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

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