Question

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

• A. $hc({\lambda }_0-\lambda )$
• B. $\frac{hc}{{\lambda }_0-\lambda }$
• C. $\frac{h}{c}[\frac{1}{\lambda }-\frac{1}{{\lambda }_0}]$
• D. $hc\left(\frac{{\lambda }_0-\lambda }{{\lambda }_0\lambda }\right)$

Answer 1

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

$hv=h{v}_0+EE=hv-h{v}_0$
But $v=\frac{c}{\lambda }E=hc[\frac{1}{\lambda }-\frac{1}{{\lambda }_0}]=hc\left[\frac{{\lambda }_0-\lambda }{{\lambda }_0\lambda }\right]$