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

The Correct Option is A

To determine which of the given processes is allowed by momentum and energy conservation, we will examine each option with respect to the laws of conservation of momentum and energy:

1. $n + p \to d + \gamma$:
   • A neutron ($n$) and a proton ($p$) can combine to form a deuteron ($d$) and emit a photon ($\gamma$).
   • The mass of a neutron is $1.0087 \, u$ and that of a proton is $1.0072 \, u$. Thus, the total mass of the reactants is $1.0087 + 1.0072 = 2.0159 \, u$.
   • The mass of a deuteron is $2.0141 \, u$. The emitted photon carries away any excess energy. Thus, the mass difference $2.0159 - 2.0141 = 0.0018 \, u$ is converted to the energy of the photon, consistent with energy conservation.
   • This process is allowed as both energy and momentum can be conserved.
2. $e^{+} + e^{-} \to \gamma$:
   • An electron ($e^{-}$) and a positron ($e^{+}$) can annihilate and generally produce two photons to conserve both energy and momentum. A process resulting in a single photon would violate momentum conservation.
   • Thus, typically, $e^{+} + e^{-} \to 2\gamma$ is observed instead.
   • This process, as written, is not allowed due to violation of momentum conservation.
3. $n + n \to$ deuterium atom (electron bound to the nucleus):
   • Two neutrons cannot directly form a deuterium atom, as a deuterium atom consists of a proton and a neutron bound together, with an electron orbiting them.
   • This process is not feasible based on the particle interaction described.
4. $p \to n + e^{+} + \overline{v}$:
   • This reaction represents the decay of a proton into a neutron, a positron, and an antineutrino. This is called inverse beta decay.
   • This process is energetically not allowed under normal conditions because the neutron is heavier than the proton, requiring an input of energy rather than releasing energy.

Based on these examinations, the correct and allowed process is $n+p \to d+\gamma$, which conserves both energy and momentum.