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)])^(½)`