Assuming an electron is confined to a 1 nm wide region, find the uncertainty in momentum using Heisenberg uncertainty principle $(\Delta x \times \Delta p \approx h)$. You can assume the uncertainty in position $\Delta x$ as 1 nm . Assuming $p \approx \Delta p$, find the energy of the electron in electronvolts.
Here, $\Delta x=1 \mathrm{~nm}=10^{-9} \mathrm{~m}, \Delta p=$ ?
As $\Delta x \Delta p \approx h$
$$\therefore\quad\Delta p=\frac{h}{\Delta x}=\frac{h}{2 \pi \Delta x}$$
$$\begin{aligned} \Rightarrow \quad & =\frac{6.62 \times 10^{-34} \mathrm{Js}}{2 \times(22 / 7)\left(10^{-9}\right) \mathrm{m}} \\ & =1.05 \times 10^{-25} \mathrm{~kg} \mathrm{~m} / \mathrm{s} \end{aligned}$$
$$\begin{aligned} &\text { Energy, }\quad E=\frac{p^2}{2 m}=\frac{(\Delta p)^2}{2 m} \quad[\because p \approx \Delta p] \end{aligned}$$
$$ \begin{aligned} & =\frac{\left(1.05 \times 10^{-25}\right)^2}{2 \times 9.1 \times 10^{-31}} \mathrm{~J} \\ \Rightarrow\quad & =\frac{\left(1.05 \times 10^{-25}\right)^2}{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19}} \mathrm{eV} \\ & =3.8 \times 10^{-2} \mathrm{eV} \end{aligned}$$
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.