Particle A of mass mA=m2  moving along the x-axis with velocity  v0 collides elastically with another particle B at rest having mass  mB=m3. If the particles move along the x-axis after the collision, the change Δ λ  in de-Broglie wavelength of particle A, in terms of its de-Broglie wavelength λ0  before collision is :

Conservation of momentum, ​m2V0+m30=m2VA+m3VB​⇒3VA+2VB=3V0.....................1​Coefficient of restitution​,e=1=VA−VB0−V0​⇒VA−VB=−V0.......................2Solving 1 and 2​VA=V05​Now, de broglie wavelength =λ=hmvSo, Initial wavelength of A=λi=hm2V0=λ0and Final wavelength of A=λf=hm2V05=5λ0Thus, Change in wavelength =△λ=λf−λi=4λ0In ground frame,  m2 moves forward with  3V05−2V05=V05So  λf=5λ0⇒Δ λ=4λ0