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If lambda(0) is the threshold wavelengt...

If `lambda_(0)` is the threshold wavelength for photoelectric emission. `lambda` wavelength of light falling on the surface on the surface of metal, and `m` mass of electron. Then de Broglie wavelength of emitted electron is :-

A

`[(2h)/(m) (lamda_(0) - lamda)]^(1//2)`

B

`[(2hc)/(m) (lamda_(0) - lamda)]^(1//2)`

C

`[(2hc)/(m) xx (lamda_(0) - lamda)/(lamda_(0)lamda)]^(1//2)`

D

`[(2hc)/(m) ((1)/(lamda_(0)) -(1)/(lamda))]^(1//2)`

Text Solution

Verified by Experts

The correct Answer is:
C
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