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(3/5)^(1//2)`
B. `gt v(4/3)^(1//2)`
C. `lt v(4/3)^(1//2)`
D. `=v(4/3)^(1//2)`

Answer

B

Explanation

`E=(KE)_("max")+f`
`[(hc)/lambda=(KE)_("max")+phi]` …(1)
`4/3 (hc)/lambda=(4/3 KE_("max")+phi/3)+phi`
`(KE)_("max")` for faster emitted electron`=1/2 mV'^(2)+phi`
`1/2 mV'^(2)=4/3 (1/2 mV^(2))+phi/3`
`V' gt V(4/3)^(1//2)`