A nucleus {}^{220}X at rest decays by emittin | vedclass

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Question

(C) The decay reaction is ${}^{220}X \rightarrow {}^{216}D + {}^{4}\alpha$.The kinetic energy of the daughter nucleus $(K_D)$ is given by the formula:

Vedclass · Accepted Answer

$11$

Vedclass · Answer

(C) The decay reaction is ${}^{220}X ightarrow {}^{216}D + {}^{4}\alpha$.The kinetic energy of the daughter nucleus $(K_D)$ is given by the formula:$K_D = \left( \frac{m_{\alpha}}{m_{\alpha} + m_D} ight) Q$Here,$m_{\alpha} = 4$ and $m_D = 216$. The total mass $m_{\alpha} + m_D = 220$.Given $K_D = 0.2 \, MeV$,we substitute the values:$0.2 = \left( \frac{4}{220} ight) Q$Solving for $Q$:$Q = 0.2 	imes \left( \frac{220}{4} ight)$$Q = 0.2 	imes 55$$Q = 11 \, MeV$.

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

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