A surface irradiated with light of wavelength 480 nm gives out electrons with maximum velocity `v m s^(-1)`, the cut off wavelength being 600 nm. The same surface would release electrons with maximum velocity `2v ms^(-1)` if it is irradiated by light of wavelength

Options

A. 325 nm
B. 360nm
C. 384 nm
D. 300 nm

Answer

D

Explanation

By photoelectric equation `=(1)/(2)mv^(2)=(hc)/(lamda)-(hc)/(lamda_0)` where `lamda` is the wavelength of incident radiation and `lamda_0` is the threshold wavelength
`(1)/(2)m(2v)^(2)=hc((1)/(lamda')-(1)/(lamda_0))`
Solving for `lamda'`, we get `lamda'=300nm`