In the nuclear fission `._(1)H^(2) + ._(1)H^(2) rarr ._(2)He^(4)` the masses of `._(1)H^(2) and ._(2)he^(4)` are 2.014 mu and 4.003 mu respectively. The energy released/atom of helium formed is ....MeV

Options

A. 16.76
B. 26.38
C. 13.26
D. 23.275

Explanation

To solve the problem of calculating the energy released per atom of helium formed in the nuclear fusion reaction \( _{1}^{2}H + _{1}^{2}H \rightarrow _{2}^{4}He \), we will follow these steps:

### Step 1: Determine the mass defect
The mass defect (\( \Delta m \)) is calculated by subtracting the mass of the products from the mass of the reactants.

1. **Mass of reactants**:
- There are 2 deuterium nuclei (each with a mass of 2.014 mu).
- Total mass of reactants = \( 2 \times 2.014 \, \text{mu} = 4.028 \, \text{mu} \).

2. **Mass of products**:
- The mass of the helium nucleus = 4.003 mu.

3. **Calculate the mass defect**:
\[ \Delta m = \text{mass of reactants} - \text{mass of products} = 4.028 \, \text{mu} - 4.003 \, \text{mu} = 0.025 \, \text{mu}. \]

### Step 2: Convert mass defect to energy
The energy equivalent of the mass defect can be calculated using Einstein's equation \( E = \Delta m c^2 \). However, in nuclear physics, we often use the conversion factor \( 1 \, \text{mu} \approx 931 \, \text{MeV} \).

1. **Calculate the energy released**:
\[ E = \Delta m \times 931 \, \text{MeV} = 0.025 \, \text{mu} \times 931 \, \text{MeV/mu} = 23.275 \, \text{MeV}. \]

### Conclusion
The energy released per atom of helium formed in the reaction is approximately **23.275 MeV**.

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