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

(d) Apply Einstein's photoelectric equation.
`hv - W_(0) = (1)/(2) mv^(2)_(max)`
`implies (hc)/(lambda) - (hc)/(lambda_(0)) = (1)/(2) mv^(2)_(max) implies hc ((lambda_(0) - lambda)/(lambda lambda_(0))) = (1)/(2) mv^(2)_(max)`
`v_(max) = sqrt((2 hc)/(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))))`
When wavelength is `(3 lambda)/(4)` and velocity is `v'`, then 4
`v' = sqrt((2hc)/(m) [(lambda_(0) - (3 lambda//4))/((3 lambda // 4) xx lambda_(0))])`
Dividing Eqs. (ii) by (i), we get
`(v')/(v) = sqrt(([lambda_(0) - (3 lambda//4)])/((3)/(4) lambda lambda_(0)) xx (lambda lambda_(0))/(lambda_(0) - lambda))`
`v' = v ((4)/(3))^(1//2) sqrt(([lambda_(0) - (3lambda // 4)])/(lambda_(0) - lambda))`
i.e., `v' gt v ((4)/(3))^(1//2)`