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

Options

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

Answer

D

Explanation

According to Einstein.s photoelectric equation
`1/2 mv_(max)^2 = hupsilon-hv_0 " or " 1/2 mv_(max)^2 = (hc)/(lamda)-(hc)/(lamda_0)`
where `lamda` is the wavelength of incident radiation and `lamda_0` is threshold wavelength .
`:. 1/2 mv^(2) = hc (1/lamda -1/lamda_0) " "...(i)`
`1/2 m(2v)^2 =hc (1/ lamda. -1/lamda_(0)) " "...(ii)`
Divide (i) by (ii) , we get
`1/4=(1/lamda-1/lamda_0)/(1/lamda.-1/lamda_0)" or " 1/4 =(1/480-1/600)/(1/lamda.-1/600)`
Solving for `lamda. , ` we get `lamda. = 300 nm `