Question

A nucleus with mass number A=220 decays by a-emission. the energy released is 5.5 MeV, a god estimate for the kinetic energy of the a-particle will be-

1. 4.4 MeV
2. 5.4 MeV
3. ​5.6 MeV
4. 6.5 MeV

Answer 1

$$\begin{gathered} suppose \\ {A}^{220}\to B^{216}+\alpha \\ nownetenrgy \\ \frac{m_B{v^2}_B}{2}+\frac{m_{\alpha }{v^2}_{\alpha }}{2}=5.5MeV \\ u\sin gthemomentumconservationrelation \\ {m}_B{v}_B=m_{\alpha }{v}_{\alpha }wecanget \\ (1+\frac{m_{\alpha }}{m_B})\times {\left(K.E\right)}_{\alpha }=5.5MeV \\ {\left(K.E\right)}_{\alpha }=\frac{5.5}{(1+\frac{m_{\alpha }}{m_B})}=\frac{5.5}{1+{\displaystyle \frac{4}{216}}}\approx 5.4MeV\left[ANS\right] \end{gathered}$$