Question

The work function of caesium is 2.14 eV. Find 

(i) the threshold frequency for caesium 

(ii) the wavelength of the incident light if photocurrent is brought to zero by a stopping potential of 0.60 V.

Answer 1

Data : Φ = 2.14eV = 2.14 × 1.6 × 10^-19 J = 3.424 × 10^-19 J, 

V₀ = 0.60 V,h = 6.63 × 10^-34 J∙s, 

c = 3 × 10⁸ m/s, 

e = 1.6 × 10^-19 C

(i) Φ = hv₀

∴ The threshold frequency, v₀ = \(\cfrac ϕh\)

(ii) V₀ e = 0.6 × 1.6 × 10^-19 = 0.96 × 10^-19 J

\(\cfrac{hc}λ\)  – Φ = V₀e

∴ \(\cfrac{hc}λ\) = Φ + V₀e

∴ λ = \(\cfrac{hc}{ϕ+V_Oe}\)

This is the required wavelength of the incident light.