In the reaction `._1^2 H + ._1^3 H rarr ._2^4 He + ._0^1 n`, if the binding energies of `._1^2 H, ._1^3 H` and `._2^4 He` are respectively `a,b` and `c` (in MeV), then the energy (in MeV) released in this reaction is.

Options

A. `c + a - b`
B. `c - a - b`
C. `a + b + c`
D. `a + b - c`

Explanation

To find the energy released in the nuclear reaction given by:

\[ _1^2 H + _1^3 H \rightarrow _2^4 He + _0^1 n \]

we will use the concept of binding energy. The binding energies of the nuclei involved in the reaction are given as follows:

- Binding energy of \( _1^2 H \) (Deuterium) = \( a \) MeV
- Binding energy of \( _1^3 H \) (Tritium) = \( b \) MeV
- Binding energy of \( _2^4 He \) (Helium) = \( c \) MeV

### Step-by-step Solution:

1. **Identify the Initial and Final States:**
- Initial state: \( _1^2 H + _1^3 H \)
- Final state: \( _2^4 He + _0^1 n \)

2. **Calculate the Binding Energy of Initial Nuclei:**
- The total binding energy of the initial nuclei (Deuterium and Tritium) is:
\[ \text{Total Initial Binding Energy} = a + b \]

3. **Calculate the Binding Energy of Final Nuclei:**
- The total binding energy of the final nuclei (Helium and Neutron) is:
\[ \text{Total Final Binding Energy} = c + 0 = c \]
(The neutron has no binding energy since it is not bound in a nucleus).

4. **Calculate the Energy Released:**
- The energy released in the reaction can be calculated by finding the difference between the total binding energy of the initial state and the final state:
\[ \text{Energy Released} = \text{Total Initial Binding Energy} - \text{Total Final Binding Energy} \]
- Substituting the values we calculated:
\[ \text{Energy Released} = (a + b) - c \]

5. **Final Expression:**
- Therefore, the energy released in the reaction is:
\[ \text{Energy Released} = c - a - b \text{ MeV} \]

### Conclusion:
The energy released in the reaction \( _1^2 H + _1^3 H \rightarrow _2^4 He + _0^1 n \) is given by:

\[ \text{Energy Released} = c - a - b \text{ MeV} \]