JEE MAIN - Physics (2022 - 26th July Evening Shift - No. 9)
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 :
1 : 2
2 : 1
1 : 1
2 : 3
Explanation
Linear momentum is conserved
so, $${p_{M'/3}} = {p_{2M'/3}}$$
so, $${{{\lambda _{M'/3}}} \over {{\lambda _{2M'/3}}}} = {1 \over 1}$$
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