Question

If ${\lambda }_0$ and $\lambda$ be the threshold wavelength and wavelength of incident light, the velocity of photoelectron ejected from the metal surface is :

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

Answer 1

Answer: (C)

$E=W+\frac{1}{2}m{v}^2$

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

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

$\Rightarrow v=\sqrt{\frac{2hc}{m}\left[\frac{{\lambda }_0-\lambda }{\lambda {\lambda }_0}\right]}$