Given the masses of various atomic particles
Question:

Given the masses of various atomic particles $\mathrm{m}_{\mathrm{p}}=1.0072 \mathrm{u}, \mathrm{m}_{\mathrm{n}}=1.0087 \mathrm{u}$, $\mathrm{m}_{\mathrm{e}}=0.000548 \mathrm{u}, \mathrm{m}_{\overline{\mathrm{v}}}=0, \mathrm{~m}_{\mathrm{d}}=2.0141 \mathrm{u}$,

where $\mathrm{p} \equiv$ proton, $\mathrm{n} \equiv$ neutron, $\mathrm{e} \equiv$ electron, $\bar{v} \equiv$ antineutrino and $d \equiv$ deuteron. Which of the following process is allowed by momentum and energy conservation?

1. $\mathrm{n}+\mathrm{p} \rightarrow \mathrm{d}+\gamma$

2. $\mathrm{e}^{+}+\mathrm{e}^{-} \rightarrow \gamma$

3. $\mathrm{n}+\mathrm{n} \rightarrow$ deuterium atom

(electron bound to the nucleus)

4. $\mathrm{p} \rightarrow \mathrm{n}+\mathrm{e}^{+}+\overline{\mathrm{v}}$

Correct Option: 1

Solution:

Only in case-I, $\mathrm{M}_{\mathrm{LHS}}>\mathrm{M}_{\mathrm{RHS}}$ i.e.

total mass on reactant side is greater then that on the product side. Hence it will only be allowed.