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When light of frequency v1 incident on a...

When light of frequency `v_1` incident on a metal with work function `W_0` (where `hv_1 gt W_0),` the photocurrent falls to zero at a stopping potential of `V_1.` If the frequency of light is increased to `v_2,` the stopping potential changes to `V_2.` Therefore, the charge of an electron is given by

A

`(W(upsilon_2+upsilon_1))/(upsilon_1V_2+upsilon_1V_1)`

B

`(W(upsilon_2 + upsilon_1))/(upsilon_1V_1 +upsilon_2V_2)`

C

`(W(upsilon_2-upsilon_1))/(upsilon_1V_2-upsilon_2V_1)`

D

`(W(upsilon_2-upsilon_1))/(upsilon_1V_2-upsilon_1V_1)`

Text Solution

Verified by Experts

The correct Answer is:
C
Using Einstein.s equation, `h upsilon = W+ eV_0`
When the incident light frequency is `upsilon_2` then stopping potential is `V_2`
`therefore h upsilon_2 = W + eV_2`
from eqns. (i) and (ii) , `(upsilon_1)/(upsilon_2) = (W+eV_1)/(W +eV_2)`
On solving `W upsilon_(1) + eV_(2)upsilon_1 = W upsilon_(2) + eV_(1) upsilon_2`
`rArr e = (W(upsilo_2 - upsilon_1))/(upsilon_1 V_2 -upsilon_2V_1)`
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