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

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$n + n \to$ deuterium atom (electron bound to the nucleus)

$n + p \to d + γ$

$e^{+} + e^{-} \to γ$

$p \to n + e^{+} + \bar{v}$

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$n + n \to$ deuterium atom is not energetically feasible as mass on product side is more than mass on reactant side.
$p \to n + e^{+} + ν^{-}$ is also not energetically feasible.
$e^{+} + e^{-} \to γ$ is not possible as momentum conservation does not hold.
rather $e^{+} + e^{-} \to 2 γ$ would be correct.
As $n + p \to d + γ$ is energetically feasible as mass on product side is less than mass on reactant side, it could be possible.