Question

Given the masses of various atomic particles $m _{ p }=1.0072 u$ , $m _{ n }=1.0087 \,u$ $m _{ e }=0.000548 u$ , $m _{\overline{ v }}=0$, $m _{ d }=2.0141 u$ where $p \equiv$ proton, $n \equiv$ neutron, $e \equiv$ electron, $\overline{ v } \equiv$ antineutrino and $d \equiv$ deuteron. Which of the following process is allowed by momentum and energy conservation ?

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

Answer 1

Answer: (A)

$n+p \to d+\gamma$

Answer 2

Only in case-I, $M _{ LHS }>\, M _{ RHS }$ i.e.
total mass on reactant side is greater then that on the product side. Hence it will only be allowed