Question

A particle $A$ of mass $m$ and initial velocity $v$ collides with a particle $B$ of mass $\frac{ m }{2}$ which is at rest. The collision is head on, and elastic. The ratio of the de-Broglie wavelengths $\lambda_{ B }$ to $\lambda_{ A }$ after the collision is _______.

Answer 1

According to law of conservation of energy,
$\therefore m v+\frac{m}{2}(0)=m v_{A}+\frac{m}{2} v_{B}$
$\therefore m v=m v_{A}+\frac{m}{2} v_{B} $
$\therefore v=v_{A}+\frac{v_{B}}{2}$...(i)
As the collision is elastic,
$ e=\frac{v_{B}-v_{A}}{v}=1 $
$\therefore v=-v_{A}+v_{B}$....(ii)
Equating (i) and (ii),
$v _{ A }+\frac{ v _{ B }}{2}=- v _{ A }+ v _{ B }$
$2 v _{ A }=\frac{ v _{ B }}{2}$
$\therefore v _{ A }=\frac{ v _{ B }}{4}$....(iii)
de-Broglie wavelength is given by,
$ \lambda=\frac{ h }{ mv } $
$\therefore \frac{\lambda_{ B }}{\lambda_{ A }}=\frac{ mv _{ A }}{\left(\frac{ m }{2}\right) v _{ B }}=\frac{2 v _{ A }}{ v _{ B }} $
$=\frac{2\left(\frac{ v _{ B }}{4}\right)}{ v _{ B }} $ ....[From (iii)]
$\therefore \frac{\lambda_{ B }}{\lambda_{ A }}=\frac{1}{2}=0.5$