Question

A nucleus of mass number 220 decays by $\alpha$ decay. The energy released in the reaction is 5 MeV. The kinetic energy of an $\alpha$-particle is:

• A. $\frac{1}{54}MeV$
• B. $\frac{27}{11}MeV$
• C. $\frac{54}{11}MeV$
• D. $\frac{55}{54}MeV$

Answer 1

Answer: (C)

The decay proceeds as ${}^{220}X{\to }^{216}Y{+}^4\alpha$

Since the momentum after the decay is conserved,

$m_{\alpha }{v}_{\alpha }=m_Y{v}_Y$

Thus $\frac{v_{\alpha }}{v_Y}=\frac{m_Y}{m_{\alpha }}=\frac{216}{4}=54$

The ratio of their kinetic energies=$\frac{K{E}_Y}{K{E}_{\alpha }}=\frac{\frac{1}{2}{m}_Y{v}_Y^2}{\frac{1}{2}{m}_{\alpha }{v}_{\alpha }^2}=\frac{1}{54}$

$Q=K{E}_Y+K{E}_{\alpha }=5MeV$

$⟹\frac{55}{54}K{E}_{\alpha }=5MeV$

$⟹K{E}_{\alpha }=\frac{54}{11}MeV$