Question

If ${\lambda }_0$ is threshold wave lenght for photo electric emission, $\lambda$ the wave length of light falling on the surface of a metal and $m$ is the mass of electron, the velocity of ejected electron is given by.

• A. ${\left[\frac{2h}{m}\right(\lambda -{\lambda }_0\left)\right]}^{1/2}$
• B. ${\left[\frac{2hc}{m}\right({\lambda }_0-\lambda \left)\right]}^{1/2}$
• C. ${\left[\frac{2hc}{m}\left(\frac{{\lambda }_0-\lambda }{\lambda {\lambda }_0}\right)\right]}^{1/2}$
• D. ${\left[\frac{2hc}{m}\left(\frac{1}{{\lambda }_0}\frac{1}{\lambda }\right)\right]}^{1/2}$

Answer 1

The correct option is A ${\left[\frac{2h}{m}\right(\lambda -{\lambda }_0\left)\right]}^{1/2}$

$h\lambda =h{\lambda }_0+\frac{1}{2}m{v}^2$

$\lambda =\frac{h}{\sqrt{2mK.E}}$

$K.E=h(\lambda -{\lambda }_0)$

So $v=\sqrt{\frac{2h}{m}(\lambda -{\lambda }_0)}$