Question

A tritum gas target is bombarded with a beam of mono-energetic protons of $K.E.3$. $MeV$. What is the $K.E.$ of the neutrons which are emitted at an angle ${30}^o$ with the incident beam?
Atomic masses are: $H^1=1.007276amu{n}^1=1.008665amu$
$H_1^3=3.06056amu$ and $H_e^3=3.016030amu.$

Answer 1

The nuclear reaction in question can be written as-
$P_1^1+H_3^1\to H{e}_3^2+n_1^0+Q$
when $Q$ is the energy released during the reaction given by
$Q=\left[\right(m_p)-(m_H)-(m_{He}+m_n\left)\right]amu$
$=(1.0072765+3.06056)-(3.016030-1.008665)$
$-0.001369amu=-1.2745MeV$
From conservation of energy, we get
$k_p=k_n+k_{He}+Q$
$\Rightarrow$ $k_n=k_H-k_{He}-Q=3-k_{He}+1.2745=4.2745-k_{He}$ . . . ($1$)
From conservation of linear moment along and perpendicular to the direction of proton beam, we get
$p_H=p_n\cos {30}^o+p_{He}\cos \theta \Rightarrow$ $p_H=\frac{\sqrt{3}}{2}{p}_n+p_{He}\cos \theta$ . . . ($2$)
and $0=p_n\sin {30}^o-p_{He}\sin \theta$
$\Rightarrow$ $p_{He}\sin \theta =\frac{p_n}{2}$ . . . ($3$)
Using the formula for kinetic energy $k=\frac{p^2}{2m}$ in the above equation $\left(1\right)-\left(3\right)$, and putting the values of masses from the given data, we get,
$\Rightarrow k_n=1.444MeV$