Question

If the momentum of an electron is increased by $\Delta p$, then the de-Broglie wavelength associated with it changes by 0.50%. The initial momentum of the electron will be

• A. $\frac{\Delta p}{200}$
• B. $\frac{\Delta p}{199}$
• C. $199\Delta p$
• D. $400\Delta p$

Answer 1

The correct option is C $199\Delta p$

$\lambda =\frac{h}{p}\Rightarrow \lambda -\frac{0.5}{100}\lambda =\frac{h}{p+\Delta p}\Rightarrow \frac{199\lambda }{200}=\frac{h}{p+\Delta p}=\frac{199}{200}\frac{h}{p}$
$\Rightarrow p+\Delta p=\frac{200}{199}p\Rightarrow p=199\Delta p$