Question

Photoemission of electrons occurs from a metal (Φ₀ = 1.96 eV) when light of frequency 64 × 10¹⁴ Hz is incident on it. Calculate:

1. Energy of a photon in the incident light,
2. The maximum kinetic energy of the emitted electrons, and
3. The stopping potential.

Answer 1

a. Photoelectric equation (E) = hν

= Φ₀ + K_max

Stopping potential (K_max) = eV₀

h = 6.63 × 10⁻³⁴ J . s

1 eV = 1.6 × 10⁻¹⁹ J

E = hν

= 6.63 × 10⁻³⁴ × 6.4 × 10¹⁴

= 4.24 × 10⁻¹⁹ J

Convert to eV:

E = `(4.24 xx 10^-19)/(1.6 xx 10^-19)`

= 2.65 eV

b. Maximum kinetic energy (K_max) = E − Φ₀

= 2.65 − 1.96

= 0.69 eV

c. Stopping potential (K_max) = eV₀

Since kinetic energy is in eV:

V₀ = 0.69 V