Discussion Forum

A particle A of mass m and intial 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 λ_A to λ_B after the collision is 

Answer:

Explanation:

For elastic collision

  P_{before collision} =P_{after collsion}

$mv=mv_{A}+\frac{m}{2}v_{B}$

$2v=2v_{A}+2v_{B}$         .........(i)

Now, coefffficent of restitution,

$e= \frac{v_{B}-v_{A}}{u_{A}-v_{B}}$

Here u_B = 0 ( particle at rest) and for elastic collisione= 1

$\therefore$     $1= \frac{v_{B}-v_{A}}{v}\Rightarrow v =v_{B}-v_{A}$              .............(ii)

From Eq .(i) and Eq .(ii)

                     $v_{A}=\frac{v}{3}$   and $v_{B}=\frac{4v}{3}$

              Hence,  $\frac{\lambda_{A}}{\lambda_{B}}=\frac{(\frac{h}{mV_{A}})}{\frac{h}{\frac{m}{2}V_{B}}}=\frac{V_{B}}{2V_{A}}=\frac{4/3}{2/3}=2$