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

Question:2

The wavelength of a photon needed to remove a proton from a nucleus which is bound to the nucleus with 1 MeV energy is nearly
(a) 1.2 nm
(b) 1.2 x 10^-3 nm
(c) 1.2 x 10^-6 nm
(d). 1.2 x 10 nm

Answer:

The answer is the option (b)
Energy of the photon should be equal to the binding energy of the proton
So, the energy of photon=
$1MeV=10^{6}\ast 1.6\ast 10^{-19}$
$\lambda=\frac{hc}{E}=6.63 \ast 10^{-34} \ast 3 \ast\frac{10^{8}}{1.6 \ast 10^{-13}}\\ =1.24 \ast 10^{-3}nm$

Question:3

Consider a beam of electrons (each electron with energy E₀) incident on a metal surface kept in an evacuated chamber. Then
(a) no electrons will be emitted as only photons can emit electrons
(b) electrons can be emitted but all with an energy, E₀
(c) electrons can be emitted with any energy, with a maximum of E₀ – $\phi$ ($\phi$ is the work function)
(d) electrons can be emitted with any energy, with a maximum of E₀

Answer:

The answer is the option (d) When a beam of electrons (each electron with energy E₀ ) incident on a metal surface is kept in an evacuated chamber, energy of incident electrons will be transferred to the emitted electrons due to elastic collisions. A part of E₀ of incident electrons is consumed against work function to emit the electrons. Thus, the maximum energy of emitted electrons is E₀

Question:4

Consider Fig. 11.7 in the NCERT textbook of physics for Class XII. Suppose the voltage applied to A is increased. The diffracted beam will have the maximum at a value of θ that
(a) will be larger than the earlier value
(b) will be the same as the earlier value
(c) will be less than the earlier value
(d) will depend on the target

Answer:

The answer is the option (c)
The de-Broglie wavelength of diffracted beam of electrons is given by
$\lambda_{d}=\frac{12.27}{\sqrt{V}}A^{\circ}..............(i)$
If there is a maxima of the diffracted electrons at an angle $\theta$, then $2d \sin \theta =\lambda$
From equation (i), we get that on increasing the voltage, $\lambda_{d}$ decreases which in turn
decreases the value of $\theta$

Question:5

A proton, a neutron, an electron and an α-particle have the same energy. Then their de Broglie wavelengths compare as
$(a) \lambda _{p} = \lambda _{n} > \lambda _{e} > \lambda _{\alpha} $
$ (b) \lambda _{\alpha} < \lambda _{p} = \lambda _{n} < \lambda _{e}$
$ (c) \lambda _{e} < \lambda _{p} = \lambda _{n} > \lambda _{\alpha} $
$ (d) \lambda _{e} = \lambda _{p} = \lambda _{n} = \lambda _{\alpha}$

Answer:

The answer is the option (b)
de-Broglie wavelength $\lambda$ is given by
$\begin{aligned} & \lambda=\frac{h}{p}=\frac{h}{m v} \\ & \quad 2 K=m v^2 \\ & 2 K m=m^2 v^2 \\ & 2 m K=p^2 \\ & p=\sqrt{2 m K} \\ & y_d=\frac{h}{p} \\ & y_d=\frac{h}{\sqrt{2 m K}} \\ & \text { Since } \lambda \propto \frac{1}{\sqrt{m}} \\ & M_a>m_p=m_n>m_e \\ & \lambda_a>\lambda_p=\lambda_n>\lambda_e\end{aligned}$

Question:6

An electron is moving with an initial velocity v = v0i and is in a magnetic field B = B0j. Then, its de-Broglie wavelength
(a) remains constant
(b) increases with time
(c) decreases with time
(d) increases and decreases periodically

Answer:

The answer is the option (a)
$\vec{v}=v_{0}\hat{i}\: \: \: \: \: B=B_{0}\hat{j}$
Force on the elctron in perpendicular magnetic field B is
$F=-e\left (\vec{v} \times \vec{B} \right )$
$=-ev_{0}B_{0}k$
Therefore, the force is perpendicular to v and B both. The de-Broglie wavelength will not change.

Question:7

An electron (mass m) with an initial velocity v = v₀i(v₀ > 0) is in an electric field $E=E_{0}\hat{i}$ (E₀ = constant > 0). Its de-Broglie wavelength at time t is given by
$(a) \frac{\lambda_{0}}{\left [ 1+\frac{eE_{0}}{m}\frac{t}{v_{0}} \right ]}$
$ (b)\lambda_{0}\left [ 1+\frac{eE_{0}t}{mv_{0}} \right ]$
$(c)\lambda_{0}$
$ (d)\lambda_{0}t$

Answer:

The answer is the option (a)
Initial de-Broglie wavelength $\lambda$is given by
$\lambda_{0}=\frac{h}{p}=\frac{h}{mv_{0}}$
Force on electron=F=qE
$ma=eE_{0}\hat{i}$
$ a=\frac{eE}{m}\hat{i}$
Velocity of electron after time t is v=v₀+at
$v=v_{0}i+\frac{eE}{m}i.t\\ v=\left [ v_{0}+\frac{eEt}{m} \right ]\hat{i}\\$
New de-Broglie wavelength
$\lambda=\frac{h}{mv}$
$\lambda=\frac{h}{m\left [ v_{0}+\frac{eE_{0}t}{m} \right ]}$
$ \lambda=\frac{\lambda_{0}}{\left [ 1+\frac{eE_{0}t}{mv_{0}} \right ]}\; \; \; \left ( Since \frac{h}{mv_{0}}=\lambda_{0} \right )$

Question:8

An electron (mass m) with an initial velocity $v=v_{0}\hat{i}$ is in an electric field $E=E_{0}\hat{i}$ IF $\lambda_{0}=\frac{h}{mv_{0}}$ its de-broglie wavelength at time t is given by
$(a)\lambda_{0}$
$(b)\lambda_{0}\sqrt{1+\frac{e^{2}E_{0}^{2}t^{0}}{m^{2}v_{0}^{2}}}$
$(c)\frac{\lambda_{0}}{\sqrt{1+\frac{e^{2}E_{0}^{2}t^{0}}{m^{2}v_{0}^{2}}}}$
$(d)\frac{\lambda_{0}}{\left (1+\frac{e^{2}E_{0}^{2}t^{0}}{m^{2}v_{0}^{2}} \right )}$

Answer:

The correct answer is option (c)
$\frac{\lambda_{0}}{\sqrt{1+\frac{e^{2}E_{0}^{2}t^{0}}{m^{2}v_{0}^{2}}}}$

Question:10

Two particles $A_{1}$ and $A_{2}$ of masses $m_{1}$, $m_{2}$ ($m_{1}$ >$ m_{2}$) have the same de Broglie wavelength. Then
(a) their momenta are the same
(b) their energies are the same
(c) energy of $A_1$ is less than the energy of $A_2$
(d) energy of $A_1$ is more than the energy of $A_2$

Answer:

The answer is the option (a. c)
$\begin{aligned} & \quad \lambda=\frac{h}{p} \\ & p=\frac{h}{\lambda} \text { or } p \propto \frac{1}{\lambda} \\ & \frac{p_1}{p_2}=\frac{\lambda_2}{\lambda_1} \\ & \lambda_1=\lambda_2=\lambda(\text { given }) p_1=p_2 \text { vertifies }(\text { a }) \\ & E_a=\frac{1}{2} m v^2=\frac{\frac{1}{2} m^2 v^2}{m}=\frac{p^2}{2 m} \\ & E \propto \frac{1}{m}\left[\text { as } p_1=p_2\right] \\ & \frac{E_1}{E_2}=\frac{m_2}{m_1} \\ & \frac{E_1}{E_2}<1\left[m_1>m_2\right] E_2>E_1\end{aligned}$
Thus c is correct.

Question:11

The de Broglie wavelength of a photon is twice the de Broglie wavelength of an electron. The speed of the electron is v_e = c/100. Then
(a) E_e/E_p = 10^-4
(b) E_e/E_p = 10^-2
(c) Pe/m_ec = 10^-2
(d) P_e/m_ec = 10^-4

Answer:

The answer is the option (b,c)
The energy of a charged particle when accelerated through a potential difference V is:
$E=\frac{1}{2}mv^{2}=qV$
De-Broglie wavelength

$\begin{gathered}
\lambda=\frac{h}{p}=\frac{h}{\sqrt{2 m E}}=\frac{h}{\sqrt{2 m q V}} \\
\lambda_{\text {Neutron }}=\frac{0.286 * 10^{-10}}{\sqrt{E}} m=\frac{0.286 * 10^{-10}}{\sqrt{E}} A^{\circ}
\end{gathered}$

At ordinary temperature, the energy of thermal neutrons
$\begin{aligned}
& \quad E=k T \Rightarrow \lambda=\frac{h}{\sqrt{2 m k T}} \\
& \lambda_{\text {Thermal } \text { Neutron }}=\frac{30.83}{\sqrt{T}} A^{\circ} \\
& \text { Mass of electron }=m_e \\
& \text { Mass of photon }=m_p
\end{aligned}$

de Broglie wavelength of the electron
$\begin{aligned}
& \quad \lambda_e=\frac{h}{m_e v_e}=\frac{100 h}{m_e c} \\
& K E_e=\frac{1}{2} m_e v_e^2 \\
& m_e v_e^2=\sqrt{2 E_e m_e} \\
& \lambda_e=\frac{h}{m_e v_e}=\frac{h}{\sqrt{2 E_e m_e}} \\
& E_e=\frac{h^2}{2 \lambda_e^2 v_e}
\end{aligned}$

de Broglie wavelength of the proton is $\lambda_p$
$\begin{aligned}
& E_p=\frac{h c}{\lambda_p}=\frac{h c}{2 \lambda_e} \\
& \frac{E_p}{E_e}=\left(\frac{h c}{\lambda_e}\right) *\left(\frac{2 \lambda_e^2 m_e}{h}\right)=100 \\
& p_e=m_e v_e=m_e * \frac{c}{100} \\
& \frac{p_e}{m_e c}=10^{-2}
\end{aligned}$

Question:12

Photons absorbed in the matter are converted to heat. A source emitting n photon/sec of frequency ν is used to convert 1kg of ice at 0°C to water at 0°C. Then, the time T taken for the conversion
(a) decreases with increasing n, with ν fixed
(b) decreases with n fixed, ν increasing
(c) remains constant with n and ν changing such that nν = constant
(d) increases when the product nν increases

Answer:

The answer is the option (a, b. c).
Heat energy required to convert 1Kg of ice at $0^{\circ}$ C to water at $0^{\circ}$ C is E=mL
If n is the number of photons incident per second and time t is taken by radiation to
melt the ice at $0^{\circ}$C.
Then E=n hv t
$t=\frac{mL}{nhv}\\$
Or $t\propto \frac{1}{nv}\\$
[m, L and h are constants ] Hence, (a), (b) and (c) are correct.
If nv increases then T decreases ;so (d) isn't correct.

Question:13

A particle moves in a closed orbit around the origin, due to a force which is directed towards the origin. The de-BrOglie wavelength of the particle varies cyclically between two values $\lambda_{1},\lambda_{2}$ with $\lambda_{1}>\lambda_{2}$ Which of the following statements are true?
(a) The particle could be moving in a circular orbit with origin as centre
(b) The particle could be moving in an elliptic orbit with origin as its focus
(c) When the de-Broglie wavetength is $\lambda_{1}$ the particle is nearer the origin than when its value is $\lambda_{2}$
(d) When the de-Broglie wavelength is $\lambda_{2}$ the particle is nearer the origin than when its value is $\lambda_{1}$

Answer:

The answer is the option (b,d)
A particle moving in an elliptic orbit with Origin as the centre can have its de-Broglie wavelength varying between $\lambda_{1}$ and $\lambda_{2}$ cyclically. When the de-Broglie wavelength is lower, the particle is closer to the origin.
$\lambda_{1}=\frac{h}{mv_{1}}\;\text{ and }\; \lambda_{2}=\frac{h}{mv_{2}}$
As,$\lambda_{1}>\lambda_{2} \implies v_{2}>v_{1}$
The particle will have a higher velocity closer to focus.

Question:15

(i) In the explanation of photoeletric effect, we assume one photon of frequency v collides with an electron and transfers its energy. This leads . to the equation for the maximum energy E_max of the emitted electron as $E_{max} = hv - \phi_{0}$
where $\phi_{0}$ is the work function of the metal. If an electron absorbs 2 photons (each of frequency v), what will be the maximum energy for the emitted electron?
(ii) Why is this fact (two photon absorption) not taken into consideration in our discussion of the stopping potential?

Answer:

(i) In the question, 2 photons transfer its energy to one electron as E=hv
E_e=E_p
hv_e=2hv
v_e=2v
v is the frequency
The maximum energy of the emitted electron is E_max=hv_e - $\phi_{0}$
=h(2v) - $\phi_{0}$
=2hv - $\phi_{0}$
(ii) Due to the mass difference, the probability that the electron absorbs 2 photons is extremely low. Therefore the chances of such emission of electrons is close to negligible.

Question:16

There are materials which absorb photons of shorter wavelength and emit photons of longer wavelength. Can there be stable substances which absorb photons of larger wavelength and emit light of shorter wavelength?

Answer:

With an increase in the wavelength of the photon, there is a decrease in the frequency or increase in the energy of the photon. Let us consider 2 possible cases:

1. The photons having a smaller wavelength emit photons of smaller energy wherein some of it is consumed against work function. The law of conservation of energy is upheld in this case.
2. Photons having a longer wavelength always emit photons with shorter wavelength. Photons having smaller energy can never emit photons of larger energy as some part of energy is consumed in work function of metal. So, it isn’t possible in stable materials.

Question:17

Do all the electrons that absorb a photon comes out as photo electrons?

Answer:

A common observation in the photoelectric effect is that most of the electrons knocked by photons are scattered into the metal by absorbing photons. In order to escape from the surface, the electron must absorb enough energy to overcome the positive ion attraction in the material of the surface.
Therefore, not all the electrons that absorb a photon comes out from the metal surface.

Question:18

There are two sources of light, each emitting with a power of 100 W. One emits X-rays of wavelength 1nm and the other visible light at 500 nm. Find the ratio of a number of photons of X-rays to the photons of visible light of the given wavelength?

Answer:

Let E_X and E_V be the energies given by one photon in X-rays and
visible rays respectively.
Then E_X=hv_X and E_V=hv_V
Let n_X and n_V be the number of photons from x-rays and visible
light are of equal energies and they emit 100 W power.
$\begin{aligned} & \quad n_X h v_X=n_V h v_V \\ & \frac{n_X}{n_V}=\frac{v_V}{v_X}=\frac{\lambda_X}{\lambda_V} \\ & \frac{n_X}{n_V}=\frac{1}{500} \\ &n_X: n_V=1: 500\end{aligned}$

Question:20

Consider a metal exposed to light of wavelength 600 nm. The maximum energy of the electron doubles when light of wavelength 400 nm is used. Find the work function in eV.

Answer:

Let the maximum energies of emitted electrons be K₁ and K₂ when 600 nm and 400 nm visible light are used according to question
$\begin{aligned} & \quad K_2=2 K_1 \\ & K_{\max }=h v-\phi=\frac{h c}{\lambda}-\phi \\ & K_1=\frac{h c}{\lambda_1}-\phi \\ & K_2=\frac{h c}{\lambda_2}-\phi=2 K_1 \\ & \frac{h c}{\lambda_2}-\phi=2\left[\frac{h c}{\lambda_1}-\phi\right]=\frac{2 h c}{\lambda_1}-2 \phi \\ & \phi=h c\left[\frac{2}{\lambda_1}-\frac{1}{\lambda_2}\right] \\ & =1240\left[\frac{2}{600}-\frac{1}{400}\right] e V \\ & =\frac{1240}{200}\left[\frac{2}{3}-\frac{1}{2}\right]=6.2\left(\frac{4-3}{6}\right) \\ & \text { Work function } \phi=\frac{6.2}{6}=1.03 \mathrm{eV}\end{aligned}$

Question:21

Assuming an electron is confined to a 1nm wide region, find the uncertainty in momentum using the Heisenberg Uncertainty principle. You can assume the uncertainty in position as 1nm. Assuming , find the energy of the electron in electron volts.

Answer:

As the electrons rotate in a circular path,
$\Delta r=1nm=10^{-9}m$
$\begin{gathered}\Delta p=\frac{h}{\Delta X} \\ \Delta p=\left(\frac{331}{314}\right) * 10^{25} \\ E=\frac{1}{2} m v^2=\frac{\Delta p^2}{2 m} \\ E=3.8 * 10^{-2} \mathrm{eV}\end{gathered}$

Question:22

Two monochromatic beams A and B of equal intensity I, hit a screen. The number of photons hitting the screen by beam A is twice that by beam B. Then what inference can you make about their frequencies?

Answer:

$\begin{aligned} & I_A=I_B(\text { Given in the question }) \\ & n_A h v_A=n_B h v_B \\ & n_A=2 n_B[\text { Given }] \\ & 2 n_B v_A=n_B v_B \\ & 2 v_A=v_B\end{aligned}$
So, the frequency of source B is twice the frequency of source A.

Question:23

Two particles A and B of de Broglie wavelengths $\lambda_{1}$ and $\lambda_{2}$ combine to form a particle C. The process conserves momentum. Find the de Broglie wavelength of the particle C. (The motion is one dimensional).

Answer:

The de-Broglie wavelength is given by
$\lambda=\frac{h}{p} \; or\; p=\frac{h}{\lambda}$
$p_{1}=\frac{h}{\lambda_{1}}, p_{2}=\frac{h}{\lambda_{2}}, p_{3}=\frac{h}{\lambda_{3}}$
By the law of conservation of momentum
$p_{1}+p_{2}= p_{3}$
$\frac{h}{\lambda_{1}}+\frac{h}{\lambda_{2}}=\frac{h}{\lambda_{3}}$ ($\lambda_{3}$ is the wavelength of particle C)

$\frac{h}{\lambda_{1}}+\frac{h}{\lambda_{2}}=\frac{h}{\lambda_{3}}\\ \frac{\lambda_{2}+\lambda_{1}}{\lambda_{1}\lambda_{2}}=\frac{h}{\lambda_{3}}$
Case I:When p₁ and p₂ are positive then $\lambda_{3}=\frac{\lambda_{1}\lambda_{2}}{\lambda_{1}+\lambda_{2}}$
Case II:When p₁ and p₂ both are negative $\lambda_{3}=\left |-\frac{\lambda_{1}\lambda_{2}}{\lambda_{1}+\lambda_{2}} \right |=\frac{\lambda_{1}\lambda_{2}}{\lambda_{1}+\lambda_{2}}$
Case III: $p_{A}>0,p_{B} <0$
$\frac{h}{\lambda_{3}}=\frac{h}{\lambda_{1}}-\frac{h}{\lambda_{2}}\; or\; \frac{h}{\lambda_{3}}=\frac{(\lambda_{2}-\lambda_{1})h}{\lambda_{1}\lambda_{2}}$
Case IV: $p_{A}<0,p_{B} >0$
$\frac{h}{\lambda_{3}}=\frac{h}{\lambda_{1}}-\frac{h}{\lambda_{2}}=\frac{\left (\lambda_{1}-\lambda_{2} \right )h}{\lambda_{1}\lambda_{2}}\\ \lambda_{3}=\frac{\lambda_{1}\lambda_{2}}{\lambda_{1}-\lambda_{2} }$

Question:24

A neutron beam of energy E scatters from atoms on a surface with a spacing d = 0.1nm. The first maximum of intensity in the reflected beam occurs at theta = 30 degree. What is the kinetic energy E of the beam in eV?

Answer:

According to the Bragg’s law of diffraction, condition for the nth maxima is given by
$\begin{aligned} & \quad 2 d \sin \theta=n \lambda \\ & n=1 \\ & \text { so } \lambda=2 d \sin \theta\left[\theta=30^{\circ}\right] \\ & =2 * 0.1 * 10^{-9} \sin 30^{\circ} \\ & p=\frac{h}{\lambda}=6.6 * \frac{10^{-34}}{10^{-10}}=6.6 * 10^{-24} \mathrm{~kg} \frac{\mathrm{~m}}{\mathrm{~s}} \\ & E=\frac{1}{2} m v^2=\frac{\frac{1}{2} m^2 v^2}{m} \\ & =\frac{p^2}{2 m}=\frac{6.6 * 6.6 * 10^{-48}}{2 * 1.6 * 10^{-27} * 1.6 * 10^{-19}} \mathrm{eV} \\ & =\frac{66 * 66 * 10^{-48+46}}{2 * 16 * 16}=33 * \frac{33}{126} * 10^{-2} \\ & =8.5 * 10^{-2}=0.085 \mathrm{eV}\end{aligned}$

Question:26

Consider an electron in front of the metallic surface at a distance d (treated as an infinite plane surface). Assume the force of attraction by the plate is given as 2 2 0 1 4 4 q $\pi\epsilon$ d Calculate work in taking the charge to an infinite distance from the plate. Taking d = 0.1nm, find the work done in electron volts. [Such a force law is not valid for d < 0.1nm].

Answer:

Let us consider the above-given figure wherein an electron is displaced slowly by a distance x by the means of an external force which is given by
$F=\frac{q^{2}}{4*4\pi\varepsilon_{0}d^{2}}\; where \;d=0.1nm=10^{-10}m$
Now work done by an external agency in taking the electron from distance d to infinity is
$\begin{aligned} W&=\int_d^{\infty} F_x d x=\int_d^{\infty} \frac{q^2 d x}{4 * 4 \pi \varepsilon_0} * \frac{1}{x^2} \\ &= \frac{q^2 d x}{4 * 4 \pi \varepsilon_0}\left(\frac{1}{d}\right)=\frac{\left(1.6 * 10^{-19}\right)^2}{1.6 * 10^{-19}} * \frac{\left(9 * 10^9\right)}{4 * 10^{-10}}=3.6 \mathrm{eV}\end{aligned}$

Question:27

A student performs an experiment on photoelectric effect, using two materials A and B. A plot of V_stop versus v is given in figure.

(i) Which material A or B has a higher work function?

(ii)Given the electric charge of an electron = 1.6 x 10^-19C, find the value of h obtained from the experiment for both A and B. Comment on whether it is consistent with Einstein's theory.

Answer:

We know that $\phi_{0}=hv_{0}$ where v₀=Threshold frequency
$\frac{\phi_{0A}}{\phi_{0B}}=\frac{hv_{0A}}{hv_{0B}}=\frac{5*10^{14}}{10*10^{14}}=\frac{1}{2}$
So, $\phi_{0B}=2\phi_{0A}$
(ii) By the differentiation of potential
$V=\frac{W}{Q}$
hv=eV
Differentiating both the sides we get h.dv=edV
$h=\frac{edV}{dv}$
Using this formula
For metal A, h=6*10^-34Js….(i)
For metal B, h=0.8*10^-33…(ii)
Since the value of plank'sconstant is not equal for both the experimental graphs, the experiment inconsistent with Einstein's theory. But the values are very close to 6.6*10^-34Js due to experimental limitation and hence can be considered consistent with Einstein theory.

Question:28

A particle A with a mass mA is moving with a velocity v and hits a particle B (mass mB) at rest (one dimensional motion). Find the change in the de Broglie wavelength of the particle A. Treat the collision as elastic.

Answer:

When the collision is elastic, law of conservation of momentum is followed.
$\begin{aligned} & m_A v+m_B 0=m_A v_1+m_B v_2 \\ & m_A\left(v-v_1\right)=m_B v_2 \\ & \frac{1}{2} m_A v_2=\frac{1}{2} m_A v_1^2+\frac{1}{2} m_B v_2^2 \\ & v_1=\frac{m_A-m_B}{\left(m_A-m_B\right)} v \quad v_2=\frac{2 m_A}{\left(m_A-m_B\right)} v \\ & \lambda_{\text {initial }}=\frac{h}{m_A v} \\ & \lambda_{\text {final }}=\frac{h}{m_A v_1} \\ & \Delta \lambda=\lambda_{\text {final }}-\lambda_{\text {initial }}=\frac{h}{m_A v}\left[\frac{m_A+m_B}{m_A-m_B}-1\right]\end{aligned}$

Question:29

Consider a 20 W bulb emitting light of wavelength 5000 $A^{\circ}$ and shining on a metal surface kept at a distance 2m. Assume that the metal surface has work function of 2 eV and that each atom on the metal surface can be treated as a circular disk of radius 1.5 $A^{\circ}$ .
(i) Estimate no. of photons emitted by the bulb per second. [Assume no other losses]
(ii) Will there be photoelectric emission?
(iii) How much time would be required by the atomic disk to receive energy equal to work function (2 eV)?
(iv) How many photons would atomic disk receive within time duration calculated in (iii) above?
(v) Can you explain how photoelectric effect was observed instantaneously?

Answer:

Given that $\mathrm{P}=20 \mathrm{~W}$

$\begin{aligned}
& \lambda=5000 A^{\circ}=5000 * 10^{-10} \mathrm{~m} \\
& d=2 \mathrm{~m} \\
& \phi=2 \mathrm{ev} \\
& r=1.5 * 10^{10} \mathrm{~m}
\end{aligned}$

i) Number of photon emitted by bulb per second= $\frac{p \lambda}{h c}=5 * 10^{19} \mathrm{sec}$

ii Energy of the incident photon= $\frac{h c}{\lambda}=2.48 \mathrm{eV}$

iii Time required by the atomic disk to receive energy is=28.4 sec
Let the time spent be $\Delta T$
$E=P * A * \Delta t=P * \pi r^2 \Delta t$

Energy transferred by the bulb in full solid angle to atoms= $ 4 \pi d^2 \phi$
$\begin{aligned}
& p * \pi r^2 \Delta t=4 \pi d^2 \phi \\
&\Delta t=\frac{4 d^2 \phi}{P r^2}=\frac{4 * 2 * 2 * 2 * 1.6 * 10^{-10}}{20 * 1.5 * 1.5 * 10^{-10} * 10^{-10}} \sec =\frac{12.8 * 10^{-19+20}}{5 * 2.25}
\end{aligned}$

(iv) Number of photons received by the atomic disk $=\mathrm{N}$
$\begin{aligned}
\frac{n_1 \pi r^2 \Delta t}{4 \pi d^2}&=\frac{n_1 r^2 \Delta t}{4 d^2} \\
&= \frac{5 * 10^{19} * 1.5 * 1.5 * 10^{-20} * 11.4}{4 * 2 * 2} \\
&\approx 0.80 \cong 1 \text { photon per atom }
\end{aligned}$

(iv) We say that photoelectric emission is instantaneous as it involves a collision between the incident photon and free electron lasting for a very short span of time, say $\leq 10^{-9} \mathrm{sec}$