Question

If the momentum of an electron is changed 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

Answer: (C)

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