Photoelectric Effect Calculator

• Pick wavelength λ or frequency f.
• If you know the metal, enter its work function φ and compute K_max and V_s.
• If you’re given stopping potential V_s or K_max, switch to Solve for work function.

• Photon energy: E = hf = hc/λ
• Einstein equation: K_max = hf − φ
• Stopping potential: eV_s = K_max
• Threshold: f₀ = φ/h, λ₀ = hc/φ

Example 1 — Sodium illuminated with 300 nm UV

Light of λ = 300 nm hits sodium with work function φ = 2.28 eV. Find K_max and V_s.

1. Photon energy: E = hc/λ ≈ 1240 eV·nm / 300 nm ≈ 4.13 eV
2. Max kinetic energy: K_max = E − φ ≈ 4.13 − 2.28 = 1.85 eV
3. Stopping potential: V_s = K_max/e ⇒ V_s ≈ 1.85 V

Example 2 — No emission (below threshold)

Light of λ = 600 nm hits a metal with work function φ = 2.50 eV. Determine whether electrons are emitted, and find K_max and V_s.

1. Photon energy: E \approx 1240 / λ(nm) ⇒ E \approx 1240 / 600 \approx 2.07 eV.
2. Compare to work function: E = 2.07 eV is less than φ = 2.50 eV. So no electrons are emitted.
3. Maximum kinetic energy: K_max = E − φ would be negative, so we report K_max = 0 eV.
4. Stopping potential: V_s = K_max(eV) ⇒ V_s = 0 V.

Key idea: Increasing intensity won’t help if E < φ. You need higher frequency (shorter λ).

Example 3 — Solve for work function from stopping potential

Light of λ = 350 nm produces a stopping potential of V_s = 1.00 V. Find the metal’s work function φ and the threshold values f₀ and λ₀.

1. Photon energy: E \approx 1240 / 350 \approx 3.54 eV.
2. Convert stopping potential to kinetic energy: K_max(eV) = V_s(V) ⇒ K_max = 1.00 eV.
3. Solve for work function: φ = E − K_max ⇒ φ \approx 3.54 − 1.00 = 2.54 eV.
4. Threshold frequency: f₀ = φ/h. (Numerically this will be around f₀ \approx 6.1 × 10¹⁴ Hz for φ \approx 2.54 eV.)
5. Threshold wavelength: λ₀ = hc/φ. Using hc \approx 1240 eV·nm: λ₀ \approx 1240 / 2.54 \approx 488 nm.

Nice shortcut: when using eV and nm, E(eV) \approx 1240/λ(nm) and λ₀(nm) \approx 1240/φ(eV).

Example 4 — Solve for work function from frequency and Kmax

Light with frequency f = 800 THz produces a maximum kinetic energy of K_max = 1.50 eV. Find the metal’s work function φ, and the threshold values f₀ and λ₀.

1. Convert frequency to Hz: 800 THz = 800 × 10¹² Hz = 8.00 × 10¹⁴ Hz.
2. Photon energy: E = hf. Using h = 4.136 × 10⁻¹⁵ eV·s: E \approx (4.136 × 10⁻¹⁵)(8.00 × 10¹⁴) \approx 3.31 eV.
3. Solve for work function: φ = E − K_max ⇒ φ \approx 3.31 − 1.50 = 1.81 eV.
4. Threshold frequency: f₀ = φ/h. Using h = 4.136 × 10⁻¹⁵ eV·s: f₀ \approx 1.81 / (4.136 × 10⁻¹⁵) \approx 4.37 × 10¹⁴ Hz (≈ 437 THz).
5. Threshold wavelength: λ₀ = hc/φ. Using hc \approx 1240 eV·nm: λ₀ \approx 1240 / 1.81 \approx 686 nm.

Shortcut for energy from frequency: E(eV) \approx (4.136 × 10⁻¹⁵) · f(Hz). For THz: E(eV) \approx 0.004136 · f(THz).