Single Choice

A particle $$A$$ of mass $$m$$ and initial velocity $$v$$ collides with a particle $$B$$ of mass $$\dfrac{m}{2}$$ 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$$\dfrac{{\lambda}_A}{{\lambda}_B}=\dfrac{1}{2}$$
B$$\dfrac{{\lambda}_A}{{\lambda}_B}=\dfrac{1}{3}$$
C$$\dfrac{{\lambda}_A}{{\lambda}_B}=2$$
Correct Answer
D$$\dfrac{{\lambda}_A}{{\lambda}_B}=\dfrac{2}{3}$$

Solution

From conservation of linear momentum, $$mv= mv_B + \dfrac{m}{2} v_B$$

$$v= v_A+ \dfrac{v_B}{2}$$

Also $$ v^2 = v_a^2 + \dfrac{v_B^2}{2}$$

$${(v_A+ \dfrac{v_B}{2})}^2= v_A^2 + {(\dfrac{v_B}{2})}^2$$

$$v_Av_B = \dfrac{v_B}{4}$$

$$v_B= 4 v_A$$ and $$m_B= \dfrac{m_A}{2}$$

$$P_A=m_Av_A$$

$$P_B= \dfrac{m_A}{2} \times 4 v_A= 2 P_A$$

$$\dfrac{\lambda_A}{\lambda_B}= \dfrac{P_B}{P_A}= \dfrac{2 P_A}{P_A}= 2$$

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