Consider a thin target $\left(10^{-2} \mathrm{~m}\right.$ square, $10^{-3} \mathrm{~m}$ thickness) of sodium, which produces a photocurrent of $100 \propto \mathrm{~A}$ when a light of intensity 100 $\mathrm{W} / \mathrm{m}^2(\lambda=660 \mathrm{~nm})$ falls on it. Find the probability that a photoelectron is produced when a photon strikes a sodium atom. [Take density of $\mathrm{Na}=0.97 \mathrm{~kg} / \mathrm{m}^3$ ]
$$\begin{aligned} \text{Given,}\quad A & =10^{-2} \mathrm{~m}^2=10^{-2} \times 10^{-2} \mathrm{~m}^2 \\ \Rightarrow\quad & =10^{-4} \mathrm{~m}^2 \\ d & =10^{-3} \mathrm{~m} \\ i & =100 \times 10^{-6} \mathrm{~A}=10^{-4} \mathrm{~A} \\ \text{Intensity,}\quad I & =100 \mathrm{~W} / \mathrm{m}^2 \\ \Rightarrow\quad \lambda & =660 \mathrm{~nm}=660 \times 10^{-9} \mathrm{~m} \\ \rho_{\mathrm{Na}} & =0.97 \mathrm{~kg} / \mathrm{m}^3 \end{aligned}$$
Avogadro's number $=6 \times 10^{26} \mathrm{~kg}$ atom
Volume of sodium target $=A \times d$
$$\begin{aligned} & =10^{-4} \times 10^{-3} \\ \Rightarrow\quad & =10^{-7} \mathrm{~m}^3 \end{aligned}$$
We know that $6 \times 10^{26}$ atoms of Na weights $=23 \mathrm{~kg}$
So, volume of $6 \times 10^6 \mathrm{Na}$ atoms $=\frac{23}{0.97} \mathrm{~m}^3$.
Volume occupied by one Na atom $=\frac{23}{0.97 \times\left(6 \times 10^{26}\right)}=3.95 \times 10^{-26} \mathrm{~m}^3$
Number of Na atoms in target $\left({ }^n \mathrm{Na}\right)$
$$=\frac{10^{-7}}{3.95 \times 10^{-26}}=2.53 \times 10^{18}$$
Let $n$ be the number of photons falling per second on the target.
Energy of each photon $=h c / \lambda$
Total energy falling per second on target $=\frac{n h c}{\lambda}=I A$
$\therefore \quad n=\frac{I A \lambda}{h c}$
$$\Rightarrow \quad=\frac{100 \times 10^{-4} \times\left(660 \times 10^{-9}\right)}{\left(6.62 \times 10^{-34}\right) \times\left(3 \times 10^8\right)}=3.3 \times 10^{16}$$
Let $P$ be the probability of emission per atom per photon.
The number of photoelectrons emitted per second
$$\begin{aligned} N & =P \times n \times\left({ }^n \mathrm{Na}\right) \\ & =P \times\left(3.3 \times 10^{16}\right) \times\left(2.53 \times 10^{18}\right) \end{aligned}$$
Now, according to question,
$$i=100 \propto \mathrm{~A}=100 \times 10^{-6}=10^{-4} \mathrm{~A}.$$
$$\begin{aligned} \text{Current,}\quad & i=\mathrm{Ne} \\ \end{aligned}$$
$$\begin{array}{ll} \therefore & 10^{-4}=P \times\left(3.3 \times 10^{16}\right) \times\left(2.53 \times 10^{18}\right) \times\left(1.6 \times 10^{-19}\right) \\ \Rightarrow & P=\frac{10^{-4}}{\left(3.3 \times 10^{16}\right) \times\left(2.53 \times 10^{18}\right) \times\left(1.6 \times 10^{-19}\right)}\\ &=7.48 \times 10^{-21} \end{array}$$
Thus, the probability of emission by a single photon on a single atom is very much less than 1. It is due to this reason, the absorption of two photons by an atom is negligible.
Consider an electron in front of metallic surface at a distance $d$ (treated as an infinite plane surface). Assume the force of attraction by the plate is given as $\frac{1}{4} \frac{q^2}{4 \pi \varepsilon_0 d^2}$. Calculate work in taking the charge to an infinite distance from the plate. Taking $d=0.1 \mathrm{~nm}$, find the work done in electron volts. [Such a force law is not valid for $d<0.1 \mathrm{~nm}$ ]
According to question, consider the figure given below From figure, $d=0.1 \mathrm{~nm}=10^{-10} \mathrm{~m}$,
$$F=\frac{q^2}{4 \times 4 \pi \varepsilon_0 d^2}$$
Let the electron be at distance $x$ from metallic surface. Then, force of attraction on it is
$$F_x=\frac{q^2}{4 \times 4 \pi \varepsilon_0 x^2}$$
Work done by external agency in taking the electron from distance $d$ to infinity is
$$\begin{aligned} W & =\int_d^{\infty} F_x d x=\int_d^{\infty} \frac{q^2 d x}{4 \times 4 \pi \varepsilon_0} \frac{1}{x^2} \\ & =\frac{q^2}{4 \times 4 \pi \varepsilon_0}\left[\frac{1}{d}\right] \\ & =\frac{\left(1.6 \times 10^{-19}\right)^2 \times 9 \times 10^9}{4 \times 10^{-10}} \mathrm{~J} \\ & =\frac{\left(1.6 \times 10^{-19}\right)^2 \times\left(9 \times 10^9\right)}{\left(4 \times 10^{-10}\right) \times\left(1.6 \times 10^{-19}\right)} \mathrm{eV}=3.6 \mathrm{eV} \end{aligned}$$
A student performs an experiment on photoelectric effect, using two materials $A$ and $B$. A plot of $V_{\text {stop }}$ 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 $=1.6 \times 10^{-19} C$, find the value of $h$ obtained from the experiment for both $A$ and $B$.
Comment on whether it is consistent with Einstein's theory.
(i) Given, thresholed frequency of $A$ is given by $v_{O A}=5 \times 10^{14} \mathrm{~Hz}$ and
For $B$, $$v_{O B}=10 \times 10^{14} \mathrm{~Hz}$$
We know that
Work function, $$\phi=h v_0 \text { or } \phi_0 \propto v_0$$
$$\Rightarrow \quad \phi_0 \propto v_0$$
$$\begin{array}{ll} \text { So, } & \frac{\phi_{O A}}{\phi_{O B}}=\frac{5 \times 10^{14}}{10 \times 10^{14}}<1 \\ \Rightarrow & \phi_{O A}<\phi_{O B} \end{array}$$
Thus, work function of B is higher than A.
(ii) For metal $A$, slope $=\frac{h}{e}=\frac{2}{(10-5) 10^{14}}$
$$\begin{aligned} \text{or}\quad h & =\frac{2 e}{5 \times 10^{14}}=\frac{2 \times 1.6 \times 10^{-19}}{5 \times 10^{14}} \\ & =6.4 \times 10^{-34} \mathrm{Js} \end{aligned}$$
For metal $B$, slope $=\frac{h}{e}=\frac{2.5}{(15-10) 10^{14}}$
$$\begin{aligned} \text{or}\quad h & =\frac{2.5 \times e}{5 \times 10^{14}}=\frac{2.5 \times 1.6 \times 10^{-19}}{5 \times 10^{14}} \\ & =8 \times 10^{-34} \mathrm{Js} \end{aligned}$$
Since, the value of $h$ from experiment for metals $A$ and $B$ is different. Hence, experiment is not consistent with theory.
A particle $A$ with a mass $m_A$ is moving with a velocity $v$ and hits a particle $B$ (mass $m_B$ ) at rest (one dimensional motion). Find the change in the de-Broglie wavelength of the particle $A$. Treat the collision as elastic.
As collision is elastic, hence laws of conservation of momentum and kinetic energy are obeyed.
According to law of conservation of momentum,
$$\begin{aligned} m_A v+m_B 0 & =m_A v_1+m_B v_2 \\ m_A\left(v-v_1\right) & =m_B v_2 \end{aligned}$$
According to law of conservation of kinetic energy,
$$\begin{aligned} & \frac{1}{2} m_A v^2=\frac{1}{2} m_A v_1^2+\frac{1}{2} m_B v_2^2 \quad\text{.... (i)}\\ & \Rightarrow \quad m_A\left(v-v_1^2\right)=m_B v_2^2 \\ & \Rightarrow \quad m_A\left(v-v_1\right)\left(v+v_1\right)=m_B v_2{ }^2\quad\text{.... (ii)} \end{aligned}$$
Dividing Eq. (ii) by Eq. (i),
we get, $v+v_1=v_2 \quad$ or $\quad v=v_2-v_1\quad\text{.... (iii)}$
Solving Eqs. (i) and (iii), we get
$$\begin{aligned} v_1 & =\left(\frac{m_A-m_B}{m_A+m_B}\right) v \text { and } v_2=\left(\frac{2 m_A}{m_A+m_B}\right) v \\ \lambda_{\text {initial }} & =\frac{h}{m_A v} \\ \lambda_{\text {final }} & =\frac{h}{m_A v_1}=\frac{h\left(m_A+m_B\right)}{m_A\left(m_A-m_B\right) v} \\ \Delta \lambda & =\lambda_{\text {final }}-\lambda_{\text {initial }}=\frac{h}{m_A v}\left[\frac{m_A+m_B}{m_A-m_B}-1\right] \end{aligned}$$
Consider a 20 W bulb emitting light of wavelength $5000 \mathop A\limits^o$ and shining on a metal surface kept at a distance 2 m . 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 $1.5 \mathop A\limits^o$.
(i) Estimate number 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?
Given,
$$\begin{aligned} P & =20 \mathrm{~W}, \lambda=5000 \mathop A\limits^o=5000 \times 10^{-10} \mathrm{~m} \\ d & =2 \mathrm{~m}, \phi_0=2 \mathrm{eV}, r=1.5 \mathrm{~A}=1.5 \times 10^{-10} \mathrm{~m} \end{aligned}$$
$$\begin{aligned} &\text { (i) Number of photon emitted by bulb per second is } n^{\prime}=\frac{p}{h c / \lambda}=\frac{p \lambda}{h c}\\ &\begin{aligned} & =\frac{20 \times\left(5000 \times 10^{-10}\right)}{\left(6.62 \times 10^{-34}\right) \times\left(3 \times 10^8\right)} \\ \Rightarrow\quad & =5 \times 10^{19} \mathrm{~s}^{-1} \end{aligned} \end{aligned}$$
$$\begin{aligned} \text {(ii) Energy of the incident photon } & =\frac{h c}{\lambda}=\frac{\left(6.62 \times 10^{-34}\right)\left(3 \times 10^8\right)}{5000 \times 10^{-10} \times 1.6 \times 10^{-19}} \\ & =2.48 \mathrm{eV} \end{aligned}$$
As this energy is greater than 2 eV (i.e., work function of metal surface), hence photoelectric emission takes place.
(iii) Let $\Delta t$ be the time spent in getting the energy $\phi=$ (work function of metal). Consider the figure,
$$\begin{aligned} \frac{P}{4 \pi d^2} \times \pi r^2 \Delta t & =\phi_0 \\ \Rightarrow\quad\Delta t & =\frac{4 \phi_0 d^2}{P r^2} \\ & =\frac{4 \times\left(2 \times 1.6 \times 10^{-19}\right) \times 2^2}{20 \times\left(1.5 \times 10^{-10}\right)^2} \approx 28.4 \mathrm{~s} \end{aligned}$$
$$\begin{aligned} &\text { (iv) Number of photons received by atomic disc in time } \Delta t \text { is }\\ &\begin{aligned} N & =\frac{n^{\prime} \times \pi r^2}{4 \pi d^2} \times \Delta t \\ \Rightarrow\quad & =\frac{n^{\prime} r^2 \Delta t}{4 d^2} \\ & =\frac{\left(5 \times 10^{19}\right) \times\left(1.5 \times 10^{-10}\right)^2 \times 28.4}{4 \times(2)^2} \approx 2 \end{aligned} \end{aligned}$$
(v) As time of emission of electrons is 11.04 s.
Hence, the photoelectric emission is not instantaneous in this problem.
In photoelectric emission, there is an collision between incident photon and free electron of the metal surface, which lasts for very very short interval of time ( $\approx 10^{-9} \mathrm{~S}$ ), hence we say photoelectric emission is instantaneous.