1.

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

Answer»

`hc(lamda_(0)-lamda)`
`(hc)/(lamda_(0)-lamda)`
`(h)/(c)((lamda_(0)-lamda)/(lamdalamda_(0)))`
`hc((lamda_(0)-lamda)/(lamdalamda_(0)))`

Solution :`hc[(1)/(LAMBDA)-(1)/(lambda_(0))]=HCP[(lambda_(0)-lambda)/(lambda lambda_(0))]`


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