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. $\sqrt{\frac{2h}{m}({\lambda }_0-\lambda )}$
• 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 $\sqrt{\frac{2hc}{m}\left(\frac{{\lambda }_0-\lambda }{\lambda {\lambda }_0}\right)}$

If ${\lambda }_0$ and $\lambda$ be the threshold wavelength and wavelength of incident light then according to the equation
$\frac{hc}{\lambda }=\frac{hc}{{\lambda }_0}+\frac{1}{2}m{v}^2$
$v=\sqrt{\frac{2hc}{m}(\frac{1}{\lambda }-\frac{1}{{\lambda }_0})}$
$v=\sqrt{\frac{2hc}{m}\left(\frac{{\lambda }_0-\lambda }{\lambda {\lambda }_0}\right)}$