Question

give the correct answer​

Answer 1

Given:

$\lambda• = threshold \: wavelength$

$\lambda = wavelength \: of \: light \: incident \: on \: surface$

given velocity of ejected electron

$\implies{v = \sqrt{ \frac{2h}{m} ( \lambda• - \lambda)K} ...(1)}$

Solution :

we know that ,

$\implies{E = W + KE }$

$\implies{ \frac{hc}{ \lambda} = \frac{hc}{ \lambda• } + \frac{m {v}^{2} }{2} }$

$\implies{v = \sqrt{ \frac{2hc}{m} ( \frac{1}{\lambda} - \frac{1}{\lambda•} )} }$

$\implies{v = \sqrt{ \frac{2hc}{m} ( \frac{ \lambda• - \lambda}{\lambda\lambda•} )}...(2)}$

Now equating equation (1) and (2) we get ,

$\implies{ \sqrt{ \frac{2hc}{m} ( \frac{ \lambda• - \lambda}{\lambda\lambda•} )} = \sqrt{ \frac{2h}{m} ( \lambda• - \lambda)K}}$

$\implies{\sqrt{ \frac{2h}{m} ( { \lambda• - \lambda} ) \frac{c}{ \lambda \lambda• } } = \sqrt{ \frac{2h}{m} ( \lambda• - \lambda)K}}$

$\boxed{K = \frac{c}{ \lambda \lambda• } }$