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?
Suppose $n_A$ is the number of photons falling per second of beam $A$ and $n_B$ is the number of photons falling per second of beam $B$.
$$\begin{aligned} \text{Thus,}\quad n_A & =2 n_B \\ \text{Energy of falling photon of beam}\quad A & =h v_A \\ \text{Energy of falling photon of beam }\quad B & =h v_B \end{aligned}$$
Now, according to question,
$$\begin{aligned} &\text { intensity of } A=\text { intensity of } B\\ &\begin{aligned} & \therefore \quad n_A h \nu_A=n_B h \nu_B \\ \Rightarrow\quad & \frac{\nu_A}{\nu_B}=\frac{n_B}{n_A}=\frac{n_B}{2 n_B}=\frac{1}{2} \\ \Rightarrow\quad & \nu_B=2 \nu_A \end{aligned} \end{aligned}$$
Thus, from this relation we can infer that frequency of beam B is twice of beam A.
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)
Given from conservation of momentum,
$$\left|\mathbf{p}_C\right|=\left|\mathbf{p}_A\right|+\left|\mathbf{p}_B\right|$$
$$\begin{aligned} \Rightarrow\quad & \frac{h}{\lambda_C}=\frac{h}{\lambda_A}+\frac{h}{\lambda_B} \quad\left[\because \lambda=\frac{h}{m v}=\frac{h}{p} \Rightarrow p=\frac{h}{\lambda}\right] \\ \Rightarrow\quad & \frac{h}{\lambda_C}=\frac{h \lambda_B+h \lambda_A}{\lambda_A \lambda_B} \\ \Rightarrow\quad & \frac{\lambda_C}{h}=\frac{\lambda_A \lambda_B}{h \lambda_A+h \lambda_B} \Rightarrow \lambda_C=\frac{\lambda_A \lambda_B}{\lambda_A+\lambda_B} \end{aligned}$$
Case I Suppose both $p_A$ and $p_B$ are positive, then
$$\lambda_C=\frac{\lambda_A \lambda_B}{\lambda_A+\lambda_B}$$
Case II When both $p_A$ and $p_B$ are negative, then
$$\lambda_C=\frac{\lambda_A \lambda_B}{\lambda_A+\lambda_B}$$
Case III When $p_A>0, p_B<0$ i.e., $p_A$ is positive and $p_B$ is negative,
$$ \begin{aligned} & \frac{h}{\lambda_C}=\frac{h}{\lambda_A}-\frac{h}{\lambda_B}=\frac{\left(\lambda_B-\lambda_A\right) h}{\lambda_A \lambda_B} \\ \Rightarrow\quad & \lambda_C=\frac{\lambda_A \lambda_B}{\lambda_B-\lambda_A} \end{aligned}$$
$$\begin{aligned} &\text { Case IV } p_A<0, p_B>0 \text {, i.e., } p_A \text { is negative and } p_B \text { is positive, }\\ &\begin{array}{rlrl} \therefore & \frac{h}{\lambda_C} =\frac{-h}{\lambda_A}+\frac{h}{\lambda_B} \\ \Rightarrow & =\frac{\left(\lambda_A-\lambda_B\right) h}{\lambda_A \lambda_B} \Rightarrow \lambda_C=\frac{\lambda_A \lambda_B}{\lambda_A-\lambda_B} \end{array} \end{aligned}$$
A neutron beam of energy $E$ scatters from atoms on a surface with a spacing $d=0.1 \mathrm{~nm}$. The first maximum of intensity in the reflected beam occurs at $\theta=30 \Upsilon$. What is the kinetic energy $E$ of the beam in eV ?
$$\begin{aligned} &\text { Given, } d=0.1 \mathrm{~nm} \text {, }\\ &\theta=30 \Upsilon \Rightarrow n=1 \end{aligned}$$
$$\begin{aligned} &\text { Now, according to Bragg's law }\\ &\begin{array}{rlrl} & 2 d \sin \theta =n \lambda \Rightarrow 2 \times 0.1 \times \sin 30=1 \lambda \\ \Rightarrow \quad \lambda & =0.1 \mathrm{~nm} \Rightarrow=10^{-10} \mathrm{~m} \end{array} \end{aligned}$$
$$\begin{array}{lrl} \text { Now, } & \lambda & =\frac{h}{m v}=\frac{h}{p} \\ \Rightarrow & p=\frac{h}{\lambda} & =\frac{6.62 \times 10^{-34}}{10^{-10}} \\ \Rightarrow & & =6.62 \times 10^{-24} \mathrm{~kg}-\mathrm{m} / \mathrm{s} \\ \text { Now, } & \mathrm{KE} & =\frac{1}{2} m v^2=\frac{1}{2} \frac{m^2 v^2}{m}=\frac{1}{2} \frac{p^2}{\mathrm{~m}} \\ & & =\frac{1}{2} \times \frac{\left(6.62 \times 10^{-24}\right)^2}{1.67 \times 10^{-27}} \mathrm{~J} \\ & & =0.21 \mathrm{eV} \end{array}$$
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
$$\begin{aligned} W & =\int_d^{\infty} F_x d x=\int_d^{\infty} \frac{q^2 d x}{4 \times 4 \pi \varepsilon_0} \frac{1}{x^2} \\ & =\frac{q^2}{4 \times 4 \pi \varepsilon_0}\left[\frac{1}{d}\right] \\ & =\frac{\left(1.6 \times 10^{-19}\right)^2 \times 9 \times 10^9}{4 \times 10^{-10}} \mathrm{~J} \\ & =\frac{\left(1.6 \times 10^{-19}\right)^2 \times\left(9 \times 10^9\right)}{\left(4 \times 10^{-10}\right) \times\left(1.6 \times 10^{-19}\right)} \mathrm{eV}=3.6 \mathrm{eV} \end{aligned}$$