Question

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 :-

Answer 1

`(1)/(2) mv^(2)=h(v-v_(0))=h((c)/(lambda)-(c)/(lambda_(0)))`
`=hc((1)/(lambda)-(c)/(lambda_(0)))=hc((lambda_(0)-lambda)/(lambda_(0)lambda))`
`v^(2)=(2hc)/(m)((lambda_(0)-lambda)/(lambda_(0)lambda)) or v= [(2hc)/(m)((lambda_(0)-lambda)/(lambda_(0)lambda))]^(1/2)`