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. a) a+b+c
B. b) c+a-b
C. c) c-a-b
D. d) a+b-c

Explanation

To find the energy released in the reaction \( _1^2 H + _1^3 H \rightarrow _2^4 He + _0^1 n \), we will use the concept of binding energy. The binding energy is the energy required to disassemble a nucleus into its constituent protons and neutrons. A higher binding energy indicates a more stable nucleus.

### Step-by-Step Solution:

1. **Identify the Binding Energies:**
- Let the binding energy of \( _1^2 H \) (Deuterium) be \( a \) MeV.
- Let the binding energy of \( _1^3 H \) (Tritium) be \( b \) MeV.
- Let the binding energy of \( _2^4 He \) (Helium-4) be \( c \) MeV.

2. **Calculate Total Binding Energy of Reactants:**
- The total binding energy of the reactants (Deuterium and Tritium) is the sum of their individual binding energies:
\[ \text{Total Binding Energy of Reactants} = a + b \]

3. **Calculate Total Binding Energy of Products:**
- The total binding energy of the products (Helium-4 and a neutron) is simply the binding energy of Helium-4, since the neutron has no binding energy:
\[ \text{Total Binding Energy of Products} = c \]

4. **Determine Energy Released:**
- The energy released in the reaction is equal to the difference in binding energy between the products and the reactants. Since the products are more stable (higher binding energy), the energy released is given by:
\[ \text{Energy Released} = \text{Total Binding Energy of Products} - \text{Total Binding Energy of Reactants} \]
- Substituting the values we calculated:
\[ \text{Energy Released} = c - (a + b) \]
- This simplifies to:
\[ \text{Energy Released} = c - a - b \]

### Final Answer:
The energy released in the reaction is \( c - a - b \) MeV.

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