Question

If ${\lambda }_0$ is the threshold wavelength for photoelectric emission, $\lambda$ the wavelength of light falling on the surface of a metal and m is the mass of the electron, then the velocity of ejected electron is given by:

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

Answer 1

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

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

$hc[\frac{1}{\lambda }-\frac{1}{{\lambda }_0}]=\frac{1}{2}m{v}^2$

$v=\sqrt{\left(\frac{2hc}{m}\left[\frac{{\lambda }_0-\lambda }{{\lambda }_0\lambda }\right]\right)}$
Option C is the correct answer.