Solution Q10: Force on Mirror
Analysis: Light travels in $+x$ direction ($\hat{i}$). The mirror geometry reflects it into $+y$ direction (Based on Area vector $A(-\hat{i}+\hat{j})$). Therefore, the reflected momentum is along $\hat{j}$.
$\Delta \vec{p}_{photon} = \vec{p}_f – \vec{p}_i = \frac{h}{\lambda}(\hat{j}) – \frac{h}{\lambda}(\hat{i}) = \frac{h}{\lambda}(-\hat{i} + \hat{j})$. (Matches Option a)
Force on Mirror = Rate of momentum transfer = $\frac{P}{c} (\vec{p}_i – \vec{p}_f)_{unit} = \frac{P}{c} (\hat{i} – \hat{j})$. (Matches Option c)
Solution Q11: Electric Field
Charge accumulates on the plate due to photoemission. $Q = i \Delta t$.
Current $i = \eta \times \frac{P}{hc/\lambda} \times e$.
Surface Charge Density $\sigma = \frac{Q}{A}$. Electric Field $E = \frac{\sigma}{2\epsilon_0}$.
$$ E = \frac{\eta P \lambda e \Delta t}{hc A \cdot 2\epsilon_0} $$Solution Q12: Photoelectric Parameters
(a) Range of KE is always 0 to $K_{max}$. Correct.
(b) $K_{max} = \frac{12420}{6000} – 1.9 = 2.07 – 1.9 = 0.17$ eV. Correct.
(c) If $\phi = 2.1$ eV and $E_{ph} = 2.07$ eV, no emission. Correct.
(d) Rate $= \eta \frac{P}{E_{ph}} = 10^{-4} \frac{60}{3.31 \times 10^{-19}} \approx 1.8 \times 10^{16}$. Correct.