Numerical 1: Find KE_max, $V_{0}$, and threshold wavelength

Problem: Light of wavelength 300 nm falls on a metal surface with work function 3.0 eV. Find: (a) photon energy in eV, (b) KE_max in eV and joules, (c) stopping potential $V_{0}$, (d) threshold wavelength.

Given:

• λ = 300 nm = $300 \times 10^{-9}$ m
• φ = 3.0 eV = 3.0 × $1.6 \times 10^{-19}$ J = $4.8 \times 10^{-19}$ J
• h = $6.63 \times 10^{-34}$ J·s; c = $3 \times 10^{8}$ m/s; hc = 1240 eV·nm

(a) Photon energy:

$E = \frac{hc}{\lambda} = \frac{1240 \text{ eV·nm}}{300 \text{ nm}} = 4.13 \text{ eV}$

(b) KE_max:

$KE_{\max} = E - \phi = 4.13 \text{ eV} - 3.0 \text{ eV} = 1.13 \text{ eV}$

$= 1.13 \times 1.6 \times 10^{-19} \text{ J} = 1.81 \times 10^{-19} \text{ J}$

Since E (4.13 eV) > φ (3.0 eV), emission occurs. ✓

(c) Stopping potential:

$eV_0 = KE_{\max} \implies V_0 = \frac{KE_{\max}}{e} = \frac{1.13 \text{ eV}}{e} = 1.13 \text{ V}$

(d) Threshold wavelength:

$\lambda_0 = \frac{hc}{\phi} = \frac{1240 \text{ eV·nm}}{3.0 \text{ eV}} = 413.3 \text{ nm}$

Since λ (300 nm) < λ_{0} (413 nm), frequency is above threshold — consistent with emission occurring. ✓