Question

The photo electric 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. $hc\left(\frac{{\lambda }_0-\lambda }{\lambda {\lambda }_0}\right)$
• B. $\frac{hc}{e}\left(\frac{{\lambda }_0-\lambda }{\lambda {\lambda }_0}\right)$
• C. $\frac{hc}{{\lambda }_0-\lambda }$
• D. $hc({\lambda }_0-\lambda )$

Answer 1

The correct option is A $hc\left(\frac{{\lambda }_0-\lambda }{\lambda {\lambda }_0}\right)$

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