Question

Given the masses of various atomic particles $m_p=1.0072u,$ $m_n=1.0087u,m_e=0.000548u,m_{\overline{v}}=0,m_d=2.01241u,$ where $p=\text{proton,}n=\text{neutron,}e=\text{electron,}$
$\overline{v}=\text{antineutrino and}d=\text{deuteron.}$ Which of the following process is allowed by momentum and energy conservation?

• A. $n+n\to$ deuterium atom (electron bound to the nucleus)
• B. $p\to n+e^{+}+\overline{v}$
• C. $n+p\to d+\gamma$
• D. $e^{+}+e^{-}\to \gamma$

Answer 1

The correct option is C $n+p\to d+\gamma$

For the momentum and energy conservation, mass defect $\left(\Delta m\right)$ should be positive. Since some energy is lost in every process.
Here only option $\left(C\right)$ has positive mass defect because $(m_p+m_n)>m_d$.

Hence, option $\left(C\right)$ is correct.