Q. An electron of mass $m$ when accelerated through a potential difference $V$ has de-Broglie wavelength $\lambda.$ The de-Broglie wavelength associated with a proton of mass $M$ accelerated through the same potential difference will be
A) $\lambda\frac{m}{M}$
B) $\lambda\sqrt{\frac{m}{M}}$
C) $\lambda\frac{M}{m}$
D) $\lambda\sqrt{\frac{M}{m}}$
Explanation
(b) $\lambda = \frac{h}{\sqrt{2mE}} \Rightarrow \lambda \propto \frac{1}{\sqrt{m}}\ \ \ \lbrack E = same\rbrack$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.