Question

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. $100p_0$
• C. $400p_0$
• D. $4p_0$

Answer 1

The correct option is C $400p_0$

$\lambda \propto \frac{1}{p}\Rightarrow \frac{\Delta p}{p}=-\frac{\Delta \lambda }{\lambda }\Rightarrow \left|\frac{\Delta p}{p}\right|=\left|\frac{\Delta \lambda }{\lambda }\right|\Rightarrow \frac{p_0}{p}=\frac{0.25}{100}=\frac{1}{400}\Rightarrow p=400p_0.$
Note: The above method is applicable only when the changes in wavelength or momenta are small.