Total photons per sec $n = P/E = 10^{-3} / (5 \times 1.6 \times 10^{-19}) = 1.25 \times 10^{15}$.
Intensity at surface $I' = I A_{surf} / 4\pi R^2 = \frac{P \times 10^{-4}}{4\pi (0.5)^2} = \frac{10^{-7}}{\pi}$.
Photons reaching surface $n_{reach} = n \frac{A}{4\pi R^2} = 1.25 \times 10^{15} \times \frac{10^{-4}}{4\pi (0.5)^2} \approx 3.98 \times 10^{10} / \text{s}$.
At 1% efficiency, electrons emitted $\approx \mathbf{3.98 \times 10^8 / \text{s}}$.
Initial freq $\nu_1 = 1.5\nu_0$. Halved freq $\nu_2 = 0.75\nu_0$.
Since $\nu_2 < \nu_0$ (threshold), photoelectric current will strictly drop to **Zero**, regardless of the intensity increase.
$\lambda_1 = 4000 \text{ \AA} \implies E_1 \approx 3.1 \text{ eV}$. $\lambda_2 = 4800 \text{ \AA} \implies E_2 \approx 2.58 \text{ eV}$.
Work function $\Phi = 3.7 \text{ eV}$. Since both $E_1$ and $E_2$ are less than $\Phi$, the number of photoelectrons is **Zero**.
$V_0 = (h/e)\nu - (\Phi/e)$. Intercept on freq axis is $\nu_0 = 5 \times 10^{14} \text{ Hz}$.
Intercept on potential axis is $-\Phi/e = -2.07 \text{ V} \implies \Phi = 2.07 \text{ eV}$.
$h = \Phi/\nu_0 = (2.07 \times 1.6 \times 10^{-19}) / 5 \times 10^{14} \approx \mathbf{6.62 \times 10^{-34} \text{ Js}}$. Consistent.
$\lambda_e = h/\sqrt{2m_eE}$. $\lambda_p = hc/E$.
Ratio $\lambda_e/\lambda_p = \frac{h}{\sqrt{2m_eE}} \cdot \frac{E}{hc} = \frac{\sqrt{E}}{c\sqrt{2m_e}} = \sqrt{\frac{E}{2m_ec^2}}$. Proved.
(i) $p = \sqrt{2mqV}$. $p_p = \sqrt{2m_p eV}$, $p_\alpha = \sqrt{2(4m_p)(2e)V} = \sqrt{8} p_p$. Ratio is **$1:2\sqrt{2}$**.
(ii) $\lambda \propto 1/p$. Ratio is **$2\sqrt{2}:1$**.
$\theta = 90 - 25 = 65^\circ$. $\lambda = 2(0.91)\sin 65^\circ = 1.65 \text{ \AA}$.
Theoretical $\lambda = 12.27/\sqrt{54} = 1.67 \text{ \AA}$. Error $\approx 1.2\%$.
$eV = hc/\lambda - hc/\lambda_0$ (1). $e(V/4) = hc/2\lambda - hc/\lambda_0$ (2).
Multiply (2) by 4: $eV = 2hc/\lambda - 4hc/\lambda_0$.
Equating: $hc/\lambda - hc/\lambda_0 = 2hc/\lambda - 4hc/\lambda_0 \implies 3hc/\lambda_0 = hc/\lambda$.
$\mathbf{\lambda_0 = 3\lambda}$.
$2\pi r = n\lambda$. For ground state $n=1$, $\lambda = 2\pi(0.53) \approx \mathbf{3.33 \text{ \AA}}$.
This wavelength exactly matches $h/mv$ for the electron in that orbit, ensuring a stationary wave pattern, supporting Bohr's quantized orbit postulate.
$P = (1+r-a)I/c$. For perfect reflection $r=1, a=0, P=2I/c$. For absorption $a=1, r=0, P=I/c$.
$n\lambda/2 = L$. For ground state $n=1, \lambda = 2L = 2 \text{ \AA}$.
$E = h^2 / 2m\lambda^2 = (12.27/2)^2 \approx \mathbf{37.6 \text{ eV}}$.
$K = (12.27/10)^2 \approx 1.5 \text{ eV}$.
$E_{photon} = \Phi + K = 2.5 + 1.5 = 4.0 \text{ eV}$.
$\lambda = 1242/4 \approx \mathbf{310.5 \text{ nm}}$.
$h = e(V_1 - V_2) / (\nu_1 - \nu_2)$. $\Phi = h\nu_1 - eV_1$.
$v_g = d\omega/dk = v_{particle}$. $v_p = \omega/k = c^2/v_{particle}$. Since $v < c$, $v_p > c$.
$I = \eta e (P\lambda/hc)$.
$\lambda = h/\sqrt{2mK} \implies K \propto 1/m$. $K_\alpha/K_d = m_d/m_\alpha = 2/4 = \mathbf{1/2}$.
Momentum is conserved: $p_1 = -p_2$. Since $\lambda = h/p$, magnitudes are same. Ratio is **$1:1$**.
$K = h^2/2m\lambda^2 = (12.27/0.01)^2 \approx \mathbf{1.5 \text{ MeV}}$. Relativistic mass should be used for better accuracy.
