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Question
When monochromatic light of frequency v1 falls on a metal surface, the stopping potential required is found to be V1. If the radiation of frequency v2 is incident on the surface, the stopping potential required V2 is ______. (v2 > v1)
Options
V1 - `"h"/"e"` (v2 + v1)
V1 + `"h"/"e"` (v1 - v2)
V1 + `"h"/"e"` (v2 - v1)
V1 - `"h"/(2"e")` (v1 - v2)
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Solution
When monochromatic light of frequency v1 falls on a metal surface, the stopping potential required is found to be V1. If the radiation of frequency v2 is incident on the surface, the stopping potential required V2 is `bbunderline("V"_1 + "h"/"e" ("v"_2 - "v"_1))`.
Explanation:
Let the work function of the metal be ϕ
Stopping potential of the metal is equal to the maximum kinetic energy of the photoelectrons emitted.
∴ eV = K.Emax = hv - ϕ
For frequency v1 :
eV1 = hv1 - ϕ ...(i)
For frequency v2 :
eV2 = hv2 - ϕ ...(ii)
From equation (2) - (1) we get,
e(V2 − V1) = hv2 − hv1
⇒ V2 = V1 + `"e"/"h"` (V2 − V1)
