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 λA to λB after the collision is :

 

Option 1)

\frac{\lambda _{A}}{\lambda _{B}}= \frac{1}{3}

Option 2)

\frac{\lambda _{A}}{\lambda _{B}}= 2

Option 3)

\frac{\lambda _{A}}{\lambda _{B}}= \frac{2}{3}

Option 4)

\frac{\lambda _{A}}{\lambda _{B}}= \frac{1}{2}

Answers (1)

As we learnt in

De - Broglie wavelength -

\lambda = \frac{h}{p}= \frac{h}{mv}= \frac{h}{\sqrt{2mE}}

- wherein

h= plank's\: constant

m= mass \: of\: particle

v= speed \: of\: the \: particle

E= Kinetic \: energy \: of \: particle

 After collision 

v_{f1}=\frac{m_{1}-m_{2}}{m_{1}+m_{2}}.v_{i1}+\frac{2m_{2}v_{i2}}{m_{1}+m_{2}}=\frac{v}{3}

v_{f2}=\frac{m_{2}-m_{1}}{m_{2}+m_{1}}.v_{i2}+\frac{2m_{1}v_{i1}}{m_{1}+m_{2}}=\frac{4v}{3}

\frac{\lambda_{A}}{\lambda_{B}}=\frac{P_{B}}{P_{A}}=\frac{(\frac{m}{2}).\frac{4v}{3}}{m.\frac{v}{3}}=2:1

Correct answer is 2.


Option 1)

\frac{\lambda _{A}}{\lambda _{B}}= \frac{1}{3}

This is an incorrect option.

Option 2)

\frac{\lambda _{A}}{\lambda _{B}}= 2

This is the correct option.

Option 3)

\frac{\lambda _{A}}{\lambda _{B}}= \frac{2}{3}

This is an incorrect option.

Option 4)

\frac{\lambda _{A}}{\lambda _{B}}= \frac{1}{2}

This is an incorrect option.

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