A neutron is absorbed by a `._3Li^6` nucleus with subsequent emission of an alpha particle. Write the corresponding nuclear reaction. Calculate the energy released in this reaction.
`m(._3Li^6)=6.015126u, m(._2He^4)=4.0026044u`
`m(._0n^1)=1.0086654u, m(._1He^3)=3.016049u`
Take `1u=931MeV`.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write the Nuclear Reaction
The nuclear reaction involves a lithium-6 nucleus absorbing a neutron and emitting an alpha particle. The reaction can be written as follows:

\[ \ _{3}^{6}\text{Li} + \ _{0}^{1}\text{n} \rightarrow \ _{2}^{4}\text{He} + \ _{1}^{3}\text{He} + Q \]
...