Question

When the momentum of a proton is changed by an amount $p_0$, then 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. $100p_0$
• C. $400p_0$
• D. $4p_0$

Answer 1

Answer: (C)

$400p_0$

We have $\lambda =\frac{h}{p}$
$\lambda \propto \frac{1}{p}$
$\Rightarrow d\frac{\Delta p}{p}=\frac{\Delta \lambda }{\lambda }$
$\Rightarrow \left|\frac{\Delta p}{p}\right|=\left|\frac{\Delta \lambda }{\lambda }\right|$
$\frac{p_0}{p}=\frac{0.25}{100}$
$\frac{p_0}{p}=\frac{1}{400}$
$p=400p_0$