Question

An alpha particle with kinetic energy $T=5.3MeV$ initiates a nuclear reaction
$B{e}_4^9+{\alpha }_2^4\to C_6^{12}+n_0^1$with energy yield $Q=5.7Mev$. The kinetic enrgy of the neutron going at right angles to the direction of motion of the alpha particle will be (in $MeV$)

Answer 1

It is understood that ${\mathrm{Be}}_4^9$ is initially at rest.
The momentum of incoming alpha particle is $\sqrt{2m_{\alpha }{T}_{\alpha }}$ and momentum of the outgoing neutron is $\sqrt{2m_n{T}_n}\hat{j}$.
Then by conservation of momentum,
Momentum of $C_6^{12}$ $=\sqrt{2m_{\alpha }{T}_{\alpha }}\hat{i}-\sqrt{2m_n{T}_n}\hat{j}$
Kinetic energy of carbon $T_C=\frac{{p_c}^2}{2m_c}=\frac{2m_{\alpha }{T}_{\alpha }+2m_n{T}_n}{2m_c}$
By energy conservation,

$T_{\alpha }+Q=T_n+T_c=T_n+\frac{2m_{\alpha }{T}_{\alpha }+2m_n{T}_n}{2m_{\alpha }}$
Thus,
$T_n=\frac{m_c(T+Q)-m_{\alpha }T}{m_c+m_n}=\frac{(m_c-m_{\alpha })T+m_{c}Q}{m_c+m_n}=\frac{Q+(1-\frac{m_{\alpha }}{m_c})T}{m_c+m_n}\text{Substituting the values, we get,}{T}_n=8.52MeV$