A neutron with kinetic energy $K = 10 M e V$ activates a nuclear reaction
$n +^{12} C \to^{9} B e + α$
whose threshold $E_{t h} = 6.17 M e V$ . Find the kinetic energy of the alpha particles outgoing at right angles to the direction of the incoming neutron.
[Take $u = 931.1 M e V$ ]

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Solution

Conservation of energy gives,
$m_{1} c^{2} + m_{2} c^{2} + K_{1} = m_{3} c^{2} + m_{4} c^{2} + K_{3} + K_{4}$
or $Q + K_{1} = K_{3} + K_{4}$ ... ( $1$ )
Momentum conservation along $x -$ axis.
$\sqrt{2 m_{1} K_{1}} = \sqrt{2 m_{4} K_{4}} cos θ$
or, $m_{1} K_{1} = m_{4} K_{4} {cos}^{2} θ$ ...( $2$ )
Along $y -$ axis
$\sqrt{2 m_{3} K_{3}} = \sqrt{2 m_{4} K_{4}} sin θ$
or, $m_{3} K_{3} = m_{4} K_{4} {sin}^{2} θ$ ( $3$ )
Adding ( $2$ ) and ( $3$ )
$m_{1} K_{1} + m_{3} K_{3} = m_{4} K_{4}$
or, $K_{4} = \frac{m_{1}}{m_{4}} K_{1} + \frac{m_{3}}{m_{4}} K_{3}$
Substituting the value of $K_{4}$ in equation( $1$ ), we get
$Q + (1 - \frac{m_{1}}{m_{4}}) K_{1} = (1 + \frac{m_{3}}{m_{4}}) K_{3}$
Here, $Q = \frac{{- E}_{t h}}{(1 + \frac{m_{1}}{m_{2}})} = \frac{- 6.17}{(1 + \frac{1}{12})} = - 6.17 (\frac{12}{13}) = - 5.69 M e V$
Thus $(1 + \frac{4}{9}) K_{3} = - 5.69 + (1 - \frac{1}{9}) (10)$
$K_{3} = \frac{9}{13} [- 5.69 + 8.89] = 2.21 M e V$

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