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

Answer: 0.5

According to law of conservation of energy,
$\therefore mv+\frac{m}{2}(0)=m{v}_A+\frac{m}{2}{v}_B$
$\therefore mv=m{v}_A+\frac{m}{2}{v}_B$
$\therefore v=v_A+\frac{v_B}{2}$...(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),
$v_A+\frac{v_B}{2}=-v_A+v_B$
$2v_A=\frac{v_B}{2}$
$\therefore v_A=\frac{v_B}{4}$....(iii)
de-Broglie wavelength is given by,
$\lambda =\frac{h}{mv}$
$\therefore \frac{{\lambda }_B}{{\lambda }_A}=\frac{m{v}_A}{\left(\frac{m}{2}\right)v_B}=\frac{2v_A}{v_B}$
$=\frac{2\left(\frac{v_B}{4}\right)}{v_B}$ ....[From (iii)]
$\therefore \frac{{\lambda }_B}{{\lambda }_A}=\frac{1}{2}=0.5$