A particle A of mass m and initial velocity v collides with a particle of B of mass `m/2`which is at rest. The collision is head-on and elastic. The ratio of the de Broglie wavelengths `lamda_(A) "to" lamda_(B)` after collision is

Answer

B

Explanation

Mass of particle A = m
Mass of particle `B = m/2`
Let the velocities of the particles A and B just after the collision be `v_1 and v_2`, respectively.
Applying law of conservation of momentum,
`mv=mv_1+(m/2)v_2`
or `2v=2v_1+v_2" "...(i)`
For the elastic collision of coefficient of restitution, e = 1
Thus,
`e=(v_2-v_1)/(v-0)=1`
or, `v=v_2-v_1" "...(ii)`
[`:.` collision is elastic, e = 1 ]
Solving equations (i) and (ii)
`v_1=v/3and v_2=(4v)/3`
The de Broglie wavelength of a particle is given by `lamda=h/p`
where p is the momentum of the particle
`:.lamda_A/lamda_B=P_B/P_A=(m/2xx4/3v)/(mxxv/3)=2/1`