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 $\lambda \left(\lambda <{\lambda }_0\right)$ will be

• A. $hc\left({\lambda }_0-\lambda \right)$
• 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)

$E=W+KE$
$KE=E-W$
$=\frac{hc}{\lambda }-\frac{hc}{{\lambda }_0}$
$=hc\left[\frac{1}{\lambda }-\frac{1}{{\lambda }_0}\right]=hc\left[\frac{{\lambda }_0-\lambda }{\lambda {\lambda }_0}\right]$