Question

If the momentum of an electron is increased by $p_m$ then the de Broglie wavelength associated with it changes by $0.5\%$. Then the initial momentum of the electron is ______times $p_m$.

Answer 1

Answer: 199

Let $p$ be initial momentum of electron,
$p^{\prime }=p+p_m$
$\therefore {\lambda }^{\prime }=\lambda -\frac{0.5}{100}\lambda =\left(1-\frac{0.5}{100}\right)\lambda$
$\therefore \frac{{\lambda }^{\prime }}{\lambda }=\frac{p}{p^{\prime }}$
$\therefore \left(1-\frac{0.5}{100}\right)=\frac{p}{p+p_m}$
$\therefore \left(\frac{99.5}{100}\right)=\frac{p}{p+p_m}$
$\therefore 99.5p+99.5p_m=100p$
$\therefore 0.5p=99.5p_m$
$\therefore p=199p_m$