Question

The work function of a metal is 2.31 eV. Photoelectric emission occurs when the light of frequency 6.4 × 10¹⁴ Hz is incident on the metal surface. Calculate

1. the energy of the incident radiation,
2. the maximum kinetic energy of the emitted electron and
3. the stopping potential of the surface.

Answer 1

(i) Frequency of incident radiation = ν = 6.4 × 10¹⁴ Hz

Energy of incident radiation = E = hν = 6.6 × 10⁻³⁴ × 6.4 × 10¹⁴ = 42.24 × 10⁻²⁰ J

(ii) KE_max = hν − Φ₀

∴ KE_max = 42.24 × 10⁻²⁰ − 2.31 × 1.6 × 10⁻¹⁹ = 5.28 × 10⁻²⁰ J

(iii) If stopping potential = V_S, then

eV_S = KE_max

∴ V_S = `"KE"_"max"/"e"`

= `(5.28 xx 10^-20)/(1.6 xx 10^-19)`

= 3.3 × 10⁻¹

= 0.33 V