$(a)\;4.2 MeV\qquad(b)\;4.116 MeV\qquad(c)\;3.96 MeV\qquad(d)\;1.9 MeV$

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Answer : (b) $\;4.166 MeV$

Explanation :

Given that $\;K_{1}+K_{2}=4.2 MeV----(1)$

From conservation of linear momentum

$P_{1}=P_{2}$

or $\; \sqrt{2 K_{1} (196 m)}=\sqrt{2 K_{2} (4m)} \qquad \; [P=\sqrt{2km}]$

$K_{2}=49K_{1}---(2)$

Solving (1) & (2)

We get

$K_{2}=KE \;of\;\alpha-particle=4.116MeV\;.$

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