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 ${\lambda }_A$ to ${\lambda }_B$ after the collision is

• A. $\frac{{\lambda }_A}{{\lambda }_B}=\frac{1}{2}$
• B. $\frac{{\lambda }_A}{{\lambda }_B}=\frac{1}{3}$
• C. $\frac{{\lambda }_A}{{\lambda }_B}=2$
• D. $\frac{{\lambda }_A}{{\lambda }_B}=\frac{2}{3}$

Answer 1

The correct option is C $\frac{{\lambda }_A}{{\lambda }_B}=2$

From conservation of linear momentum, $mv=m{v}_A+\frac{m}{2}{v}_B$

$v=v_A+\frac{v_B}{2}$

Also, by conservation of K.E,
$v^2=v_A^2+\frac{v_B^2}{2}$
${(v_A+\frac{v_B}{2})}^2=v_A^2+{\left(\frac{v_B}{2}\right)}^2$

$v_A{v}_B=\frac{v_B^2}{4}$

$\Rightarrow v_B=4v_A$ and $m_B=\frac{m_A}{2}$

$P_A=m_A{v}_A$

$P_B=\frac{m_A}{2}\times 4v_A=2P_A$

$\Rightarrow \frac{{\lambda }_A}{{\lambda }_B}=\frac{P_B}{P_A}=\frac{2P_A}{P_A}=2$
Alternative:
Velocity of particle A after collision,
$v_A=\frac{m_A-m_B}{m_A+m_B}v=\frac{m-m/2}{3m/2}v=v/3$
Momentum of particle A,
$P_A=m\times \frac{v}{3}=\frac{mv}{3}$
Velocity of particle B after collision,
$v_B=\frac{2m_A}{m_A+m_B}v=\frac{2m}{3m/2}v=\frac{4v}{3}$
Momentum of particle B,
$P_B=m\times \frac{4v}{3}=\frac{4mv}{3}$
$\Rightarrow \frac{{\lambda }_A}{{\lambda }_B}=\frac{P_B}{P_A}=\frac{2P_A}{P_A}=2$