Question 205 of 239

A nucleus with mass number $184$ initially at rest emits an $\alpha -$particle. If the $Q$ value of the reaction is $5.5MeV,$ calculate the kinetic energy of the $\alpha -$ particle.

A$5.0MeV$B$5.5MeV$C$0.12MeV$D$5.38MeV$

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Answer: (D) $5.38MeV$

$\dfrac{1}{2}\left(4m\right){v}^{2}+\dfrac{1}{2}\left(180m\right){\left(\dfrac{4v}{180}\right)}^{2}=5.5MeV$
$\Rightarrow \dfrac{1}{2}4{mv}^{2}\left[1+45{\left(\dfrac{4}{180}\right)}^{2}\right]=5.5MeV$
$\Rightarrow K.E{.}_{\alpha }=\dfrac{5.5}{1+45\cdot {\left(\dfrac{4}{180}\right)}^{2}}MeV=5.38MeV$