A nucleus of mass $M$ at rest splits into two parts having masses $\frac{M^{\prime}}{3}$ and ${{2M'} \over 3}(M' < M)$. The ratio of de Broglie wavelength of two parts will be :
Solution
<p>Linear momentum is conserved</p>
<p>so, ${p_{M'/3}} = {p_{2M'/3}}$</p>
<p>so, ${{{\lambda _{M'/3}}} \over {{\lambda _{2M'/3}}}} = {1 \over 1}$</p>
About this question
Subject: Physics · Chapter: Atoms and Nuclei · Topic: Nuclear Fission and Fusion
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