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Photoelectric emission is observed from ...

Photoelectric emission is observed from a surface for frequencies `v_1` and `v_2` of the incident radiation `(v_1 gt v_2)`. If the maximum kinetic energies of the photoelectrons in two cases are in ratio `1 : K` then the threshold frequency `v_0` is given by.

A

`(v_2 - v_1)/(K-1)`

B

`(Kv_1 - v_2)/(K-1)`

C

`(Kv_2 - v_2)/(K-1)`

D

`(v_2 - v_2)/(K)`

Text Solution

Verified by Experts

The correct Answer is:
B
`hv_1 = hv_0 + 1 rArr h [v_1 - v_0] = 1 ------(1)`
`hv_2 = hv_0 + k rArr h[v_2 - v_0]=k------(2)`
`((1))/((2)) = (1)/(K) = (v_1 - v_0)/(v_2 - v_0)`
`v_2 - v_0 = Kv_1 - Kv_0 " "Kv_0 - v_0 = Kv_1 - v_2`
`v_0 [K-1] = Kv_1 - v_2 " "v_0 = (Kv_1 - v_2)/(K-1)`
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