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The atoms with same atomic number but different atomic mass are called isotopes. On the other hand the nuclei with same atomic mass but different atomic numbers (therefor different elements) are called isobars.

Some of the nuclei are stable and some are not. The unstable nuclei undergo radioactive decay. There are three basic types of radioactive decay- alpha, gamma and beta. Beta decay undergoes radioactive decay by two mechanisms:

Beta-plus decay is ${}^A{X}_z→{}^{A}X_{z-4}+e^{+}+energy$and the Beta-minus decay islocalid="1651151833754" ${}^A{X}_z→{}^{A}X_{z+1}+e^{-}+energy$.

Here, $X$is the element that goes beta decay, $Z$is the atomic number, and$A$is the atomic mass. In this problem we will determine the minimum mass threshold for the beta-plus decay.

In beta-plus decay a proton decays into a neutron and a positron $p^{+}→n+e^{+}$.

The positron escapes the nucleus and the daughter nucleus becomes a charged ion. So the minimum threshold mass for the beta-plus decay must have two electron mass (of positron and of electron) in addition to the mass of the daughter nucleus.

$m_x=m_y+2m_e$

Here, the mass of the nucleus is $m_x$mass of the daughter nucleus is $m_y$, and the mass of an electron is $m_{e^{-}}$. Therefore, the energy released in beta-plus decay is,

$∆E=(m_y-m_x+2m_e)931.49MeV/u$
Here, the conversion factor of mass (u) to energy (MeV) is$931.49MeV/u$.

The ${}^{13}N$can undergo beta-plus decay to ${}^{13}C$

role="math" ${}^{13}{N}_7→{}^{13}C_6+e^{+}+energy$

The mass of ${}^{13}N$.${}^{13}C$and an electron are respectively $13.0057386u,13.0033548u$and role="math" width="92" height="20">0.0005486u

Substitute $13.0057386u$for $m_x$and $0.0005486u$for $m_e$in equation (1)

role="math" $$\begin{gathered} ∆E=(0.0012866u)(931.49MeV/u) \\ =1.1985MeV \end{gathered}$$

Therefore, the energy released in this process is$1.1985Mev$