If `I_(0)` is the threshold wavelength for photoelectric emission, 1 the wavelength of light falling on the surface of a metal and m is the mass of the electron, then the velocity of ejected electron is given by

Options

A. `[(2h)/(m) (lambda- lambda)]^(1//2)`
B. `[(2hc)/(m) (lambda_0- lambda)]^(1//2)`
C. `[(2hc)/(m) {(lambda_0- lambda)/(lambda lambda_0)}]^(1//2)`
D. `[(2hc)/(m) {(1)/ (lambda_0)- (1)/(lambda)}]^(1//2)`

Answer

C

Explanation

`(hc)/(lambda)=h( c)/(lambda_0) + (1)/(2) mv^2 " "hc [ (1)/(lambda) - (1)/(lambda_0) ] = (1)/(2) mv^2`
`V = ((2hc)/(m) [ (lambda_0 - lambda)/(lambda_0 lambda)])^(1//2)`
`hv=w+K.E, w= hv - K.E`