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

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

Explore More

Similar Questions

Deuteron is a bound state of a neutron and a proton with a binding energy $B = 2.2 \, MeV$. $A$ $\gamma$-ray of energy $E$ is aimed at a deuteron nucleus to try to break it into a (neutron + proton) such that the $n$ and $p$ move in the direction of the incident $\gamma$-ray. If $E = B$,show that this cannot happen. Hence,calculate how much bigger than $B$ must $E$ be for such a process to happen.

Difficult
View Solution

If the binding energy of $N^{14}$ is $7.5 \text{ MeV}$ per nucleon and that of $N^{15}$ is $7.7 \text{ MeV}$ per nucleon, then the energy required to remove a neutron from $N^{15}$ is (in $\text{ MeV}$)

As the mass number increases, the binding energy per nucleon .....

What is nuclear energy? Explain how nuclear energy is released from the curve of binding energy.

Difficult
View Solution

If the binding energy per nucleon in $Li^7$ and $He^4$ nuclei are respectively $5.60 \ MeV$ and $7.06 \ MeV$, then the energy of the reaction $Li^7 + p \to 2He^4$ is ......... $MeV$.

Difficult
View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo