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# A tritium gas target is bombareled with a beam of mono-energetic protons of K.E 3MeV . What is K.E of neutrons emitted at an angle $\;30^{0}\;$ with incident beam ?  Atomic masses are $\;H^{1}=1.007276 amu$  $\;m^{1}=1.008665 amu$ $\;H_{1}^{3}=3.06056 amu$ and $\;He^{2}=3.016030 amu$

$(a)\;1.44 MeV\qquad(b)\;2.36 MeV\qquad(c)\;1.92 MeV\qquad(d)\;2.6 MeV$

Answer : (a) $\;1.44 MeV$
Explanation :
The nuclear reaction in question can be written as $\;P_{1}^{1}+H_{1}^{3} \to He_{2}^{3}+n_{0}^{1}+Q$
When Q is energy released during reaction given by
$Q=[(m_{P}+m_{H})-(m_{He}+m_{m})]\;amu$
$=(1.0072765+30.16050)-(3.016030-1.008665)$
$=-0.001369 amu=-1.2745 MeV$
The kinetic energy of associated emitted motion is given by
$K_{n}=[u_{n}\pm (U_{n}^{2}+v^2)]---(1)$
where $\;u_{n}=\large\frac{(m_{p} m_{n} K_{p})^{\large\frac{1}{2}}}{m_{He}+m_{n}} cos \theta$
$u_{n}=\large\frac{(1.007276\times1.008665\times3)^{\large\frac{1}{2}}}{3.016030+1.008665}\times\large\frac{\sqrt{3}}{2}$
$u_{n} \approx =0.3753$
and $\;V=\large\frac{m_{He}Q+K_{p}(m_{He}-m_{p})}{m_{He}+m_{n}}$
$V \approx 0.5424$
$K_{n}=[0.3753\pm 0.8266]$
$K_{n}=(1.2019)^2$
$K_{n}=1.44 MeV\;.$