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Q. A nucleus ${ }^{220} X$ at rest decays emitting an $\alpha$-particle. If energy of daughter nucleus is $0.2\, MeV , Q$ value of the reaction is

Nuclei

Solution:

Energy of daughter nucleus $=0.2 \,MeV$
$0.2\, MeV =\frac{m_{\alpha}}{m_{\alpha}+m_{D}} Q $
$0.2\, MeV =\frac{4}{220} Q$
$\frac{0.2 \times 220}{4} \,M e V=Q $
$\frac{2 \times 55}{10}=Q $
$Q=11 \,MeV$