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The photoelectric threshold wavelength f...

The photoelectric threshold wavelength for silver is `lamda_(0)`. The energy of the electron ejected from the surface of silver by an incident wavelength `lamda(lamdaltlamda_(0))` will be

A

`hc((lambda_(0)-lambda)/(lambda lambda_(0)))`

B

`hc(lambda_(0)-lambda)`

C

`(hc)/(lambda_(0)-lambda)`

D

`h/c((lambda_(0)-lambda)/(lambda lambda_(0)))`

Text Solution

Verified by Experts

The correct Answer is:
A
`E=W_(0)+K`
`:.K=E-W_(0)`
`=(hC)/(lambda)-(hC)/(lambda_(0))`
`=hC(1/lambda -1/lambda_(0))=hC((lambda_(0)-lambda)/(lambda lambda_(0)))`
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MARVEL PUBLICATION-ELECTRONS AND PHOTONS -TEST YOUR GRASP - 17
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