Consider figure for photoemission. How would you reconcile with momentum-conservation? Note light (photons) have momentum in a different direction than the emitted electrons.
During photoelectric emission, the momentum of incident photon is transferred to the metal. At microscopic level, atoms of a metal absorb the photon and its momentum is transferred mainly to the nucleus and electrons.
The excited electron is emitted. Therefore, the conservation of momentum is to be considered as the momentum of incident photon transferred to the nucleus and electrons.
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.
Given,
For the first condition, Wavelength of light $\lambda=600 \mathrm{~nm}$ and for the second condition, Wavelength of light $\lambda^{\prime}=400 \mathrm{~nm}$
Also, maximum kinetic energy for the second condition is equal to the twice of the kinetic energy in first condition.
i.e., $$K_{\max }^{\prime}=2 K_{\max }$$
$$\begin{aligned} & \begin{aligned} \text { Here, } \quad K_{\max }^{\prime} & =\frac{h c}{\lambda}-\phi \\ \Rightarrow \quad 2 K_{\max } =\frac{h c}{\lambda^{\prime}}-\phi_0 \\ \Rightarrow \quad 2\left(\frac{1230}{600}-\phi\right) & =\left(\frac{1230}{400}-\phi\right) \quad[\because h c \approx 1240 \mathrm{eVnm}] \\ \Rightarrow \quad \phi =\frac{1230}{1200}=1.02 \mathrm{eV} \end{aligned} \end{aligned}$$
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,
$$ \begin{aligned} & \frac{h}{\lambda_C}=\frac{h}{\lambda_A}-\frac{h}{\lambda_B}=\frac{\left(\lambda_B-\lambda_A\right) h}{\lambda_A \lambda_B} \\ \Rightarrow\quad & \lambda_C=\frac{\lambda_A \lambda_B}{\lambda_B-\lambda_A} \end{aligned}$$
$$\begin{aligned} &\text { Case IV } p_A<0, p_B>0 \text {, i.e., } p_A \text { is negative and } p_B \text { is positive, }\\ &\begin{array}{rlrl} \therefore & \frac{h}{\lambda_C} =\frac{-h}{\lambda_A}+\frac{h}{\lambda_B} \\ \Rightarrow & =\frac{\left(\lambda_A-\lambda_B\right) h}{\lambda_A \lambda_B} \Rightarrow \lambda_C=\frac{\lambda_A \lambda_B}{\lambda_A-\lambda_B} \end{array} \end{aligned}$$