Question

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

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

Answer 1

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

As per Einstein's equation of photoelectric effect,
$h\nu =h{\nu }_0+K.E.$
$\therefore$ $\frac{1}{2}m{v}^2=(h\nu -h{\nu }_0)=\frac{hc}{\lambda }-\frac{hc}{{\lambda }_0}$
$\Rightarrow {\nu }^2=\frac{2hc}{m}(\frac{1}{\lambda }-\frac{1}{{\lambda }_0})$
$\Rightarrow \nu ={\left[\frac{2hc}{m}(\frac{1}{\lambda }-\frac{1}{{\lambda }_0})\right]}^{\frac{1}{2}}$

$\Rightarrow \nu =\sqrt{\frac{2hc}{m}\left(\frac{{\lambda }_0-\lambda }{\lambda {\lambda }_0}\right)}$