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The K.E. of photoelecttron emitted from ...

The K.E. of photoelecttron emitted from a metal are `K_(1)` and `K_(2)` ,when it is irradiated with lights of wavelength `lambda_(1)` and `lambda_(2)` respectively .The work function of metal is …..

A

`(K_(1)lambda_(1)-K_(2)lambda_(2))/(lambda_(2)-lambda_(1))`

B

`(K_(1)lambda_(2)-K_(2)lambda_(1))/(lambda_(2)-lambda_(1))`

C

`(K_(1)lambda_(1)+K_(2)lambda_(2))/(lambda_(2)+lambda_(1))`

D

`(K_(1)lambda_(2)+K_(2)+lambda_(1))/(lambda_(2)+lambda_(1))`

Text Solution

Verified by Experts

The correct Answer is:
A
`(hc)/(lambda)-phi=K`
`therefore hc=philambda_(1)+K_(1)lambda_(1)` and `hc=philambda_(2)+K_(2)lambda_(2)`
`therefore philambda_(1)+K_(1)lambda_(1)=philambda_(2)+K_(2)lambda_(2)`
`therefore phi=(K_(1)lambda_(1)-K_(2)lambda_(2))/(lambda_(2)-lambda_(1))`
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