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

Options

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

Answer

D

Explanation

`hv - W_(0) = (1)/(2) mv_(max)^(2)`
`rArr (hc)/(lambda) - (hc)/(lambda_(0)) = (1)/(2) mv_(max)^(2)`
`rArr hc((lambda_(0) - lambda)/(lambda lambda_(0))) = (1)/(2) mv_(max)^(2) v_(max) = sqrt((2hc)/(m) ((lambda_(0) - lambda)/(lambda . lambda_(0))))`
When lavelength is `lambda` and velocity is `v` , then
`v = sqrt((2hc)/(m)((lambda_(0) - lambda)/(lambdalambda_(0))))` .....(i)
When wavelength is `3 lambda//4` and velocity is `v'` , then `4`
`v' = sqrt(( 2hc)/(m) [(lambda_(0) - ( 3lambda//4))/((3 lambda //4) xx lambda_(0))])` .....(ii)
Dividing Eqs. (ii) by (i) , we get
`(v')/(v) = sqrt(([lambda_(0) - ( 3lambda//4)])/((3)/(4) lambda lambda_(0)) xx(lambda lambda_(0))/(lambda_(0) - lambda))`
`v' = v((4)/(3))^(1//2) sqrt(([lambda_(0) - (3 lambda//4)])/(lambda_(0) - lambda))`
i.e., `v' gt v((4)/(3))^(1//2)`