Question

By how much would the stopping potential for a given photosensitive surface go up if the frequency of the incident radiations were to be increased from 4 × 10¹⁵ Hz to 8 × 10¹⁵ Hz?

Answer 1

Stopping potential V_S is given by

eV_S = hv – W ⇒ V_S = h/W v- w/e

when v₁ = 4 × 10¹⁵ Hz, V_s = S₁ V (say)

when v₂ = 8 × 10¹⁵ Hz, V_s = V_s2 (say)

∴ V_s1 = h/e V₁ - w/e

V_s2 = h/e v₂ - w/e

Subtracting V_s2 - V_s1 = h/e (v₂ -v₁)

= (6.4/10^-34/1.6 x10^-19) (8 x 10^-15 -4 x 10¹⁵)

= 16 volt.