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# Q-value of the decay ${}_{11}^{22}Na \to {}_{10}^{22}Ne + {e^ + } + \nu$ is (A) $Q = \left[ {m\left( {{}_{11}^{22}Na} \right) - m\left( {{}_{10}^{22}Ne} \right)} \right]{c^2}$(B) $Q = \left[ {m\left( {{}_{11}^{22}Na} \right) - m\left( {{}_{10}^{22}Ne} \right) - {m_e}} \right]{c^2}$(C) $Q = \left[ {m\left( {{}_{11}^{22}Na} \right) - m\left( {{}_{10}^{22}Ne} \right) - 2{m_e}} \right]{c^2}$(D) $Q = \left[ {m\left( {{}_{11}^{22}Na} \right) - m\left( {{}_{10}^{22}Ne} \right) - 3{m_e}} \right]{c^2}$

Last updated date: 13th Jun 2024
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Hint The Q-value for a beta decay is the energy due to the difference in the combined mass of the daughter nuclei and that of the beta particle, and mass of the parent nuclei. All particles with mass are not considered as part of the Q-value
In this solution we will be using the following formula;
$\Rightarrow Q = {c^2}\left[ {{M_p} - ({M_d} + {m_e})} \right]$ where is the Q-value for beta decay, ${M_d}$ is the mass of the daughter nucleus, ${M_p}$ is mass of the parent, and ${m_e}$ is the mass of the electron.

In a nucleus undergoing beta decay, one of the protons in the nucleus is converted into an electron or positron and ejected from the nucleus of the atom. The process is usually accompanied by a particle called a neutrino (symbolized $\nu$) with almost zero mass. ,
$\Rightarrow Q = {c^2}\left[ {{M_p} - ({M_d} + {m_e})} \right]$${M_d}$ is the rest mass of the daughter nucleus, ${M_p}$ is rest mass of the parent, and ${m_e}$ is the rest mass of the electron
$\Rightarrow Q = \left[ {m\left( {{}_{11}^{22}Na} \right) - m\left( {{}_{10}^{22}Ne} \right) - {m_e}} \right]{c^2}$