An alpha particle with kinetic energy $T = 5.3 M e V$ initiates a nuclear reaction
$B e_{4}^{9} + α_{2}^{4} \to C_{6}^{12} + n_{0}^{1}$ with energy yield $Q = 5.7 M e v$ . The kinetic enrgy of the neutron going at right angles to the direction of motion of the alpha particle will be (in $M e V$ )

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Solution

It is understood that ${Be}_{4}^{9}$ is initially at rest.
The momentum of incoming alpha particle is $\sqrt{2 m_{α} T_{α}}$ and momentum of the outgoing neutron is $\sqrt{2 m_{n} T_{n}} ˆ j$ .
Then by conservation of momentum,
Momentum of $C_{6}^{12}$ $= \sqrt{2 m_{α} T_{α}} ˆ i - \sqrt{2 m_{n} T_{n}} ˆ j$
Kinetic energy of carbon $T_{C} = \frac{{p_{c}}_{c}^{2}}{2 m_{c}} = \frac{2 m_{α} T_{α} + 2 m_{n} T_{n}}{2 m_{c}}$
By energy conservation,

$T_{α} + Q = T_{n} + T_{c} = T_{n} + \frac{2 m_{α} T_{α} + 2 m_{n} T_{n}}{2 m_{α}}$
Thus,
$T_{n} = \frac{m_{c} (T + Q) - m_{α} T}{m_{c} + m_{n}} = \frac{(m_{c} - m_{α}) T + m_{c} Q}{m_{c} + m_{n}} = \frac{Q + (1 - \frac{m_{α}}{m_{c}}) T}{m_{c} + m_{n}} Substituting the values, we get, T_{n} = 8.52 M e V$

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T = 5.3 M e V

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