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
$$\begin{aligned} W & =\int_d^{\infty} F_x d x=\int_d^{\infty} \frac{q^2 d x}{4 \times 4 \pi \varepsilon_0} \frac{1}{x^2} \\ & =\frac{q^2}{4 \times 4 \pi \varepsilon_0}\left[\frac{1}{d}\right] \\ & =\frac{\left(1.6 \times 10^{-19}\right)^2 \times 9 \times 10^9}{4 \times 10^{-10}} \mathrm{~J} \\ & =\frac{\left(1.6 \times 10^{-19}\right)^2 \times\left(9 \times 10^9\right)}{\left(4 \times 10^{-10}\right) \times\left(1.6 \times 10^{-19}\right)} \mathrm{eV}=3.6 \mathrm{eV} \end{aligned}$$
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.