Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
In a beta-plus decay, a nucleus $ ^AX_Z$ (with atomic number $ Z $ and mass number $ A $) decays into a nucleus $ ^AY_{Z-1}$ by emitting a positron ($ e^+ $) and a neutrino ($ \nu $):
$$
^AX_Z\rightarrow \;^AY_{Z-1}+ e^+ + \nu
$$
The initial mass of the $ X $ atom must exceed the combined mass of the $ Y $ atom and the positron.
We also consider that atomic masses include the mass of electrons associated with each element.
Noting that:
$\Rightarrow$ The nucleus $ X $ contains $ Z $ protons and $ N $ neutrons.
$\Rightarrow$ After decay, the nucleus $ Y $ has $ Z-1 $ protons and $ N+1 $ neutrons.
$\Rightarrow$ The energy condition for beta-plus decay is that the mass of $ X $ must be greater than the mass of $ Y $ plus the positron's mass, allowing the decay to proceed and release energy.
So,
$$Z m_p + N m_n \gt (Z - 1) m_p + (N + 1) m_n + m_e$$
$\Rightarrow$ In atomic mass terms, we need to add back the electron masses because atomic masses include the electron cloud:
$\Rightarrow$ $ X $ has $ Z $ electrons, while $ Y $ has $ Z-1 $ electrons.
Thus, we need to add $Zm_e$ for both sides,
$$Z m_p+Z m_e + N m_n \gt (Z - 1) m_p +Z m_e+ (N + 1) m_n + m_e$$
$$(m_p + m_e) + N m_n \gt [(Z - 1) (m_p + m_e) + (N + 1) m_n] + 2 m_e$$
The expression of $(m_p + m_e)$ represents the atomic mass of the atom $^A_ZX$, and the term of $ [(Z - 1) (m_p + m_e) + (N + 1) m_n] $ represents the atomic mass of the resulting atom $^A_{Z-1}Y$.
Thus, the threshold condition for beta-plus decay:
$$ \boxed{ m_{(^AX_Z) }\gt m_{(^AY_{Z-1})} + 2 m_e }$$
where $ 2 m_e $ accounts for the positron emitted and the difference in the electron count between $ X $ and $ Y $.
This inequality means that for beta-plus decay to occur, the atomic mass of $ X $ must exceed the atomic mass of $ Y $ by at least the mass of two electrons.
$$\color{blue}{\bf [b]}$$
Now, we need to check if $ ^{13} \rm{N} $ can decay into $ ^{13} \rm{C} $ via beta-plus decay and calculate the energy released if this decay is possible.
From Appendix C;
$ m_{(^{13}\text{N})} = 13.005738 \, \text{u} $
$ m_{(^{13}\text{C})} = 13.003355 \, \text{u} $
The mass of two electrons is $ 2 m_e = 0.001096 \, \text{u} $.
Now Let's Check the Threshold Condition, so we need to Calculate $ m_{(^{13}\text{C})} + 2 m_e $:
$$
m_{(^{13}\text{C})} + 2 m_e = 13.003355 + 0.001096 \\
= \bf 13.004451 \, \rm {u}
$$
Now it is obvious that
$$ m_{(^{13}\text{N})} \gt m_{(^{13}\text{C})} + 2 m_e $$
So the condition is satisfied, allowing $^{13}\text{N}$ to decay to $^{13}\text{C}$ via beta-plus decay.
Thus, the energy released $ E $ in the decay is given by
$$
E=\Delta m c^2
$$
$$
E=(m_{(^{13}\text{N})} -[m_{(^{13}\text{C})} + 2 m_e ]) c^2
$$
where $ c^2 = 931.49 \, \text{MeV}/\text{u} $
$$
E=(13.005738 - 13.004451 )(931.49 )=\color{red}{\bf 1.20}\;\rm MeV
$$
