A particle of mass 4M at rest disintegrates i | vedclass

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(D) According to the law of conservation of linear momentum,since the initial particle of mass $4M$ is at rest,the total initial momentum is $0$.When

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$1: 1$

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(D) According to the law of conservation of linear momentum,since the initial particle of mass $4M$ is at rest,the total initial momentum is $0$.When it disintegrates into two particles of masses $M$ and $3M$,their final momenta must be equal in magnitude and opposite in direction to keep the total momentum zero.Let $p_1$ be the momentum of the particle of mass $M$ and $p_2$ be the momentum of the particle of mass $3M$. Then,$|p_1| = |p_2| = p$.The de-Broglie wavelength $\lambda$ is given by the relation $\lambda = \frac{h}{p}$,where $h$ is Planck's constant.Since both particles have the same magnitude of momentum $p$,their de-Broglie wavelengths will be:$\lambda_1 = \frac{h}{p}$ and $\lambda_2 = \frac{h}{p}$.Therefore,the ratio of their wavelengths is $\frac{\lambda_1}{\lambda_2} = \frac{h/p}{h/p} = 1:1$.

$A$ particle of mass $4M$ at rest disintegrates into two particles of mass $M$ and $3M$ respectively,having non-zero velocities. The ratio of the de-Broglie wavelength of the particle of mass $M$ to that of mass $3M$ will be:

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