A neutron is absorbed by a `._(3)^(6)Li` nucleus with the subsequent emission of an alpha particle.
(i) Write the corresponding nuclear reaction.
(ii) Calculate the energy released, in `MeV`, in this reaction.
[Given: mass `._(3)^(6)Li=6.015126u`, mass (neutron)`=1.0086654 u` mass (alpha particle) `=4.0026044 u` and mass (triton) `=3.010000 u`. Take `i u=931 MeV//c^(2)`]

Options

A. `10.92 MeV`
B. `10.415 MeV`
C. `9.791 MeV`
D. `8.73 MeV`

Answer

To solve the problem step by step, we will address both parts of the question: writing the nuclear reaction and calculating the energy released.

### Part (i): Write the corresponding nuclear reaction.

1. **Identify the reactants and products**:
- The reactant is a lithium-6 nucleus (`._(3)^(6)Li`) and a neutron (`n`).
- The products are an alpha particle (`._(2)^(4)He`) and a triton (`._(1)^(3)H`).

2. **Write the nuclear reaction**:
The reaction can be represented as follows:
\[ _{3}^{6}\text{Li} + n \rightarrow _{2}^{4}\text{He} + _{1}^{3}\text{H} \]

### Part (ii): Calculate the energy released in this reaction.

1. **Calculate the mass defect**:
- The mass of the reactants:
- Mass of lithium-6: \(6.015126 \, u\)
- Mass of neutron: \(1.0086654 \, u\)
- Total mass of reactants:
\[ \text{Mass of reactants} = 6.015126 + 1.0086654 = 7.0237914 \, u \]

- The mass of the products:
- Mass of alpha particle: \(4.0026044 \, u\)
- Mass of triton: \(3.010000 \, u\)
- Total mass of products:
\[ \text{Mass of products} = 4.0026044 + 3.010000 = 7.0126044 \, u \]

- Calculate the mass defect (\(\Delta m\)):
\[ \Delta m = \text{Mass of reactants} - \text{Mass of products} = 7.0237914 - 7.0126044 = 0.011187 \, u \]

2. **Convert mass defect to energy**:
- Use the mass-energy equivalence relation \(E = \Delta mc^2\).
- Given that \(1 \, u = 931 \, \text{MeV}/c^2\), we can calculate the energy released:
\[ E = 0.011187 \, u \times 931 \, \text{MeV}/c^2 = 10.415 \, \text{MeV} \]

### Final Answers:
(i) The nuclear reaction is:
\[ _{3}^{6}\text{Li} + n \rightarrow _{2}^{4}\text{He} + _{1}^{3}\text{H} \]

(ii) The energy released in the reaction is approximately \(10.415 \, \text{MeV}\).