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# A nucleus with mass number 200 initially at rest emits and $\;\alpha-\;$ particle . If Q value of reaction is 4.2MeV .Then kinetic energy of $\;\alpha-\;$ particle will be

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

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