Question

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) $Q = \left[ {m\left( {{}_{11}^{22}Na} \right) - m\left( {{}_{10}^{22}Ne} \right) - 3{m_e}} \right]{c^2}$

Answer 1

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.

Complete step by step answer
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. ,
Due to the conversion of the proton to an electron, the nucleus thus now has one proton less than the initial amount of proton. Since the element of an atom is determined by the number of protons, the nucleus hence, becomes a new element. This new nucleus is called the daughter nucleus
In nuclear science, we write all products on the right hand side of an arrow and the decayed nucleus on the left hand side as done in the question.
Usually, when the decay happens, it is accompanied by a release of energy called Q-value, which is manifested in the kinetic energy of the parent nucleus, the beta particle (electron or positron) and the neutrino, and it is given by
$\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
Hence, for the particular decay, we have
$\Rightarrow Q = \left[ {m\left( {{}_{11}^{22}Na} \right) - m\left( {{}_{10}^{22}Ne} \right) - {m_e}} \right]{c^2}$
Hence, the correct answer is B.

Note
To avoid any confusion, the neutrino of the beta decay reaction isn’t considered in the formula because its mass is extremely small and thus negligible compared to the other nuclei. However, some of the Q-value energy is actually deposited in the neutrino accelerating it to a very high velocity.