Question

The stopping potential, when a metal with work function 0.6 eV is illuminated with the radiation of energy 2 eV will be 

(a) 3 V 

(b) 2.4 V 

(c) 1.8 V 

(d) 1.4 V

Answer 1

Correct option: (d) 1.4 V

As, we know, K_max = hv - ϕ₀

= 2 eV - 0.6eV = 1.4 eV

∴ Stopping potential,

V0 = \(\frac{K_{max}}{e} = \frac{1.4\,eV}{e} = 1.4\,V\)