Q. If the momentum of an electron is changed by $\mathrm{\Delta}p,$ then the de-Broglie wavelength associated with it changes by 0.50%. The initial momentum of the electron will be
A) $\frac{\mathrm{\Delta}p}{200}$
B) $\frac{\mathrm{\Delta}p}{199}$
C) $199\ \mathrm{\Delta}p$
D) $400\ \mathrm{\Delta}p$
Explanation
(c) $\lambda = \frac{h}{p} \Rightarrow \lambda - \frac{0.5}{100}\lambda = \frac{h}{p + \Delta p} \Rightarrow \frac{199\lambda}{200} = \frac{h}{p + \Delta p} = \frac{199}{200}\frac{h}{p}$
$\Rightarrow p + \Delta p = \frac{200}{199}p \Rightarrow p = 199\ \Delta p$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.