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A particle of mass $3 m$ at rest decays into two particles of masses $m$ and $2 m$ having non-zero velocities. The ratio of the de-Broglie wavelengths of the particles $(\frac{λ_{1}}{λ_{2}})$ is

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$\frac{1}{2}$

$\frac{1}{4}$

answer is C.

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Detailed Solution

By Law of Conservation of Momentum, we have

$0 = m v_{1} + 2 m (v_{2}) \Rightarrow m v_{1} = - 2 m v_{2} \Rightarrow |m v_{1}| = |2 m v_{2}| \Rightarrow |p_{1}| = |p_{2}|$

Now, according to de-Broglie relation, we have

$λ = \frac{h}{|p|}$

$\Rightarrow \frac{λ_{1}}{λ_{2}} = \frac{|p_{2}|}{|p_{1}|} = 1$

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