Q. A nucleus with mass number 220 initially at rest emits an $\alpha -$particle.If the Q value of the reaction is 5.5 MeV, calculate the kinetic energy of the $\alpha -$particle
A) 4.4MeV
B) 5.4MeV
C) 5.6MeV
D) 6.5MeV
Explanation
(b) Given that,
$K_{1} + K_{2} = 5.5\ MeV\ \ \ \ \ldots(i)$
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From conservation of linear Momentum
Or $\ \ \ \ \ \ \ \ \ \sqrt{2K_{1}(216m)} = \sqrt{2k_{2}(4m)}$
Or $k_{2} = 54\ K_{1}\ \ \ \ \ \ \ldots(ii)$
Solving Eq.(i)& (ii),we get
$k_{2} = KE\ $of $\alpha - particle = 5.4Me\ V.$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.