Q. When the momentum$\ $of a proton is changed by an amount $P_{0},$ the corresponding change in the de-Broglie wavelength is found to be $0.25\%.$ Then, the original momentum of the proton was
A) $p_{0}$
B) $100\ p_{0}$
C) $400\ p_{0}$
D) $4\ p_{0}$
Explanation
(c) $\lambda \propto \frac{1}{p} \Rightarrow \frac{\mathrm{\Delta}p}{p} = - \frac{\mathrm{\Delta}\lambda}{\lambda} \Rightarrow \left| \frac{\Delta p}{p} \right| = \left| \frac{\Delta\lambda}{\lambda} \right|$
$\Rightarrow \frac{p_{0}}{p} = \frac{0.25}{100} = \frac{1}{400}\ \Rightarrow p = 400\ p_{0}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.