The Q-value of the decay { }_{11}^{22} Na rig | vedclass

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Question

(C) In $\beta^+$ decay,a proton inside the nucleus converts into a neutron,a positron $(e^+)$,and a neutrino $(v)$. The reaction is: ${ }_{11}^{22} Na

Vedclass · Accepted Answer

$[m({ }_{11}^{22} Na) - m({ }_{10}^{22} Ne) - 2m_e] c^2$

Vedclass · Answer

(C) In $\beta^+$ decay,a proton inside the nucleus converts into a neutron,a positron $(e^+)$,and a neutrino $(v)$. The reaction is: ${ }_{11}^{22} Na \rightarrow{ }_{10}^{22} Ne + e^+ + v$. Here,$m({ }_{11}^{22} Na)$ and $m({ }_{10}^{22} Ne)$ represent the masses of the neutral atoms. The mass of the ${ }_{11}^{22} Na$ atom includes $11$ electrons,while the mass of the ${ }_{10}^{22} Ne$ atom includes $10$ electrons. Writing the mass balance for the nuclei: $M_{nuc}({ }_{11}^{22} Na) = m({ }_{11}^{22} Na) - 11m_e$ and $M_{nuc}({ }_{10}^{22} Ne) = m({ }_{10}^{22} Ne) - 10m_e$. The $Q$-value is given by: $Q = [M_{nuc}({ }_{11}^{22} Na) - M_{nuc}({ }_{10}^{22} Ne) - m_e] c^2$. Substituting the atomic masses: $Q = [(m({ }_{11}^{22} Na) - 11m_e) - (m({ }_{10}^{22} Ne) - 10m_e) - m_e] c^2$. $Q = [m({ }_{11}^{22} Na) - m({ }_{10}^{22} Ne) - 11m_e + 10m_e - m_e] c^2$. $Q = [m({ }_{11}^{22} Na) - m({ }_{10}^{22} Ne) - 2m_e] c^2$.

The $Q$-value of the decay ${ }_{11}^{22} Na \rightarrow{ }_{10}^{22} Ne + e ^{+}+v$ is

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