Given the masses of various atomic particles
mp = 1.0072 u, mn = 1.0087 u, me = 0.000548 u,
${m_{\overline v }}$ = 0, md = 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?
Solution
n + n $\to$ deuterium atom (This is incorrect)
<br><br>Correct is n + p $\to$ d + $\gamma$
<br><br>p $\to$ n + e<sup>+</sup> + $\overline v$ <br>(This is incorrect as mass is increasing)
<br><br>e<sup>+</sup> + e<sup>-</sup> $\to$ $\gamma$ (This is incorrect)
<br><br>Correct is e<sup>+</sup> + e<sup>-</sup> $\to$ 2$\gamma$
About this question
Subject: Physics · Chapter: Atoms and Nuclei · Topic: Nuclear Binding Energy
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