

#### Answer:

(a) move with the same speed

(c) move with the same kinetic energy

(d) have fallen through the same height

Let *m*_{1} be the mass of the heavier particle and *m*_{2} be the mass of the lighter particle.

If both the particles are moving with the same speed *v*, de Broglie wavelength of the heavier particle,

${\lambda}_{1}=\frac{h}{{m}_{1}v}$ ...(1)

de Broglie wavelength of the lighter particle,

${\lambda}_{2}=\frac{h}{{m}_{2}v}$ ...(2)

Thus, from equations (1) and (2), we find that if the particles are moving with the same speed *v*, then ${\lambda}_{1}<{\lambda}_{2}$.

Hence, option (a) is correct.

If they are moving with the same linear momentum, then using the de Broglie relation

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

We find that both the bodies will have the same wavelength. Hence, option (b) is incorrect.

If *K* is the kinetic energy of both the particles, then de Broglie wavelength of the heavier particle,

${\lambda}_{1}=\frac{h}{\sqrt{2{m}_{1}K}}$

de Broglie wavelength of the lighter particle,

${\lambda}_{2}=\frac{h}{\sqrt{2{m}_{2}K}}$

It is clear from the above equation that if ${m}_{1}>{m}_{2}$, then ${\lambda}_{1}<{\lambda}_{2}$.

Hence, option (c) is correct.

When they have fallen through the same height *h*, then velocity of both the bodies,

*v* = $\sqrt{2gh}$

Now,

${\lambda}_{1}=\frac{h}{{m}_{1}\sqrt{2gh}}$

${\lambda}_{2}=\frac{h}{{m}_{2}\sqrt{2gh}}$

*m*_{1}>*m*_{2}

$\therefore $ ${\lambda}_{1}<{\lambda}_{2}$

Hence, option (d) is correct.

#### Page No 365:

#### Question 1:

Visible light has wavelengths in the range of 400 nm to 780 nm. Calculate the range of energy of the photons of visible light.

#### Answer:

Given:

Range of wavelengths, ${\lambda}_{1}$ = 400 nm to ${\lambda}_{2}$ = 780 nm

Planck's constant, *h* = 6.63$\times $10^{$-$34} Js

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

Energy of photon,

$E=hv$

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

$\therefore E=h\nu =\frac{hc}{\lambda}$

Energy *$\left({E}_{1}\right)$* of a photon of wavelength $\left({\lambda}_{1}\right)$:

$\phantom{\rule{0ex}{0ex}}{E}_{1}=\frac{hc}{{\lambda}_{1}}\phantom{\rule{0ex}{0ex}}=\frac{6.63\times {10}^{-34}\times 3\times {10}^{8}}{400\times {10}^{-9}}\phantom{\rule{0ex}{0ex}}=\frac{6.63\times 3}{4}\times {10}^{-9}\phantom{\rule{0ex}{0ex}}=4.97725\times {10}^{-19}\phantom{\rule{0ex}{0ex}}=5\times {10}^{-19}\mathrm{J}\phantom{\rule{0ex}{0ex}}$

Energy (E_{2}) of a photon of wavelength (${\lambda}_{2}$):

$\phantom{\rule{0ex}{0ex}}{E}_{2}=\frac{6.63\times 3}{7.8}\times {10}^{-19}\phantom{\rule{0ex}{0ex}}=2.55\times {10}^{-9}\mathrm{J}$

So, the range of energy is 2.55 × 10^{−19} J to 5 × 10^{−19} J.

#### Page No 365:

#### Question 2:

Calculate the momentum of a photon of light of wavelength 500 nm.

#### Answer:

Given:

Wavelength of light, $\lambda $ = 500 nm

Planck's constant, *h = *6.63$\times $10^{$-34$} J-s

Momentum of a photon of light,

$p=\frac{h}{\lambda}\phantom{\rule{0ex}{0ex}}=\frac{6.63\times {10}^{-34}}{500\times {10}^{-9}}\phantom{\rule{0ex}{0ex}}=\frac{6.63}{5}\times {10}^{-27}\phantom{\rule{0ex}{0ex}}=1.326\times {10}^{-27}\phantom{\rule{0ex}{0ex}}=1.33\times {10}^{-27}\mathrm{kg}-m/\mathrm{s}$

#### Page No 365:

#### Question 3:

An atom absorbs a photon of wavelength 500 nm and emits another photon of wavelength 700 nm. Find the net energy absorbed by the atom in the process.

#### Answer:

Given:

Wavelength of absorbed photon, ${\lambda}_{1}$ = 500 nm

Wavelength of emitted photon, ${\lambda}_{2}$ = 700 nm

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

Planck's constant, *h* = 6.63$\times $10^{$-34$} Js

Energy of absorbed photon,

${E}_{1}=\frac{hc}{{\lambda}_{1}}=\frac{h\times 3\times {10}^{8}}{500\times {10}^{-9}}$

Energy of emitted photon,

${E}_{2}=\frac{hc}{{\lambda}_{2}}=\frac{h\times 3\times {10}^{8}}{700\times {10}^{-9}}$

Energy absorbed by the atom in the process:

${E}_{1}-{E}_{2}=hc\left[\frac{1}{{\lambda}_{1}}-\frac{1}{{\lambda}_{2}}\right]\phantom{\rule{0ex}{0ex}}=6.63\times 3\left[\frac{1}{5}-\frac{1}{7}\right]\times {10}^{-19}\phantom{\rule{0ex}{0ex}}=6.63\times 3\times \frac{2}{35}\times {10}^{-19}\phantom{\rule{0ex}{0ex}}=1.136\times {10}^{-19}\mathrm{J}$

#### Page No 365:

#### Question 4:

Calculate the number of photons emitted per second by a 10 W sodium vapour lamp. Assume that 60% of the consumed energy is converted into light. Wavelength of sodium light = 590 nm

#### Answer:

Given:

Power of the* *sodium vapour lamp,* P* = 10 W

Wavelength of sodium light, $\lambda $ = 590 nm

Electric energy consumed by the bulb in one second = 10 J

Amount of energy converted into light = 60 %

∴ Energy converted into light = $\frac{60}{100}\times 10=6\mathrm{J}$

Energy needed to emit a photon from the sodium atom,

$E\text{'}=\frac{hc}{\lambda}\phantom{\rule{0ex}{0ex}}E\text{'}=\frac{6.63\times {10}^{-34}\times 3\times {10}^{8}}{590\times {10}^{-9}}\phantom{\rule{0ex}{0ex}}E\text{'}=\frac{6.63\times 3}{590}\times {10}^{-17}\mathrm{J}$

Number of photons emitted,

$n=\frac{6}{{\displaystyle \frac{6.63\times 3}{590}}\times {10}^{-17}}\phantom{\rule{0ex}{0ex}}n=\frac{6\times 590}{6.63\times 3}\times {10}^{17}$

*n* = ^{19}

#### Page No 365:

#### Question 5:

When the sun is directly overhead, the surface of the earth receives 1.4 × 10^{3} W m^{−2} of sunlight. Assume that the light is monochromatic with average wavelength 500 nm and that no light is absorbed in between the sun and the earth's surface. The distance between the sun and the earth is 1.5 × 10^{11} m. (a) Calculate the number of photons falling per second on each square metre of earth's surface directly below the sun. (b) How many photons are there in each cubic metre near the earth's surface at any instant? (c) How many photons does the sun emit per second?

#### Answer:

Here,

Intensity of light, *I *= 1.4 × 10^{3} W/m^{2},

Wavelength of light, $\lambda $ = 500 nm = 500$\times $10^{$-9$} m

Distance between the Sun and Earth, *l* = 1.5$\times $10^{11} m

Intensity,

$I=\frac{\mathrm{Power}}{\mathrm{Area}}=1.4\times {10}^{3}\mathrm{W}/{\mathrm{m}}^{2}$

Let *n* be the number of photons emitted per second.

∴ Power, *P* = Energy emitted/second

$P=\frac{nhc}{\lambda}$,

where $\lambda $ = wavelength of light

*h* = Planck's constant

c = speed of light

Number of photons/m^{2} = $\frac{nhc}{\lambda \times A}=\frac{nhc}{\lambda \times 1}$ = *I*

$\therefore n=\frac{I\times \lambda}{hc}\phantom{\rule{0ex}{0ex}}=\frac{1.4\times {10}^{3}\times 500\times {10}^{-9}}{6.63\times {10}^{-34}\times 3\times {10}^{8}}\phantom{\rule{0ex}{0ex}}=3.5\times {10}^{21}$

(b) Consider number of two parts at a distance *r* and *r* + *dr* from the source.

Let *dt*' be the time interval* *in which the photon travels from one part to another*.*

Total number of photons emitted in this time interval,

$N=ndt=\left(\frac{P\lambda}{hc\times A}\right)\frac{dr}{\mathrm{c}}$

These points will be between two spherical shells of radius '*r*' and *r* + *dr*. It will be the distance of the 1st point from the sources.

In this case,

$l=1.5\times {10}^{11}\mathrm{m}\phantom{\rule{0ex}{0ex}}\mathrm{Wavelength},\lambda =500\mathrm{nm}=500\times {10}^{-9}\mathrm{m}\phantom{\rule{0ex}{0ex}}\frac{P}{4\mathrm{\pi}{r}^{2}}=1.4\times {10}^{3}\phantom{\rule{0ex}{0ex}}\therefore \mathrm{No}.\mathrm{of}\mathrm{photons}/{\mathrm{m}}^{3}=\frac{P}{4\mathrm{\pi}{r}^{2}}\frac{\lambda}{h{c}^{2}}\phantom{\rule{0ex}{0ex}}=1.4\times {10}^{3}\times \frac{500\times {10}^{-9}}{6.63\times {10}^{-34}\times 9\times {10}^{16}}\phantom{\rule{0ex}{0ex}}=1.2\times {10}^{13}$

(c) Number of photons emitted = (Number of photons / s-m^{2}) × Area

$=(3.5\times {10}^{21})\times 4\pi {l}^{2}\phantom{\rule{0ex}{0ex}}=3.5\times {10}^{21}\times 4\times (3.14)\times (1.5\times {10}^{11}{)}^{2}\phantom{\rule{0ex}{0ex}}=9.9\times {10}^{44}$

#### Page No 365:

#### Question 6:

A parallel beam of monochromatic light of wavelength 663 nm is incident on a totally reflecting plane mirror. The angle of incidence is 60° and the number of photons striking the mirror per second is 1.0 × 10^{19}. Calculate the force exerted by the light beam on the mirror.

#### Answer:

Here,

$\mathrm{Wavelength}\mathrm{of}\mathrm{monochromatic}\mathrm{light},\lambda =663\times {10}^{-9}\mathrm{m}\phantom{\rule{0ex}{0ex}}\mathrm{Angle}\mathrm{of}\mathrm{incidence},\mathrm{\theta}=60\xb0\phantom{\rule{0ex}{0ex}}\mathrm{Number}\mathrm{of}\mathrm{photons}\mathrm{per}\mathrm{second},n=1\times {10}^{19}\phantom{\rule{0ex}{0ex}}\mathrm{Momentum}\mathrm{of}\mathrm{photon},\phantom{\rule{0ex}{0ex}}p=\frac{h}{\lambda},\phantom{\rule{0ex}{0ex}}\text{w}\mathrm{here}h\mathrm{is}\text{P}\mathrm{lanck}\text{'}s\mathrm{constant}.\phantom{\rule{0ex}{0ex}}p=\frac{6.63\times {10}^{-34}}{663\times {10}^{-9}}={10}^{-27}$

Force exerted on the wall,

$F=n\times \left[p\mathrm{cos\theta}-(-p\mathrm{cos\theta})\right]=2np\mathrm{cos\theta}\phantom{\rule{0ex}{0ex}}=2\times 1\times {10}^{19}\times {10}^{-27}\times \frac{1}{2}\phantom{\rule{0ex}{0ex}}=1.0\times {10}^{-8}\mathrm{N}$

#### Page No 365:

#### Question 7:

A beam of white light is incident normally on a plane surface absorbing 70% of the light and reflecting the rest. If the incident beam carries 10 W of power, find the force exerted by it on the surface.

#### Answer:

Power of the incident beam,* P *= 10 watt

Relation between wavelength $\left(\lambda \right)$ and momentum (*p*):

$\lambda =\frac{h}{p},\phantom{\rule{0ex}{0ex}}\text{w}\mathrm{here}h\mathrm{is}\text{P}\mathrm{lanck}\text{'}\text{s}\mathrm{constant}\phantom{\rule{0ex}{0ex}}\Rightarrow p=\frac{h}{\lambda}\phantom{\rule{0ex}{0ex}}\mathrm{On}\mathrm{dividing}\mathrm{both}\mathrm{side}\text{s}\mathrm{by}t,\mathrm{we}\mathrm{get}:\phantom{\rule{0ex}{0ex}}\frac{p}{t}=\frac{h}{\lambda t}...\left(1\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Energy,

$E=\frac{hc}{\lambda}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{E}{t}=\frac{hc}{\lambda t}\phantom{\rule{0ex}{0ex}}$

Let *P* be the power. Then,

$P=\frac{E}{t}=\frac{hc}{\lambda t}\phantom{\rule{0ex}{0ex}}P=\frac{pc}{t}[\text{U}\mathrm{sing}\mathrm{equation}\left(1\right)]\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{P}{c}=\frac{p}{t}\phantom{\rule{0ex}{0ex}}\text{F}\mathrm{orce},\phantom{\rule{0ex}{0ex}}F=\frac{p}{t}=\frac{P}{c}\left(\mathrm{Since}F=\frac{\text{M}\mathrm{omentum}}{\text{T}\mathrm{ime}}\right)\phantom{\rule{0ex}{0ex}}\mathrm{Force},F=\frac{7}{10}\left(\mathrm{absorbed}\right)+2\times \frac{3}{10}\left(\mathrm{reflected}\right)\phantom{\rule{0ex}{0ex}}F=\frac{7}{10}\times \frac{P}{c}+2\times \frac{3}{10}\times \frac{P}{c}\phantom{\rule{0ex}{0ex}}F=\frac{7}{10}\times \frac{10}{3\times {10}^{8}}+2\times \frac{3}{10}\times \frac{10}{3\times {10}^{8}}\phantom{\rule{0ex}{0ex}}F=\frac{13}{3}\times {10}^{-8}=4.33\times {10}^{-8}\mathrm{N}$

#### Page No 365:

#### Question 8:

A totally reflecting, small plane mirror placed horizontally faces a parallel beam of light, as shown in the figure. The mass of the mirror is 20 g. Assume that there is no absorption in the lens and that 30% of the light emitted by the source goes through the lens. Find the power of the source needed to support the weight of the mirror.

Figure

#### Answer:

Given:

Mass of the mirror, *m* = 20 g = 20 × 10^{−3} kg

The weight of the mirror will be balanced if the force exerted by the photons will be equal to the weight of the mirror.

Now,

Relation between wavelength $\left(\lambda \right)$ and momentum (*p*):

$p=\frac{h}{\lambda}\phantom{\rule{0ex}{0ex}}\mathrm{On}\mathrm{dividing}\mathrm{both}\mathrm{side}\text{s}\mathrm{by}t,\mathrm{we}\mathrm{get}:\phantom{\rule{0ex}{0ex}}\frac{P}{t}=\frac{h}{\lambda t}...\left(1\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Energy,

$E=\frac{hc}{\lambda}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{E}{t}=\frac{hc}{\lambda t}\phantom{\rule{0ex}{0ex}}$

Let *P* be the power. Then,

$P=\frac{\mathrm{E}}{t}=\frac{hc}{\lambda t}\phantom{\rule{0ex}{0ex}}P=\frac{pc}{t}[\text{U}\mathrm{sing}\mathrm{equation}\left(1\right)]\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{P}{c}=\frac{p}{t}\phantom{\rule{0ex}{0ex}}\text{F}\mathrm{orce},\phantom{\rule{0ex}{0ex}}F=\frac{P}{t}=\frac{P}{c}\left(\mathrm{Since}F=\frac{\text{M}\mathrm{omentum}}{\text{T}\mathrm{ime}}\right)\phantom{\rule{0ex}{0ex}}$

Thus, rate of change of momentum = Power/*c*

As the light gets reflected normally,

Force exerted = 2 (Rate of change of momentum) = 2 × Power/*c
$30\%\mathrm{of}\left(\frac{2\times \mathrm{Power}}{c}\right)=mg\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{Power}=\frac{20\times {10}^{-3}\times 10\times 3\times {10}^{8}\times 10}{2\times 3}\phantom{\rule{0ex}{0ex}}=100\mathrm{MW}$*

#### Page No 365:

#### Question 9:

A 100 W light bulb is placed at the centre of a spherical chamber of radius 20 cm. Assume that 60% of the energy supplied to the bulb is converted into light and that the surface of the chamber is perfectly absorbing. Find the pressure exerted by the light on the surface of the chamber.

#### Answer:

Given:

Power of the light bulb*, P** *= 100 W

Radius of the spherical chamber, *R* = 20 cm = 0.2 m

It is given that 60% of the energy supplied to the bulb is converted to light.

Therefore, power of light emitted by the bulb, *P'* = 60 W

Force,

$F=\frac{P}{c}$,

where *c* is the speed of light

$\phantom{\rule{0ex}{0ex}}F=\frac{60}{3\times {10}^{8}}=2\times {10}^{-7}\mathrm{N}\phantom{\rule{0ex}{0ex}}$

Pressure = $\frac{\mathrm{Force}}{\mathrm{Area}}$

$\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{2\times {10}^{-7}}{4\times 3.14\times (0.2{)}^{2}}\left(A=4\mathrm{\pi}{r}^{2}\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{8\times 3.14}\times {10}^{-5}\phantom{\rule{0ex}{0ex}}=0.039\times {10}^{-5}\phantom{\rule{0ex}{0ex}}=3.9\times {10}^{-7}\phantom{\rule{0ex}{0ex}}=4\times {10}^{-7}\mathrm{N}/{\mathrm{m}}^{2}$

#### Page No 365:

#### Question 10:

A sphere of radius 1.00 cm is placed in the path of a parallel beam of light of large aperture. The intensity of the light is 0.5 W cm^{−2}. If the sphere completely absorbs the radiation falling on it, find the force exerted by the light beam on the sphere.

#### Answer:

Given:

Radius of the sphere, *r = *1 cm

Intensity of light, *I* = 0.5 Wcm^{−2}

Let *A* be the effective area of the sphere perpendicular to the light beam.

So, force exerted by the light beam on the sphere is given by,

$F=\frac{P}{c}=\frac{AI}{c}$

$F=\frac{\mathrm{\pi}\times (1{)}^{2}\times 0.5}{3\times {10}^{8}}\phantom{\rule{0ex}{0ex}}=\frac{3.14\times 0.5}{3\times {10}^{8}}\phantom{\rule{0ex}{0ex}}=0.523\times {10}^{-8}\phantom{\rule{0ex}{0ex}}=5.2\times {10}^{-9}\mathrm{N}$

#### Page No 365:

#### Question 11:

Consider the situation described in the previous problem. Show that the force on the sphere due to the light falling on it is the same even if the sphere is not perfectly absorbing.

#### Answer:

Consider a sphere of centre O and radius OP. As shown in the figure, the radius OP of the sphere is making an angle *θ* with OZ. Let us rotate the radius about OZ to get another circle on the sphere. The part of the sphere between the circle is a ring of area $2\mathrm{\pi}{r}^{2}\mathrm{sin}\theta d\theta $.

Consider a small part of area $\u2206A$ of the ring at point P.

Energy of the light falling on this part in time $\u2206t$,

$\u2206U=I\u2206t\left(\u2206A\mathrm{cos}\theta \right)$

As the light is reflected by the sphere along PR, the change in momentum,

$\u2206p=2\frac{\u2206U}{c}\mathrm{cos}\theta =\frac{2}{c}I\u2206t\left(\u2206A{\mathrm{cos}}^{2}\theta \right)$

Therefore, the force will be

$\frac{\u2206p}{\u2206t}=\frac{2}{c}I\u2206A{\mathrm{cos}}^{2}\theta $

The component of force on $\u2206A$, along ZO, is

$\frac{\u2206p}{\u2206t}\mathrm{cos}\theta =\frac{2}{c}I\u2206A{\mathrm{cos}}^{3}\theta $

Now, force action on the ring,

$dF=\frac{2}{c}I\left(2\mathrm{\pi}{r}^{2}\mathrm{sin}\theta d\theta \right){\mathrm{cos}}^{3}\theta $

The force on the entire sphere,

$F={\int}_{0}^{\frac{\mathrm{\pi}}{2}}\frac{4\mathrm{\pi}{r}^{2}I}{c}{\mathrm{cos}}^{3}\theta \mathrm{sin}\theta d\theta \phantom{\rule{0ex}{0ex}}=-{\int}_{0}^{\frac{\mathrm{\pi}}{2}}\frac{4\mathrm{\pi}{r}^{2}I}{c}{\mathrm{cos}}^{3}\theta d\left(\mathrm{cos}\theta \right)\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{\pi}{r}^{2}I}{c}$

This is same as given in the previous problem.

#### Page No 365:

#### Question 12:

Show that it is not possible for a photon to be completely absorbed by a free electron.

#### Answer:

When an electron undergoes an inelastic collision with a photon, we can apply the principle of conservation of energy to this collision. So,

$pc+{m}_{e}{c}^{2}=\sqrt{{p}^{2}{c}^{2}+{{m}_{e}}^{2}{c}^{4}}$ ...(i)

Here, *h =* Planck's constant

* c* = the speed of light

*m*_{e }= rest mass of electron

*pc* = energy of the photon

Squaring on both side of equation (i),

$(pc+{m}_{e}{c}^{2}{)}^{2}={p}^{2}{c}^{2}+{{m}_{e}}^{2}{c}^{4}\phantom{\rule{0ex}{0ex}}\Rightarrow {p}^{2}{c}^{2}+{{m}_{e}}^{2}{c}^{4}+2(pc\left)\right({m}_{e}{c}^{2})={p}^{2}{c}^{2}+{{m}_{e}}^{2}{c}^{4}\phantom{\rule{0ex}{0ex}}\Rightarrow 2(pc\left)\right({m}_{e}{c}^{2})=0\mathrm{or}pc=0(\mathrm{as}\mathit{}\mathit{\text{'}}m\mathit{\text{'}}\mathrm{and}\mathit{}\mathit{\text{'}}c\mathit{\text{'}}\mathit{}\mathrm{are}\mathrm{non}\mathrm{zero})$

This gives vanishing energy of photon which is not possible.

#### Page No 365:

#### Question 13:

Two neutral particles are kept 1 m apart. Suppose by some mechanism some charge is transferred from one particle to the other and the electric potential energy lost is completely converted into a photon. Calculate the longest and the next smaller wavelength of the photon possible.

#### Answer:

Given:

Distance between the two neutral particles, *r* = 1 m

Electric potential energy,

*E _{1}*= $\frac{k{q}^{2}}{r}$ = $k{q}^{2}$,

where

*k*= $\frac{1}{4{\mathrm{\pi \epsilon}}_{0}}$

Energy of photon,

*E*

_{2}= $\frac{hc}{\lambda}$,

where $\lambda $ = wavelength of light

*h*= Planck's constant

c = speed of light

Here,

*E*

_{1}=

*E*

_{2}

$\therefore k{q}^{2}=\frac{hc}{\lambda}\phantom{\rule{0ex}{0ex}}\Rightarrow \lambda =\frac{hc}{k{q}^{2}}$

For wavelength, λ, to be maximum, charge

*q*should be minimum.

$q=e=1.6\times {10}^{-19}\mathrm{C}\phantom{\rule{0ex}{0ex}}\mathrm{Maximum}\mathrm{wavelength},\phantom{\rule{0ex}{0ex}}\lambda =\frac{hc}{k{q}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{6.63\times 3\times {10}^{-34}\times {10}^{8}}{9\times {10}^{2}\times (1.6{)}^{2}\times {10}^{-38}}\phantom{\rule{0ex}{0ex}}=0.863\times {10}^{3}=863\mathrm{m}$

Next smaller wave length,

$\lambda =\frac{6.63\times 3\times {10}^{-34}\times {10}^{8}}{9\times {10}^{4}\times 4\times (1.6{)}^{2}\times {10}^{-38}}\phantom{\rule{0ex}{0ex}}=\frac{863}{4}\phantom{\rule{0ex}{0ex}}=215.74\mathrm{m}$

#### Page No 365:

#### Question 14:

Find the maximum kinetic energy of the photoelectrons ejected when light of wavelength 350 nm is incident on a cesium surface. Work function of cesium = 1.9 eV

#### Answer:

Given:

Wavelength of light, *λ *= 350 nm = 350 × 10^{−9} m

Work-function of cesium, *ϕ* = 1.9 *e*V

From Einstein's photoelectric equation,

$E=\varphi +\mathrm{Kinetic}\mathrm{energy}\mathrm{of}\mathrm{electron}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{K}\mathrm{E}=E-\varphi \phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{K}\mathrm{E}=\frac{hc}{\lambda}-\varphi ,\phantom{\rule{0ex}{0ex}}\text{w}\mathrm{here}\lambda =\mathrm{wavelength}\mathrm{of}\mathrm{light}\phantom{\rule{0ex}{0ex}}h=\text{P}\mathrm{lan}\text{ck}\text{'}\mathrm{s}\mathrm{constant}$

Maximum kinetic energy of electrons,

${E}_{\mathrm{max}}=\frac{hc}{\lambda}-\varphi \phantom{\rule{0ex}{0ex}}{E}_{\mathrm{max}}=\frac{6.63\times {10}^{-34}\times 3\times {10}^{8}}{350\times {10}^{-9}\times 1.6\times {10}^{-19}}-1.9\phantom{\rule{0ex}{0ex}}{E}_{\mathrm{max}}=\frac{6.63\times 3\times {10}^{2}}{350\times 1.6}-1.9\phantom{\rule{0ex}{0ex}}{E}_{\mathrm{max}}=1.65\mathrm{eV}=1.6\mathrm{eV}$

#### Page No 365:

#### Question 15:

The work function of a metal is 2.5 × 10^{−19} J. (a) Find the threshold frequency for photoelectric emission. (b) If the metal is exposed to a light beam of frequency 6.0 × 10^{14} Hz, what will be the stopping potential?

#### Answer:

Given:

Work function of a metal, *W _{0}* = 2.5 × 10

^{−19}J

Frequency of light beam,

*v =*6.0 × 10

^{14}Hz

(a) Work function of a metal,

*W*

_{0}=

*hv*

_{0},

where

*h*= Planck's constant

*v*

_{0}

_{ = }threshold frequency

$\therefore $

*v*

_{0}

_{ }= $\frac{{W}_{0}}{h}$

$\Rightarrow {v}_{0}=\frac{2.5\times {10}^{-19}}{6.63\times {10}^{-34}}\phantom{\rule{0ex}{0ex}}=3.77\times {10}^{14}\mathrm{Hz}\phantom{\rule{0ex}{0ex}}=3.8\times {10}^{14}\mathrm{Hz}\phantom{\rule{0ex}{0ex}}$

(b) Einstein's photoelectric equation:

$\phantom{\rule{0ex}{0ex}}e{V}_{0}=hv-{W}_{0}\phantom{\rule{0ex}{0ex}}$,

where

*v*= frequency of light

*V*

_{0}

_{ = }Stopping potential

*e*= charge on electron

$\therefore {V}_{0}=\frac{hv-{\mathrm{W}}_{0}}{e}\phantom{\rule{0ex}{0ex}}=\frac{6.63\times {10}^{-34}\times 6\times {10}^{14}-2.5\times {10}^{-19}}{1.6\times {10}^{-19}}\phantom{\rule{0ex}{0ex}}=\frac{3.97\times {10}^{-19}-2.5\times {10}^{-19}}{1.6\times {10}^{-19}}=0.91\mathrm{V}$

#### Page No 365:

#### Question 16:

The work function of a photoelectric material is 4.0 eV. (a) What is the threshold wavelength? (b) Find the wavelength of light for which the stopping potential is 2.5 V.

#### Answer:

Work function of a photoelectric material, *ϕ* = 4 *e*V = 4 × 1.6 × 10^{−19} J

Stopping potential, *V*_{0} = 2.5 V

Planck's constant, *h* = 6.63$\times {10}^{-34}\mathrm{Js}$

(a) Work function of a photoelectric material,

$\varphi =\frac{hc}{{\lambda}_{0}}$

Here, *λ*_{0} = threshold wavelength of light

*c* = speed of light

$\therefore {\lambda}_{0}=\frac{hc}{\varphi}\phantom{\rule{0ex}{0ex}}{\lambda}_{0}=\frac{6.63\times {10}^{-34}\times 3\times {10}^{8}}{4\times 1.6\times {10}^{-19}}\phantom{\rule{0ex}{0ex}}{\lambda}_{0}=\frac{6.63\times 3}{64}\times \frac{{10}^{-27}}{{10}^{-9}}\phantom{\rule{0ex}{0ex}}{\lambda}_{0}=3.1\times {10}^{-7}\mathrm{m}\phantom{\rule{0ex}{0ex}}{\mathrm{\lambda}}_{0}=310\mathrm{nm}$

(b) From Einstein's photoelectric equation,

$E=\varphi +e{V}_{0}\phantom{\rule{0ex}{0ex}}\mathrm{On}\mathrm{substituting}\mathrm{the}\mathrm{respective}\mathrm{values},\mathrm{we}\text{get:}\phantom{\rule{0ex}{0ex}}\frac{hc}{\lambda}=4\times 1.6\times {10}^{-19}+1.6\times {10}^{-19}\times 2.5\phantom{\rule{0ex}{0ex}}\Rightarrow \lambda =\frac{6.63\times {10}^{-34}\times 3\times {10}^{8}}{6.5\times 1.6\times {10}^{-19}}\phantom{\rule{0ex}{0ex}}\Rightarrow \lambda =\frac{6.63\times 3\times {10}^{-26}}{1.6\times {10}^{-19}\times 6.5}\phantom{\rule{0ex}{0ex}}\Rightarrow \lambda =1.9125\times {10}^{-7}=191\mathrm{nm}$

#### Page No 365:

#### Question 17:

Find the maximum magnitude of the linear momentum of a photoelectron emitted when a wavelength of 400 nm falls on a metal with work function 2.5 eV.

#### Answer:

Given:

Wavelength of light, $\lambda $ = 400 nm = 400$\times $10^{$-9$} m

Work function of metal, $\varphi $ = 2.5 eV

From Einstein's photoelectric equation,

$\mathrm{Kinetic}\mathrm{energy}=\frac{h\mathrm{c}}{\lambda}-\varphi \phantom{\rule{0ex}{0ex}}$

Here, c = speed of light

*h* = Planck's constant

$\therefore \mathrm{K}.\mathrm{E}.=\frac{6.63\times {10}^{-34}\times 3\times {10}^{8}}{4\times {10}^{-7}\times 1.6\times {10}^{-19}}-2.5\mathrm{eV}\phantom{\rule{0ex}{0ex}}=0.605\mathrm{eV}$

Also, K.E. = $\frac{{p}^{2}}{2m}$,

where *p* is momentum and *m* is the mass of an electron.

$\therefore {p}^{2}=2m\times \mathrm{K}.\mathrm{E}.\phantom{\rule{0ex}{0ex}}\Rightarrow {p}^{2}=2\times 9.1\times {10}^{-31}\times 0.605\times 1.6\times {10}^{-19}\phantom{\rule{0ex}{0ex}}\Rightarrow p=4.197\times {10}^{-25}\mathrm{kg}-\mathrm{m}/\mathrm{s}$

#### Page No 365:

#### Question 18:

When a metal plate is exposed to a monochromatic beam of light of wavelength 400 nm, a negative potential of 1.1 V is needed to stop the photo current. Find the threshold wavelength for the metal.

#### Answer:

Given:

Wavelength of light, $\lambda $ = 400 nm = 400$\times $10^{$-9$} m

Stopping potential,* **V*_{0} = 1.1 V

From Einstein's photoelectric equation,

$\frac{hc}{\lambda}=\frac{hc}{{\lambda}_{0}}+e{V}_{0}$,

where* h* = Planck's constant

c= speed of light

$\lambda $ = wavelength of light

${\lambda}_{0}$ = threshold wavelength

${V}_{0}$ = stopping potential

On substituting the respective values in the above formula, we get:

$\frac{6.63\times {10}^{-34}\times 3\times {10}^{8}}{400\times {10}^{-9}}=\frac{6.63\times {10}^{-34}\times 3\times {10}^{8}}{{\lambda}_{0}}+1.6\times {10}^{-19}\times 1.1\phantom{\rule{0ex}{0ex}}\Rightarrow 4.97\times {10}^{-19}=\frac{19.89\times {10}^{-26}}{{\lambda}_{0}}+1.76\times {10}^{-19}\phantom{\rule{0ex}{0ex}}\Rightarrow 4.97=\frac{19.89\times {10}^{-7}}{{\lambda}_{0}}+1.76\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{19.89\times {10}^{-7}}{{\lambda}_{0}}=4.97-1.76=3.21\phantom{\rule{0ex}{0ex}}\Rightarrow {\lambda}_{0}=\frac{19.89\times {10}^{-7}}{3.21}\phantom{\rule{0ex}{0ex}}=6.196\times {10}^{-7}\mathrm{m}=620\mathrm{nm}$

#### Page No 365:

#### Question 19:

In an experiment on photoelectric effect, the stopping potential is measured for monochromatic light beams corresponding to different wavelengths. The data collected are as follows:

Wavelength (nm): 350 400 450 500 550

Stopping potential (V): 1.45 1.00 0.66 0.38 0.16

Plot the stopping potential against inverse of wavelength (1/λ) on a graph paper and find (a) Planck's constant (b) the work function of the emitter and (c) the threshold wavelength.

#### Answer:

(a)

When *λ* = 350, *V*_{s} = 1.45

and when *λ* = 400, *V*_{s} = 1

$\therefore \frac{hc}{350}=w+1.45...\left(1\right)\phantom{\rule{0ex}{0ex}}\mathrm{and}\frac{hc}{400}=w+1...\left(2\right)$

Subtracting (2) from (1) and solving to get the value of *h*, we get:

*h* = 4.2 × 10^{−15} eV-s

(b) Now, work function,

$w=\frac{12240}{350}-1.45=2.15e\mathrm{v}\phantom{\rule{0ex}{0ex}}\left(c\right)w=\frac{nc}{\lambda}\phantom{\rule{0ex}{0ex}}\Rightarrow {\lambda}_{\mathrm{threshold}}=\frac{hc}{w}\phantom{\rule{0ex}{0ex}}\Rightarrow {\lambda}_{\mathrm{threshold}}=\frac{1240}{1.15}\phantom{\rule{0ex}{0ex}}\Rightarrow {\lambda}_{\mathrm{threshold}}=576.8\mathrm{nm}$

#### Page No 365:

#### Question 20:

The electric field associated with a monochromatic beam is 1.2 × 10^{15} times per second. Find the maximum kinetic energy of the photoelectrons when this light falls on a metal surface whose work function is 2.0 eV.

#### Answer:

Given:

Electric field of the monochromatic beam, *E* = 1.2 × 10^{15} times per second

Frequency, *v* = $\frac{1.2\times {10}^{15}}{2}=0.6\times {10}^{15}\mathrm{Hz}$

Work function of the metal surface, $\varphi $ = 2.0 eV

From Einstein's photoelectric equation, kinetic energy,

$K=hv-{\mathrm{\varphi}}_{0}\phantom{\rule{0ex}{0ex}}\Rightarrow K=\frac{6.63\times {10}^{-34}\times 0.6\times {10}^{15}}{1.6\times {10}^{-19}}-2\phantom{\rule{0ex}{0ex}}=0.486\mathrm{eV}$

Thus, the maximum kinetic energy of a photon is 0.486 eV.

#### Page No 365:

#### Question 21:

The electric field associated with a light wave is given by

$E={E}_{0}\mathrm{sin}\left[\right(1.57\times {10}^{7}{\mathrm{m}}^{-1}\left)\right(x-ct\left)\right].$

Find the stopping potential when this light is used in an experiment on photoelectric effect with the emitter having work function 1.9 eV.

#### Answer:

Given:

$\mathrm{Electric}\mathrm{field},E={E}_{0}\mathrm{sin}\left[\right(1.57\times {10}^{7}{\mathrm{m}}^{-1}\left)\right(x-ct\left)\right]$

Work function, $\varphi $ = 1.9 eV

On comparing the given equation with the standard equation, $E={E}_{0}\mathrm{sin}\left(kx-wt\right)$, we get:

$\omega =1.57\times {10}^{7}\times c\phantom{\rule{0ex}{0ex}}$

Now, frequency,

$v=\frac{1.57\times {10}^{7}\times 3\times {10}^{8}}{2\mathrm{\pi}}\mathrm{Hz}\phantom{\rule{0ex}{0ex}}$

From Einstein's photoelectric equation,

$e{\mathrm{V}}_{0}=hv-\varphi \phantom{\rule{0ex}{0ex}}$

Here,* **V*_{0} = stopping potential

* e* = charge on electron

* h *= Planck's constant

On substituting the respective values, we get:

$e{V}_{0}=6.63\times {10}^{-34}\times \frac{1.57\times 3\times {10}^{15}}{2\mathrm{\pi}\times 1.6\times {10}^{-19}}-1.9\mathrm{eV}\phantom{\rule{0ex}{0ex}}\Rightarrow e{V}_{0}=3.105-1.9=1.205\mathrm{eV}\phantom{\rule{0ex}{0ex}}\Rightarrow {V}_{0}=\frac{1.205\times 1.6\times {10}^{-19}}{1.6\times {10}^{-19}}=1.205\mathrm{V}$

Thus, the value of the stopping potential is 1.205 V.

#### Page No 366:

#### Question 22:

The electric field at a point associated with a light wave is

$E=\left(100{\mathrm{Vm}}^{-1}\right)\mathrm{sin}\left[\right(3.0\times {10}^{15}{s}^{-1}\left)t\right]\mathrm{sin}\left[\right(6.0\times {10}^{15}{s}^{-1}\left)t\right].$

If this light falls on a metal surface with a work function of 2.0 eV, what will be the maximum kinetic energy of the photoelectrons?

#### Answer:

Given:

$E=100\mathrm{sin}\left[\right(3\times {10}^{-15}{s}^{-1}\left)t\right]\mathrm{sin}\left[\right(6\times {10}^{-15}{s}^{-1}\left)t\right]\phantom{\rule{0ex}{0ex}}=100\times \frac{1}{2}\mathrm{cos}\left[\right(9\times {10}^{15}{s}^{-1}\left)t\right]-\mathrm{cos}\left[\right(3\times {10}^{15}{s}^{-1}\left)t\right]\phantom{\rule{0ex}{0ex}}$

The values of angular frequency $\omega $ are 9 × 10^{15} and 3 × 10^{15} .

Work function of the metal surface, $\varphi $ = 2 eV

Maximum frequency,

$v=\frac{{\omega}_{max}}{2\mathrm{\pi}}$=$\frac{9\times {10}^{15}}{2\mathrm{\pi}}$ Hz

From Einstein's photoelectric equation, kinetic energy,

*K* = *hv* $-$ $\varphi $

$\Rightarrow $*K = *$6.63\times {10}^{-34}\times \frac{9\times {10}^{15}}{2\mathrm{\pi}}\times \frac{1}{1.6\times {10}^{-19}}$ $-$ 2 eV

$\Rightarrow $K = 3.938 eV

#### Page No 366:

#### Question 23:

A monochromatic light source of intensity 5 mW emits 8 × 10^{15} photons per second. This light ejects photoelectrons from a metal surface. The stopping potential for this setup is 2.0 V. Calculate the work function of the metal.

#### Answer:

Given:

Intensity of light, *I* = 5 mW

Number of photons emitted per second,* n* = 8 × 10^{15}

Stopping potential, *V*_{0} = 2 V

Energy, *E* = $hv$ = $\frac{I}{n}=\frac{5\times {10}^{-3}}{8\times {10}^{15}}$

From Einstein's photoelectric equation, work function,

${W}_{0}=hv-e{V}_{0}$

Here,* h *= Planck's constant

*e* = 1.6$\times $10^{$-19$ }C

On substituting the respective values, we get:

${W}_{0}=\frac{5\times {10}^{-3}}{8\times {10}^{15}}-1.6\times {10}^{-19}\times 2\phantom{\rule{0ex}{0ex}}=6.25\times {10}^{-19}-3.2\times {10}^{-19}\phantom{\rule{0ex}{0ex}}=3.05\times {10}^{-19}\phantom{\rule{0ex}{0ex}}=\frac{3.05\times {10}^{-19}}{1.6\times {10}^{-15}}=1.906\mathrm{eV}$

#### Page No 366:

#### Question 24:

The figure is the plot of stopping potential versus the frequency of the light used in an experiment on photoelectric effect. Find (a) the ratio *h/e* and (b) the work function.

Figure

#### Answer:

We have to take two cases.

$Case\left(\mathrm{I}\right)\phantom{\rule{0ex}{0ex}}\mathrm{When}\mathrm{stopping}\mathrm{potential},{V}_{0}=1.656\mathrm{volts}\phantom{\rule{0ex}{0ex}}\mathrm{Frequency},v=5\times {10}^{14}\mathrm{Hz}\phantom{\rule{0ex}{0ex}}Case\left(\mathrm{II}\right)\mathrm{When}\mathrm{stopping}\mathrm{potential},{V}_{0}=0\phantom{\rule{0ex}{0ex}}\mathrm{Frequency},v=1\times {10}^{14}\mathrm{Hz}\phantom{\rule{0ex}{0ex}}\left(b\right)\mathrm{From}\mathrm{Einstein}\text{'s}\mathrm{equation},\phantom{\rule{0ex}{0ex}}e{V}_{0}=hv-{W}_{0}\phantom{\rule{0ex}{0ex}}\mathrm{On}\mathrm{substituting}\mathrm{the}\mathrm{values}\mathrm{of}\mathrm{case}\left(1\right)\mathrm{and}\mathrm{case}\left(2\right),\mathrm{we}\text{get:}\phantom{\rule{0ex}{0ex}}1.656\mathrm{e}=h\times 5\times {10}^{14}-{W}_{0}...\left(1\right)\phantom{\rule{0ex}{0ex}}0=5\times h\times 1\times {10}^{14}-5\times {W}_{0}...\left(2\right)\phantom{\rule{0ex}{0ex}}\mathrm{Subtracting}\mathrm{equation}\left(2\right)\mathrm{from}\left(1\right),\mathrm{we}\mathrm{get}:\phantom{\rule{0ex}{0ex}}{\mathrm{W}}_{0}=\frac{1.656}{4}\mathrm{eV}\phantom{\rule{0ex}{0ex}}=0.414\mathrm{eV}$

(a) Putting the value of *W*_{0} in equation (2), we get:

$\Rightarrow 5{\mathrm{W}}_{0}=5h\times {10}^{14}\phantom{\rule{0ex}{0ex}}\Rightarrow 5\times 0.414=5\times h\times {10}^{14}\phantom{\rule{0ex}{0ex}}\Rightarrow h=4.414\times {10}^{-15}\mathrm{eVs}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Or}\phantom{\rule{0ex}{0ex}}\frac{h}{e}=4.414\times {10}^{-15}\mathrm{Vs}$

#### Page No 366:

#### Question 25:

A photographic film is coated with a silver bromide layer. When light falls on this film, silver bromide molecules dissociate and the film records the light there. A minimum of 0.6 eV is needed to dissociate a silver bromide molecule. Find the maximum wavelength of light that can be recorded by the film.

#### Answer:

Given:

Work function, *W*_{0} = 0.6 eV

Now, work function,

${W}_{0}=\frac{hc}{\lambda}\phantom{\rule{0ex}{0ex}}$,

where, *h* = Planck's constant

$\lambda $ = wavelength of light

*c* = speed of light

$\therefore \lambda =\frac{hc}{{W}_{0}}\phantom{\rule{0ex}{0ex}}=\frac{6.63\times {10}^{-34}\times 3\times {10}^{8}}{0.6\times 1.6\times {10}^{-19}}\phantom{\rule{0ex}{0ex}}=20.71\times {10}^{-7}\mathrm{m}\phantom{\rule{0ex}{0ex}}=2071\mathrm{nm}$

#### Page No 366:

#### Question 26:

In an experiment on photoelectric effect, light of wavelength 400 nm is incident on a cesium plate at the rate of 5.0 W. The potential of the collector plate is made sufficiently positive with respect to the emitter, so that the current reaches its saturation value. Assuming that on average, one out of every 10^{6} photons is able to eject a photoelectron, find the photocurrent in the circuit.

#### Answer:

Given:

Wavelength of light, *λ *= 400 nm

Power, *P* = 5 W

Energy of photon,

*E* = $\frac{hc}{\lambda}=\left(\frac{1242}{400}\right)\mathrm{eV}$

Number of photons, *n* = $\frac{P}{E}$

$n=\frac{5\times 400}{1.6\times {10}^{-19}\times 1242}$

Number of electrons = 1 electron per 10^{6} photons

Number of photoelectrons emitted,

$n\text{'}=\frac{5\times 400}{1.6\times 1242\times {10}^{-19}\times {10}^{6}}$

Photo electric current,

*I* = Number of electron $\times $ Charge on electron

$I=\frac{5\times 400}{1.6\times 1242\times {10}^{-19}\times {10}^{6}}\times 1.6\times {10}^{-19}\phantom{\rule{0ex}{0ex}}=1.6\times {10}^{-6}\mathrm{A}=1.6\mathrm{\mu A}$

#### Page No 366:

#### Question 27:

A silver ball of radius 4.8 cm is suspended by a thread in a vacuum chamber. Ultraviolet light of wavelength 200 nm is incident on the ball for some time during which light energy of 1.0 × 10^{−7} J falls on the surface. Assuming that on average, one photon out of every ten thousand is able to eject a photoelectron, find the electric potential at the surface of the ball, assuming zero potential at infinity. What is the potential at the centre of the ball?

#### Answer:

Given:

Radius of the silver ball, *r* = 4.8 cm

Wavelength of the ultra violet light, *λ* = 200 nm = 2 × 10^{−7} m

Total energy of light, *E* = 1.0 × 10^{−7} J

We are given that one photon out of every ten thousand is able to eject a photoelectron.

Energy of one photon,

$E\text{'}=\frac{hc}{\lambda}\phantom{\rule{0ex}{0ex}}$ ,

where *h* = Planck's constant

c = speed of light

$\lambda $ = wavelength of light

On substituting the respective values in the above formula, we get:

$E\text{'}=\frac{6.63\times {10}^{-34}\times 3\times {10}^{8}}{2\times {10}^{-7}}\phantom{\rule{0ex}{0ex}}=9.945\times {10}^{-19}$

Number of photons,

$n=\frac{E}{E\text{'}}=\frac{1\times {10}^{-7}}{9.945\times {10}^{-19}}=1\times {10}^{11}$^{}

Number of photoelectrons

$=\frac{1\times {10}^{11}}{{10}^{4}}=1\times {10}^{7}$

The amount of positive charge developed due to the outgoing electrons,

$q=1\times {10}^{7}\times 1.6\times {10}^{-19}\phantom{\rule{0ex}{0ex}}=1.6\times {10}^{-12}\mathrm{C}$

Potential developed at the centre as well as on surface,

$V=\frac{Kq}{r}\phantom{\rule{0ex}{0ex}}$,

where *K* = $\frac{1}{4{\mathrm{\pi \epsilon}}_{0}}$

Potential inside the silver ball remains constant. Therefore, potential at the centre of the sphere is 0.3 V.

#### Page No 366:

#### Question 28:

In an experiment on photoelectric effect, the emitter and the collector plates are placed at a separation of 10 cm and are connected through an ammeter without any cell. A magnetic field *B* exists parallel to the plates. The work function of the emitter is 2.39 eV and the light incident on it has wavelengths between 400 nm and 600 nm. Find the minimum value of B for which the current registered by the ammeter is zero. Neglect any effect of space charge.

Figure

#### Answer:

Given:

Separation between the collector and emitter, *d* = 10 cm

Work function,* ϕ *= 2.39 eV

Wavelength range, *λ*_{1} = 400 nm to* **λ*_{2} = 600 nm

Magnetic field *B* will be minimum if energy is maximum.

For maximum energy, wavelength *λ* should be minimum.

Einstein's photoelectric equation:

$E=\frac{hc}{\lambda}-\varphi \phantom{\rule{0ex}{0ex}}\text{H}\mathrm{ere},h=\mathrm{Planck}\text{'s}\mathrm{constant}\phantom{\rule{0ex}{0ex}}\lambda =\mathrm{Wavelength}\mathrm{of}\mathrm{light}\phantom{\rule{0ex}{0ex}}c=\mathrm{speed}\mathrm{of}\mathrm{light}\phantom{\rule{0ex}{0ex}}\therefore E=\frac{1242}{400}-2.39\phantom{\rule{0ex}{0ex}}=3.105-2.39=0.715\mathrm{eV}$

The beam of ejected electrons will be bent by the magnetic field. If the electrons do not reach the other plates, there will be no current.

When a charged particle is sent perpendicular to a magnetic field, it moves along a circle of radius,

$r=\frac{mv}{qB}\phantom{\rule{0ex}{0ex}}$,

where *m =* mass of charge particle

*B* = magnetic field

*v* = velocity of particle

*q = *charge on the particle

Radius of the circle should be equal to *r = d, *so that no current flows in the circuit.

$\Rightarrow r=\sqrt{\frac{2mE}{q\mathrm{B}}}\left(\because mv=\sqrt{2mE}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 0.1=\sqrt{\frac{2\times 9.1\times {10}^{-31}\times 1.6\times {10}^{-19}\times 0.715}{1.6\times {10}^{-19}\times B}}\phantom{\rule{0ex}{0ex}}\Rightarrow B=\sqrt{\frac{2\times 9.1\times 1.6\times 0.715\times {10}^{-50}}{1.6\times {10}^{-20}}}\phantom{\rule{0ex}{0ex}}\Rightarrow B=\sqrt{\frac{2\times 9.1\times 1.6\times 0.715}{1.6}}\times {10}^{-5}\phantom{\rule{0ex}{0ex}}\Rightarrow B=2.85\times {10}^{-5}\mathrm{T}$

#### Page No 366:

#### Question 29:

In the arrangement shown in the figure, *y* = 1.0 mm, *d* = 0.24 mm and *D* = 1.2 m. The work function of the material of the emitter is 2.2 eV. Find the stopping potential *V* needed to stop the photocurrent.

Figrue]

#### Answer:

Given:

Fringe width, y = 1 mm$\times $2 = 2 mm

Work function, *W*_{0} = 2.2 eV

*D* = 1.2 m

*d* = 0.24 mm

Fringe width,

$y=\frac{\lambda \mathrm{D}}{d}\phantom{\rule{0ex}{0ex}}$ ,

where $\lambda $ = wavelength of light

$\therefore \lambda =\frac{2\times {10}^{-3}\times 0.24\times {10}^{-3}}{1.2}\phantom{\rule{0ex}{0ex}}=4\times {10}^{-7}\mathrm{m}\phantom{\rule{0ex}{0ex}}$

$\mathrm{Energy},E=\frac{hc}{\lambda}\phantom{\rule{0ex}{0ex}}=\frac{4.14\times {10}^{-15}\times 3\times {10}^{8}}{4\times {10}^{-7}}\phantom{\rule{0ex}{0ex}}=3.105\mathrm{eV}$

From Einstein's photoelectric equation,

$e{V}_{0}=E-{W}_{0}$,

where *V*_{0} is the stopping potential and *e *is charge of electron.

$\therefore e{V}_{0}=3.105-2.2=0.905\mathrm{eV}\phantom{\rule{0ex}{0ex}}{V}_{0}=\frac{0.905}{1.6\times {10}^{-19}}\times 1.6\times {10}^{-19}\mathrm{V}\phantom{\rule{0ex}{0ex}}=0.905\mathrm{V}$

#### Page No 366:

#### Question 30:

In a photoelectric experiment, the collector plate is at 2.0 V with respect to the emitter plate made of copper (φ = 4.5 eV). The emitter is illuminated by a source of monochromatic light of wavelength 200 nm. Find the minimum and maximum kinetic energy of the photoelectrons reaching the collector.

#### Answer:

Given:

Work function of copper,* ϕ* = 4.5 eV,

Wavelength of monochromatic light, *λ* = 200 nm

From Einstein's photoelectric equation, kinetic energy,

$K=E-\varphi =\frac{h\mathrm{c}}{\lambda}-\varphi ,\phantom{\rule{0ex}{0ex}}\text{w}\mathrm{here},h=\mathrm{Planck}\text{'s}\mathrm{constant}\phantom{\rule{0ex}{0ex}}\mathrm{c}=\mathrm{speed}\mathrm{of}\mathrm{light}\phantom{\rule{0ex}{0ex}}\therefore K=\frac{1242}{200}-4.5\phantom{\rule{0ex}{0ex}}=6.21-4.5=1.71\mathrm{eV}$

Thus, at least 1.7 eV is required to stop the electron. Therefore, minimum kinetic energy will be 2 eV.

It is given that electric potential of 2 V is required to accelerate the electron. Therefore, maximum kinetic energy

$=(2+1.7)\mathrm{eV}=3.7\mathrm{eV}$

#### Page No 366:

#### Question 31:

A small piece of cesium metal (φ = 1.9 eV) is kept at a distance of 20 cm from a large metal plate with a charge density of 1.0 × 10^{−9} C m^{−2} on the surface facing the cesium piece. A monochromatic light of wavelength 400 nm is incident on the cesium piece. Find the minimum and maximum kinetic energy of the photoelectrons reaching the large metal plate. Neglect any change in electric field due to the small piece of cesium present.

#### Answer:

Given:

Charge density of the metal plate, $\sigma $ = 1.0 × 10^{−9} Cm^{−2}

Work function of the cesium metal, φ = 1.9 eV

Wavelength of monochromatic light, $\lambda $ = 400 nm = 400$\times $10^{$-9$ } m

Distance between the metal plates,* d* = 20 cm = 0.20 m

Electric potential due to a charged plate,

*V* = *E* × *d,*

where *E*, the electric field due to the charged plate, is $\frac{\sigma}{{\epsilon}_{0}}$ and

*d* is the separation between the plates.

$\therefore V=\frac{\sigma}{{\epsilon}_{0}}\times d\phantom{\rule{0ex}{0ex}}=\frac{1\times {10}^{-9}\times 20}{8.85\times {10}^{-12}\times 100}\left(\because {\epsilon}_{0}=8.65\times {10}^{-12}{\mathrm{C}}^{2}{\mathrm{N}}^{-1}-{\mathrm{m}}^{-2}\right)\phantom{\rule{0ex}{0ex}}=22.598\mathrm{V}=22.6\mathrm{V}\phantom{\rule{0ex}{0ex}}\mathrm{From}\mathrm{Einstein}\text{'s}\mathrm{photoelectric}\mathrm{equation},\phantom{\rule{0ex}{0ex}}e{V}_{o}=hv-{\mathrm{W}}_{0}\phantom{\rule{0ex}{0ex}}=\frac{hc}{\lambda}-\mathrm{W}\phantom{\rule{0ex}{0ex}}\mathrm{On}\mathrm{substituting}\mathrm{the}\mathrm{respective}\mathrm{values},\mathrm{we}\text{get:}\phantom{\rule{0ex}{0ex}}{V}_{0}=\frac{4.14\times {10}^{-15}\times 3\times {10}^{8}}{4\times {10}^{-7}}-1.9\phantom{\rule{0ex}{0ex}}=3.105-1.9=1.205\mathrm{eV}\phantom{\rule{0ex}{0ex}}=1.205\mathrm{V}$

As *V*_{0} is much less than '*V*', the minimum energy required to reach the charged plate must be equal to 22.7*e*V.

For maximum KE, 'V' must have an accelerating value.

Hence maximum kinetic energy,

$\mathrm{K}.\mathrm{E}.={\mathrm{V}}_{0}+\mathrm{V}=1.205+22.6\phantom{\rule{0ex}{0ex}}=23.8005\mathrm{eV}$

#### Page No 366:

#### Question 32:

Consider the situation of the previous problem. Consider the faster electron emitted parallel to the large metal plate. Find the displacement of this electron parallel to its initial velocity before it strikes the large metal plate.

#### Answer:

Electric field of the metal plate,

$E=\frac{\sigma}{{\epsilon}_{0}}=\frac{1\times {10}^{-9}}{8.85\times {10}^{-12}}\phantom{\rule{0ex}{0ex}}=113\mathrm{V}/\mathrm{m}\phantom{\rule{0ex}{0ex}}\mathrm{Acceleration},\phantom{\rule{0ex}{0ex}}a=\frac{q\mathrm{E}}{m},\phantom{\rule{0ex}{0ex}}\text{w}\mathrm{here}q=\mathrm{charge}\mathrm{on}\mathrm{electron}\phantom{\rule{0ex}{0ex}}E=\mathrm{electric}\mathrm{field}\phantom{\rule{0ex}{0ex}}m=\mathrm{mass}\mathrm{of}\mathrm{electron}\phantom{\rule{0ex}{0ex}}a=\frac{1.6\times {10}^{-19}\times 113}{9.1\times {10}^{-31}}=19.87\times {10}^{12}\phantom{\rule{0ex}{0ex}}t=\sqrt{\frac{2\mathrm{y}}{a}}=\sqrt{\frac{2\times 20\times {10}^{-2}}{19.87\times {10}^{12}}}\phantom{\rule{0ex}{0ex}}=1.41\times {10}^{-7}\mathrm{s}\phantom{\rule{0ex}{0ex}}\mathrm{From}\mathrm{Einstein}\text{'}s\mathrm{photoelectric}\mathrm{equation},\phantom{\rule{0ex}{0ex}}\mathrm{K}.\mathrm{E}.=\frac{hc}{\mathrm{\lambda}}-\mathrm{W}=1.2\mathrm{eV}\phantom{\rule{0ex}{0ex}}=1.2\times 1.6\times {10}^{-19}\mathrm{J}\phantom{\rule{0ex}{0ex}}[\because \mathrm{in}\mathrm{problem}31:\mathrm{KE}=1.2\mathrm{eV}]\phantom{\rule{0ex}{0ex}}\therefore \mathrm{Velocity},v=\sqrt{\frac{2\mathrm{KE}}{m}}\phantom{\rule{0ex}{0ex}}=\sqrt{{\displaystyle \frac{2\times 1.2\times 1.6\times {10}^{-19}}{4.1\times {10}^{-31}}}}\sqrt{{\displaystyle \frac{2\times 1.2\times 1.6\times {10}^{-19}}{4.1\times {10}^{-31}}}}\phantom{\rule{0ex}{0ex}}=0.665\times {10}^{-6}\mathrm{m}/\mathrm{s}$

∴ Horizontal displacement,

$S=v\times t\phantom{\rule{0ex}{0ex}}S=0.665\times {10}^{-6}\times 1.4\times {10}^{-7}\phantom{\rule{0ex}{0ex}}S=0.092\mathrm{m}=9.2\mathrm{cm}$

#### Page No 366:

#### Question 33:

A horizontal cesium plate (φ = 1.9 eV) is moved vertically downward at a constant speed *v* in a room full of radiation of wavelength 250 nm and above. What should be the minimum value of *v* so that the vertically-upward component of velocity is non-positive for each photoelectron?

#### Answer:

Given:

Work function of the cesium plate,* φ *= 1.9 eV

Wavelength of radiation, *λ *= 250 nm

Energy of a photon,

$E=\frac{hc}{\lambda},\phantom{\rule{0ex}{0ex}}\text{w}\mathrm{here}h=\text{P}\mathrm{lanck}\text{'s}\mathrm{constant}\phantom{\rule{0ex}{0ex}}c=\mathrm{speed}\mathrm{of}\mathrm{light}\phantom{\rule{0ex}{0ex}}\therefore E=\frac{1240}{250}=4.96\mathrm{eV}\phantom{\rule{0ex}{0ex}}$

From Einstein's photoelectric equation, kinetic energy of an electron,

$K=E-\varphi \phantom{\rule{0ex}{0ex}}\Rightarrow K=\frac{hc}{\lambda}-\varphi \phantom{\rule{0ex}{0ex}}\left[\text{H}\mathrm{ere},h\text{is}\mathrm{Planck}\text{'s}\mathrm{constant}\mathrm{and}\mathrm{c}isthe\mathrm{speed}\mathrm{of}\mathrm{light}\right]\phantom{\rule{0ex}{0ex}}\Rightarrow K=4.96\mathrm{eV}-1.9\mathrm{eV}\phantom{\rule{0ex}{0ex}}=3.06\mathrm{eV}.$

For non-positive velocity of each photo electron, the velocity of a photoelectron should be equal to minimum velocity of the plate.

∴ Velocity of the photoelectron,

$v=\sqrt{\frac{2K}{m}}\left(m=\mathrm{mass}\mathrm{of}\mathrm{electron}\right)\phantom{\rule{0ex}{0ex}}\therefore v=\sqrt{\frac{2\times 3.06\times 1.6\times {10}^{-19}}{9.1\times {10}^{-31}}}\phantom{\rule{0ex}{0ex}}=1.04\times {10}^{6}{\mathrm{ms}}^{-1}$

#### Page No 366:

#### Question 34:

A small metal plate (work function φ) is kept at a distance *d* from a singly-ionised, fixed ion. A monochromatic light beam is incident on the metal plate and photoelectrons are emitted. Find the maximum wavelength of the light beam, so that some of the photoelectrons may go round the ion along a circle.

#### Answer:

From Einstein's photoelectric equation,

$e{V}_{0}=\frac{hc}{\lambda}-\mathrm{\varphi}\phantom{\rule{0ex}{0ex}}\Rightarrow {V}_{0}=\left(\frac{hc}{\lambda}-\mathrm{\varphi}\right)\frac{1}{e}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Here, *V*_{0} = stopping potential

*h *= Planck's constant

c = speed of light

$\varphi $ = work function

The particle will move in a circle when the stopping potential is equal to the potential due to the singly charged ion at that point so that the particle gets the required centripetal force for its circular motion.

$\Rightarrow \frac{\mathrm{K}e}{2d}=\left(\frac{hc}{\lambda}-\mathrm{\varphi}\right)\frac{1}{e}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\mathrm{K}{e}^{2}}{2d}=\frac{hc}{\lambda}-\mathrm{\varphi}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{hc}{\lambda}=\frac{{\mathrm{Ke}}^{2}}{2d}+\mathrm{\varphi}=\frac{\mathrm{K}{e}^{2}+2d\mathrm{\varphi}}{2d}\phantom{\rule{0ex}{0ex}}\Rightarrow \lambda =\frac{\left(hc\right)\left(2d\right)}{k{e}^{2}+2d\mathrm{\varphi}}\phantom{\rule{0ex}{0ex}}\Rightarrow \lambda =\frac{2hdc}{{\displaystyle \frac{1}{4\mathrm{\pi}{\in}_{0}}}{e}^{2}+2d\mathrm{\varphi}}\phantom{\rule{0ex}{0ex}}\Rightarrow \lambda =\frac{8\mathrm{\pi}{\in}_{0}hcd}{{e}^{2}+8\mathrm{\pi}{\in}_{0}d\mathrm{\varphi}}$

#### Page No 366:

#### Question 35:

A light beam of wavelength 400 nm is incident on a metal plate of work function 2.2 eV. (a) A particular electron absorbs a photon and makes two collisions before coming out of the metal. Assuming that 10% of the extra energy is lost to the metal in each collision, find the kinetic energy of this electron as it comes out of the metal. (b) Under the same assumptions, find the maximum number of collisions the electron can suffer before it becomes unable to come out of the metal.

#### Answer:

Given:

Wavelength of light beam, $\lambda $ = 400 nm

Work function of metal plate, $\varphi $ = 2.2 eV

Energy of the photon,

$E=\frac{hc}{\lambda},\phantom{\rule{0ex}{0ex}}\text{w}\mathrm{here}h=\mathrm{Planck}\text{'s}\mathrm{constant}\phantom{\rule{0ex}{0ex}}c=\mathrm{speed}\mathrm{of}\mathrm{light}\phantom{\rule{0ex}{0ex}}\therefore E=\frac{1240}{400}=3.1\mathrm{eV}$

This energy will be supplied to the electrons.

Energy lost by the electron in the first collision

= 3.1 eV × 10%

= 0.31 eV

Now, the energy of the electron after the first collision = 3.1 $-$ 0.31 = 2.79 eV

Energy lost by electron in the second collision

= 2.79 eV× 10%

= 0.279 eV

Total energy lost by the electron in two collisions

= 0.31 + 0.279 = 0.589 eV

Using Einstein's photoelectric equation, kinetic energy of the photoelectron when it comes out from the metal,

$K=E-\varphi -\mathrm{energy}\mathrm{lost}\mathrm{due}\mathrm{to}\mathrm{collisions}\phantom{\rule{0ex}{0ex}}$

= (3.1 − 2.2 − 0.589) eV

= 0.31 eV

(b) Similarly for the third collision, the energy lost = (2.79 $-$ 0.279) eV × 10%

= 0.2511 eV

Energy of the electron after the third collision = 2.790 $-$ 0.2511 = 2.5389

Energy lost in the fourth collision = 2.5389 × 10%

Energy of the electron after the fourth collision = 2.5389 $-$ 0.25389 = 2.28501

This value is very close to the work function of the metal plate. After the fifth collision, the energy of the electron becomes less than the work function of the metal.

Therefore, the electron can suffer maximum four collisions before it becomes unable to come out of the metal.

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