Q. An electromagnetic wave of wavelength '$\lambda$' is incident on a photosensitive surface of negligible work function. If 'm' mass is of photoelectron emitted from the surface has de-Broglie wavelength $\lambda_d$, then:
A) $\lambda_d = \left(\frac{2mc}{h}\right)\lambda^2$
B) $\lambda = \left(\frac{2mc}{h}\right)\lambda_d^2$
C) $\lambda = \left(\frac{2h}{mc}\right)\lambda_d^2$
D) $\lambda = \left(\frac{2m}{hc}\right)\lambda_d^2$
Explanation
[Solution] $\frac{hc}{\lambda} = \frac{p^2}{2m} = \frac{h^2}{2m\lambda_d^2}$, so $\lambda = \frac{2mc}{h}\lambda_d^2$. Answer (2).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.