# Nuclear physics problem – energy of reaction

**1. The problem statement, all variables and given/known data**

To activate the reaction ##(n,\alpha)## with stationary ##B^{11}## nuclei, neutrons must have the activation kinetic energy ##T_{th}=4\,MeV##. ##(n,\alpha)## means that ##n## is bombarded to obtain ##\alpha##. Find the energy of this reaction.

(Ans: ##Q=-11/12 T_{th}##)

**2. Relevant equations**

**3. The attempt at a solution**

The reaction taking place is:

$$n+B^{11}\rightarrow B^{12}\rightarrow A^{8}+\alpha$$

From conservation of energy

$$T_{th}=Q+T_A+T_{\alpha}\,\,\,\,\,(*)$$

where ##T## is used to denote the kinetic energy.

Also, from conservation of linear momentum, ##p_{A}=p_{\alpha} \Rightarrow p_{A}^2=p_{\alpha}^2 \Rightarrow m_AT_A=m_{\alpha}T_{\alpha}\,\,\,(**)##.

From (*) and (**), I get:

$$T_{th}=Q+T_A\left(1+\frac{m_A}{m_{\alpha}}\right)$$

I still need one more equation. :confused:

Any help is appreciated. Thanks!

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