Radiation of wavelength `lambda`, is incident on a photocell. The fastest emitted electron has speed v. If the wavelength is changed to `(3lambda)/(4)`, the speed of the fastest emitted electron will be :

Options

A. `v((3)/(4))^((1)/(2))`
B. `v((4)/(3))^((1)/(2))`
C. less than `v((4)/(3))^((1)/(2))`
D. greater than `v((4)/(3))^((1)/(2))`

Answer

D

Explanation

`(1)/(2)mv^2=(4hc)/(3lamda)-W_0`
Let the speed of the fastest electron be `V_1` when excitation wavelength is changed to `(3lamda)/(4)`.
`(1)/(2)mv_1^2=(4hc)/(3lamda)-W_0`
`implies(1)/(2)mv_1^2=(4)/(3)((hc)/(lamda)-W_0)+(W_0)/(3)`
`implies(1)/(2)mv_1^2=(4)/(3)((1)/(2)mv^2)+(W_0)/(3)`
`impliesv_1^2=(4v^2)/(3)+(2W_0)/(3m)`
`v_1gtsqrt((4)/(3))v`