Discussion
The following nuclear reaction is induced by bombarding neutrons on the \(^{13}\)C target.
\(^{13}\)C + n \(\rightarrow\) \(^{10}\)Be + \(^{4}\)He
If the R value of this reaction is - 3.835 MeV, the minimum neutron energy required to induce the reaction is nearly :
Answer: B
The minimum neutron energy is nearly 4.13 MeV.
For an endothermic reaction the threshold kinetic energy of the projectile exceeds |Q| because momentum must be conserved, so part of the energy goes into motion of the centre of mass:
\[ \begin{aligned} E_{th} &= -Q\left(1 + \frac{m_{n}}{M_{C}}\right) \\ &= 3.835 \times \frac{14}{13} \approx 4.13\ \text{MeV} \end{aligned} \]
The option 4.835 MeV adds 1 MeV for no reason, and 5.835 MeV adds 2 MeV; 2.13 MeV is below |Q| and could not drive the reaction at all.
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