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 VS is given by

eVS = hv – W ⇒ VS = h/W v- w/e

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

when v2 = 8 × 1015 Hz, Vs = Vs2 (say)

∴ Vs1 = h/e V1 - w/e

Vs2 = h/e v2 - w/e

Subtracting Vs2 - Vs1 = h/e (v2 -v1)

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

= 16 volt.