Question

If ${\lambda }_0$ and $\lambda$ be the threshold wavelength and the wavelength of the incident light respectively, then the maximum velocity of photo-electrons ejected from the metal surface is:

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

Answer 1

The correct option is B $\sqrt{\frac{2hc}{m}\left(\frac{{\lambda }_0-\lambda }{\lambda {\lambda }_0}\right)}$

We know that, $hv=\varphi +K.E$, where
$v=$incident frequency
$\varphi =$threshold energy
K.E $=$ kinetic energy
Given, $\varphi =\frac{hc}{{\lambda }_0}$
$⟹hv=\frac{hc}{\lambda }$
$⟹K.E=\frac{1}{2}m{V}_{max}^2$
$⟹\therefore \frac{hc}{\lambda }=\frac{hc}{{\lambda }_0}+\frac{1}{2}m{V}_{max}^2$
$⟹\therefore hc(\frac{1}{\lambda }-\frac{1}{{\lambda }_0})=\frac{1}{2}m{V}_{max}^2$
$⟹V_{max}^2=\frac{2hc}{m}(\frac{1}{\lambda }-\frac{1}{{\lambda }_0})$
$⟹V_{max}=\sqrt{\frac{2hc}{m}\left(\frac{{\lambda }_0-\lambda }{\lambda {\lambda }_0}\right)}$