If a star can convert all the He nuclei completely into oxygen nuclei. The energy released per oxygen nuclei is (Mass of the helium nucleus is `4.0026` amu and mass of oxygen nucleus is `15.9994` amu)

Options

A. `10.24 MeV`
B. `23.9 Me V`
C. `7.56 MeV`
D. 5MeV

Answer

To solve the problem of calculating the energy released when helium nuclei are converted into oxygen nuclei, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reaction**:
The reaction involves converting 4 helium nuclei (\(^4He\)) into 1 oxygen nucleus (\(^16O\)). The balanced equation can be written as:
\[ 4 \, ^4He \rightarrow \, ^{16}O \]

2. **Calculate the Mass of Reactants and Products**:
- Mass of 4 helium nuclei:
\[ \text{Mass of } 4 \, ^4He = 4 \times 4.0026 \, \text{amu} = 16.0104 \, \text{amu} \]
- Mass of 1 oxygen nucleus:
\[ \text{Mass of } ^{16}O = 15.9994 \, \text{amu} \]

3. **Calculate the Mass Defect (\(\Delta m\))**:
The mass defect is the difference between the mass of the reactants and the mass of the products:
\[ \Delta m = \text{Mass of reactants} - \text{Mass of products} = 16.0104 \, \text{amu} - 15.9994 \, \text{amu} = 0.011 \, \text{amu} \]

4. **Convert Mass Defect to Energy**:
To find the energy released, we use the relation:
\[ E = \Delta m \times c^2 \]
where \(c^2\) can be converted from amu to MeV using the conversion factor \(1 \, \text{amu} \approx 931.5 \, \text{MeV}\):
\[ E = 0.011 \, \text{amu} \times 931.5 \, \text{MeV/amu} \approx 10.24 \, \text{MeV} \]

5. **Conclusion**:
The energy released per oxygen nucleus when 4 helium nuclei are converted into 1 oxygen nucleus is approximately:
\[ \boxed{10.24 \, \text{MeV}} \]