A nucleus of mass M at rest splits into two p | vedclass

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(C) According to the law of conservation of linear momentum,since the initial nucleus is at rest,the total initial momentum is zero.Therefore,the magn

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$1:1$

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(C) According to the law of conservation of linear momentum,since the initial nucleus is at rest,the total initial momentum is zero.Therefore,the magnitudes of the momenta of the two parts must be equal and opposite:$|\vec{P}_1| = |\vec{P}_2| = P$The de Broglie wavelength $\lambda$ is given by the relation:$\lambda = \frac{h}{P}$Since both parts have the same magnitude of momentum $P$,their de Broglie wavelengths will also be equal.Therefore,the ratio of their de Broglie wavelengths is $\lambda_1 : \lambda_2 = 1 : 1$.

$A$ nucleus of mass $M$ at rest splits into two parts having masses $\frac{M'}{3}$ and $\frac{2M'}{3}$ (where $M' < M$). The ratio of the de Broglie wavelengths of the two parts will be:

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