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 ?
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 ?
Updated On: Feb 2, 2026
- $n+p \to d+\gamma$
- $e ^{+}+ e ^{-}\to \gamma$
- $n + n \to $ deuterium atom (electron bound to the nucleus)
- $p\to n + e ^{+}+\overline{ v }$
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The Correct Option is A
Solution and Explanation
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
total mass on reactant side is greater then that on the product side. Hence it will only be allowed
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