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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

`sqrt((2h(lambda_(0)-lambda))/(m_(e)))`

B

`sqrt((2hc(lambda_(0)-lambda))/(m_(e)))`

C

`sqrt((2hc)/(m_(e))((lambda_(0) - lambda)/(lambdalambda_(0))))`

D

`sqrt((2h)/(m_(e))((1)/(lambda_(0))- (1)/(lambda)))`

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The correct Answer is:
C
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