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. Less than `v(4//3)^(1//2)`
D. Greater than `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) rArr v_(max) = sqrt((2 h c)/(m)((lambda_(0) - lambda)/(lambda lambda_(0))))`
When wavelength is `lambda` and velocity is `v`, then
`v = sqrt(( 2hc)/(m) ((lambda_(0) - lambda)/(lambda lambda_(0))))` (i)
When wavelength is `( 3lambda)/(4)` and velocity is `v'` then
`v' = sqrt(( 2hc)/(m) [ (lambda_(0) - ( 3 lambda//4))/(( 3 lambda//4) xx lambda_(0))])` ....(ii)
Dividing equation (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 ((4)/(3))^(1//2)`