Question

The work function of caesium is 2.14 eV. Find (a) the threshold frequency for caesium and (b) the wavelength of incident light if the photocurrent is brought to zero by a stopping potential of 0.60 V.

Answer 1

(a) W = hv

Where W = work function

h = planck constant

v = threshold frequency

\(v = \frac Wh \)

\(= \frac{2.14\times 1.6 \times 10^{-19}}{6.6 \times 0^{-34}}\)

\(= 5.2 \times 10^{14}\) Hz

(b) Wavelength of incident photon = \(\frac{hc}{ev} + W\)

By substituting the values we get 

Wavelength = 4.5 × 10⁻⁷m

Answer 2

(a) Let v₀ be the threshold frequency. Then hv₀ = ω₀