Q. 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) $hc\left( \lambda_{0} - \lambda \right)$
B) $\frac{hc}{\lambda_{0} - \lambda}$
C) $\frac{h}{c}\left( \frac{1}{\lambda} - \frac{1}{\lambda_{0}} \right)$
D) $hc\left( \frac{\lambda_{0} - \ \lambda}{\lambda_{0}\lambda} \right)$
Explanation
(d) $E_{k} = \frac{hc}{\lambda} - \frac{hc}{\lambda_{0}} = hc\left\lbrack \frac{\lambda_{0} - \lambda}{\lambda_{0}\lambda} \right\rbrack$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.