$K_1 = \frac{1}{2}mv_1^2 = h(2\nu_0) - h\nu_0 = h\nu_0$.
$K_2 = \frac{1}{2}mv_2^2 = h(5\nu_0) - h\nu_0 = 4h\nu_0$.
Dividing: $v_1^2/v_2^2 = 1/4 \implies \mathbf{v_1/v_2 = 1/2}$.
Energy of photon $E = h\nu = 6.63 \times 10^{-34} \times 10^{15} \text{ J} \approx 4.14 \text{ eV}$. Photoelectric emission occurs only if $E \ge \Phi_0$.
Since $4.14 \text{ eV} < 5.0 \text{ eV}$, emission will **NOT occur for Surface B**.
(i) $E = 1242 / 300 \approx 4.14 \text{ eV}$. $K_{max} = E - \Phi_0 = 4.14 - 3.3 = \mathbf{0.84 \text{ eV}}$.
(ii) $V_0 = K_{max}/e = \mathbf{0.84 \text{ V}}$.
$eV_{01} = hc/\lambda_1 - \Phi_0 \implies \Phi_0 = hc/\lambda_1 - eV_{01} = (1242/400) - 1.1 = 3.105 - 1.1 = 2.005 \text{ eV}$.
For $\lambda_2 = 300 \text{ nm}$: $eV_{02} = (1242/300) - 2.005 = 4.14 - 2.005 = \mathbf{2.135 \text{ V}}$.
(Image Placeholder)
The stopping potential is directly proportional to the frequency of incident light. Higher frequency photons have more energy, producing photoelectrons with higher kinetic energy, which require a higher negative potential to stop.
Slope of $V_0$ vs $\nu$ is $h/e$.
$h = \text{Slope} \times e = 4.12 \times 10^{-15} \times 1.6 \times 10^{-19} \approx \mathbf{6.59 \times 10^{-34} \text{ Js}}$.
$K_{max} = hc(1/\lambda - 1/\lambda_0) = (6.6 \times 10^{-34} \times 3 \times 10^8) \times (10^9) \times (1/400 - 1/500)$.
$K_{max} = 1.98 \times 10^{-16} \times (0.0005) = 9.9 \times 10^{-20} \text{ J}$.
$v = \sqrt{2 \times 9.9 \times 10^{-20} / 9 \times 10^{-31}} \approx \mathbf{4.69 \times 10^5 \text{ m/s}}$.
Saturation current occurs when all photoelectrons emitted from the cathode reach the anode. It depends on the number of electrons emitted per second, which depends on intensity. It does **not** depend on frequency (provided it is above threshold).
$P = n (hc/\lambda) \implies \lambda = nhc/P$.
$\lambda = (4 \times 10^{20} \times 6.63 \times 10^{-34} \times 3 \times 10^8) / 200 \approx \mathbf{397.8 \text{ nm}}$.
For a reflecting surface, change in momentum per photon is $2p = 2h/\lambda$.
$F = \text{Rate of change of momentum} = (n \times 2h/\lambda) = 2 P / c$.
Since $P = IA$, $\mathbf{F = 2IA/c}$.
$K_{max} = \frac{1}{2}mv^2 = 0.5 \times 9.1 \times 10^{-31} \times (6 \times 10^5)^2 \approx 1.638 \times 10^{-19} \text{ J} \approx 1.02 \text{ eV}$.
$h\nu = 4.14 \times (7.21 \times 10^{14} / 10^{15}) \approx 2.98 \text{ eV}$.
$\Phi_0 = 2.98 - 1.02 = 1.96 \text{ eV}$. $\nu_0 = \Phi_0/h \approx \mathbf{4.73 \times 10^{14} \text{ Hz}}$.
$\lambda_0 = hc / \Phi_0$. Since $\lambda_0 \propto 1/\Phi_0$, the metal with the **smaller work function (Sodium)** will have a longer threshold wavelength.
$\lambda_{0,Na} = 1242 / 2.3 \approx 540 \text{ nm}$. $\lambda_{0,Ag} = 1242 / 4.73 \approx 262 \text{ nm}$.
$eV_0 = h\nu - \Phi_0 \implies V_0 = (h/e)\nu - (\Phi_0/e)$. This is in form $y = mx + c$. The intercept on the potential axis is **$-\Phi_0/e$**, which relates to the work function.
$E_p = (6.63 \times 10^{-34} \times 3 \times 10^8) / (6 \times 10^{-7}) \approx 3.315 \times 10^{-19} \text{ J}$.
$n = P / E_p = 25 / 3.315 \times 10^{-19} \approx \mathbf{7.54 \times 10^{19} \text{ photons/s}}$.
Intensity represents the number of photons. More photons mean more collisions and more electrons (current), but individual photon energy (which determines $K_{max}$) depends only on frequency.
$\lambda_e = h/mv = 6.63 \times 10^{-34} / (9.1 \times 10^{-31} \times 3 \times 10^6) \approx \mathbf{2.42 \text{ \AA}}$.
$\lambda_{photon} = 1242 / 100 \approx \mathbf{12.42 \text{ \AA}}$. The electron wavelength is much smaller.
$\lambda = h / \sqrt{2mqV}$. For same $V$: $\lambda \propto 1/\sqrt{mq}$.
$\lambda_p/\lambda_\alpha = \sqrt{(4m_p \cdot 2e) / (m_p \cdot e)} = \sqrt{8} = \mathbf{2\sqrt{2}}$.
(i) $p = h/\lambda = 6.63 \times 10^{-34} / 10^{-9} = \mathbf{6.63 \times 10^{-25} \text{ kg m/s}}$.
(ii) $K_{photon} = pc$. $K_{electron} = p^2/2m$. Ratio $K_p/K_e = 2mc/p \approx \mathbf{8.2 \times 10^2}$.
$K = \frac{3}{2}k_BT$. $\lambda = h / \sqrt{3mk_BT} \approx \mathbf{0.28 \text{ \AA}}$.
(Image Placeholder)
The slope of the $\lambda$ vs $1/p$ graph is exactly equal to Planck's constant ($h$).
$\lambda = h / \sqrt{2mqV}$. $\lambda_A / \lambda_B = \sqrt{m_B/m_A} = \sqrt{2m/m} = \mathbf{\sqrt{2}}$.
For particle: $p = \sqrt{2mK} \implies \lambda = h/p$. For photon: $E = h\nu = hc/\lambda \implies \lambda = hc/E$. Photon doesn't have rest mass $m$.
Same $p \implies$ Same $\lambda$. $K = p^2/2m$. Since $m_e$ is smallest, **electron** has highest $K$.
$K = h^2 / 2m\lambda^2 = (6.63 \times 10^{-34})^2 / (2 \times 1.675 \times 10^{-27} \times (1.4 \times 10^{-10})^2) \approx \mathbf{6.69 \times 10^{-21} \text{ J}} \approx \mathbf{0.04 \text{ eV}}$.
$\lambda \propto 1/\sqrt{V} \implies \lambda_1/\lambda_2 = \sqrt{V_2/V_1} = \sqrt{4V/V} = \mathbf{2}$.
$n\lambda = 2\pi r_n$. $r_n = n^2 r_1 = 4 \times 0.53 = 2.12 \text{ \AA}$.
$2\lambda = 2\pi \times 2.12 \implies \lambda = \mathbf{6.66 \text{ \AA}}$.
$m_e v_e = m_p v_p \implies v_e/v_p = m_p/m_e \approx 1836$. The **electron** moves much faster.
$2\pi r = n\lambda = n(h/mv) \implies mvr = n(h/2\pi)$. This is Bohr's quantization condition.
$\lambda = h / \sqrt{3mk_BT}$. Neutrons are used because their wavelength at room temp ($\sim 1 \text{ \AA}$) is comparable to interatomic distances, enabling diffraction.
$\Delta p \ge h / 4\pi \Delta x$. From $p = h/\lambda$, $|\Delta \lambda| = (\lambda^2/h) \Delta p = \lambda^2 / 4\pi \Delta x$.
It verified the wave nature of electrons by showing they undergo diffraction when scattered by a crystal lattice (Nickel).
$\theta = 90^\circ - (50^\circ/2) = \mathbf{65^\circ}$.
$\lambda_{dB} = 12.27/\sqrt{54} = \mathbf{1.67 \text{ \AA}}$. $\lambda_{Bragg} = 2(0.91) \sin 65^\circ = \mathbf{1.65 \text{ \AA}}$. They are consistent.
The peak occurs due to constructive interference of waves scattered from different atomic planes, satisfying Bragg's condition.
As $V$ increases, $\lambda$ decreases. From $\lambda = 2d\sin\theta$, $\theta$ must decrease, so the scattering angle **$\phi$ will increase**.
It ensures that the electron beam is mono-energetic (all electrons have the same velocity/wavelength) and well-collimated.
It collects scattered electrons. It was movable to measure intensity at different scattering angles to map the diffraction pattern.
The graph shows a "bump" developing at 44V, becoming most prominent at 54V at $50^\circ$, and then diminishing at 64V.
Wavelength at 54V is $1.67 \text{ \AA}$, while Nickel spacing is $0.91 \text{ \AA}$ (comparable). This makes diffraction effects observable.
It proves that electrons "behave as waves" under these conditions, confirming the wave-particle duality predicted by de Broglie.
