Q. If light of wavelength $\lambda_{1}$ is allowed to fall on a metal, then kinetic energy of photoelectrons emitted is $E_{1}.$ If wavelength of light changes to $\lambda_{2}$ then kinetic energy of electrons changes to $E_{2}.$ Then work function of the metal is
A) $\frac{E_{1}E_{2}(\lambda_{1} - \lambda_{2})}{\lambda_{1}\lambda_{2}}$
B) $\frac{E_{1}\lambda_{1} - E_{2}\lambda_{2}}{(\lambda_{1} - \lambda_{2})}$
C) $\frac{E_{1}\lambda_{1} - E_{2}\lambda_{2}}{(\lambda_{2} - \lambda_{1})}$
D) $\frac{\lambda_{1}\lambda_{2}E_{1}E_{2}}{(\lambda_{2} - \lambda_{1})}$
Explanation
(c) $E = W_{0} + K_{\max} \Rightarrow \frac{hc}{\lambda_{1}} = W_{0} + E_{1}\ and\ \frac{hc}{\lambda_{2}} = W_{0} + E_{2}$
$\Rightarrow hc = W_{0}\lambda_{1} + E_{1}\lambda_{1}\ $and$\ hc = W_{0}\lambda_{2} + E_{2}\lambda_{2}$
$\Rightarrow W_{0}\lambda_{1} + E_{1}\lambda_{1} = W_{0}\lambda_{2} + E_{2}\lambda_{2} \Rightarrow W_{0} = \frac{E_{1}\lambda_{1} - E_{2}\lambda_{2}}{(\lambda_{2} - \lambda_{1})}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.