Question

The photoelectric threshold wavelength for silver is ${\lambda }_0$. The energy of the electron ejected from the surface of silver by an incident wavelength $\lambda (\lambda <{\lambda }_0)$ will be:

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

Answer 1

Answer: (D)

$hc\left[\frac{{\lambda }_0-\lambda }{\lambda {\lambda }_0}\right]$

According to Einstein's photoelectric equation

$K=\frac{hc}{\lambda }-\frac{hc}{{\lambda }_0}=hc(\frac{1}{\lambda }-\frac{1}{{\lambda }_0})=hc\left(\frac{{\lambda }_0-\lambda }{\lambda {\lambda }_0}\right)$