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)

We have $\lambda=\frac{h}{p}$
$\lambda \propto \frac{1}{p}$
$\Rightarrow \frac{\Delta p}{p} =\frac{\Delta \lambda}{\lambda}$
$\Rightarrow \left|\frac{\Delta p}{p}\right| =\left|\frac{\Delta \lambda}{\lambda}\right|$
$\frac{p_{0}}{p} =\frac{0.25}{100}$
$\frac{p_{0}}{p} =\frac{1}{400}$
$p =400\, p_{0}$