Question
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 $${\lambda _A}$$ to $${\lambda _B}$$ after the collision is
A.
$$\frac{2}{3}$$
B.
$$\frac{1}{2}$$
C.
$$\frac{1}{3}$$
D.
2
Answer :
2
Solution :
From question, $${m_A} = M;{m_B} = \frac{m}{2}$$
$${u_A} = V,{u_B} = 0$$
Let after collision velocity of $$A = {V_1}$$ and velocity of $$B = {V_2}$$
Applying law of conservation of momentum,
$$mu = m{v_1} + \left( {\frac{m}{2}} \right){v_2}\,\,{\text{or,}}\,\,2\mu = 2{v_1} + {v_2}\,.......\left( {\text{i}} \right)$$
By law of collision
$$e = \frac{{{v_2} - {v_1}}}{{u - 0}}\,\,{\text{or,}}\,\,u = {v_2} - {v_1}\,.......\left( {{\text{ii}}} \right)\,\,\left[ {\because {\text{collision is elastic,}}\,e = 1} \right]$$
using eqns (i) and (ii)
$${v_1} = \frac{1}{3}\mu \,\,{\text{and}}\,\,{v_2} = \frac{4}{3}\mu $$
de-Broglie wavelength $$\lambda = \frac{h}{p}$$
$$\therefore \frac{{{\lambda _A}}}{{{\lambda _B}}} = \frac{{{P_B}}}{{{P_A}}} = \frac{{\frac{m}{2} \times \frac{4}{3}u}}{{m \times \frac{1}{3}}} = 2$$