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 $400p_0$

Given:

$\Delta \lambda =0.25\%;\Delta p=p_0$

Let initial momentum and wavelength be $p$ and $\lambda$ respectively.

The de-Broglie wavelength $\left(\lambda \right)$ is,

$\lambda =\frac{h}{p}\Rightarrow \lambda p=h=\text{constant}$

$\Rightarrow \frac{\Delta p}{p}=-\frac{\Delta \lambda }{\lambda }$

$\Rightarrow \left|\begin{array}{c}\frac{\Delta p}{p}\end{array}\right|=\left|\begin{array}{c}-\frac{\Delta \lambda }{\lambda }\end{array}\right|$

$\Rightarrow \frac{p_0}{p}=\frac{0.25}{100}=\frac{1}{400}$

$\Rightarrow p=400p_0$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, option $\left(C\right)$ is the correct answer.