If `lamda_o and lamda` be the threshold wavelength and wavelength of incident light , the velocity of photoelectron ejected from the metal surface is :

Options

A. `sqrt((2h)/(m)(lambda_(o) - lambda))`
B. `sqrt((2hc)/(m)(lambda_(o) - lambda))`
C. `sqrt((2hc)/(m)((lambda_(o) - lambda)/(lambda lambda_(o))))`
D. `sqrt((2h)/(m)((1)/(lambda_(o))-(1)/(lambda)))`

Answer

C

Explanation

The kinetic energy of the ejected electron is given by the equation .
`hv = hv_(o) + (1)/(2) mv^(2)" "because v = (c)/(lambda)`
or `(hc)/(lambda) = (hc)/(lambda_(o)) + (1)/(2) mv^(2)`
`(1)/(2)mv^(2) = (hc)/(lambda) - (hc)/(lambda_(o)) = hc ((lambda_(o) - lambda)/(lambda lambda_(o)))`
`therefore " " v^(2) = (2hc)/(m) ((lambda_(o) - lambda)/(lambda lambda_(o)))`
or `" " v = sqrt((2h)/(m)((lambda_(o) - lambda)/(lambda lambda_(o))))`