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