A particle A with a mass mA is moving with a velocity v and hits particle B (mass mB) at rest (one-dimensional motion). Find the change in the de-Broglie wavelength of the particle A. Treat collision as elastic.

Hint: The de-Broglie wavelength depends on the momentum of the particle.

Step 1: Conserve the momentum.

As collision is elastic, hence laws of conservation of momentum and kinetic energy are obeyed. 

According to the law of conservation of momentum, 

mAv+mB×0=mAv1+mBv2mA (v-v1)=mBv2

Step 2: Conserve the kinetic energy.

According to the law of conservation of kinetic energy,

12mAv2=12mAv12+12mBv22       ...(i)mA(vv12)=m8v22mA(vv1)(v+v1)=mBv22         ...(ii)

Step 3: Find the final velocities of A and B.

Dividing Eq. (ii) by Eq. (i) ;

we get,     v+v1=v2 or v=v2-v1                  ...(iii) 

Step 4: Find the change in wavelength of particle A.

Solving Eqs. (i) and (ii), we get;

                          v1=mA-mBmA+mBv and v2=2mAmA+mBλinitial =hmAvλfinal =hmAv1=h(mA+mB)mA(mA-mB)vλ=λfinal -λinitial =hmAvmA+mBmA-mB-1