When four hydrogen nuclei combine to form a helium nucleus in the interior of sun, the loss in mass is 0.0265 a.m.u. How much energy is released ?

Answer

To solve the problem of how much energy is released when four hydrogen nuclei combine to form a helium nucleus, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Mass Defect**:
We are given that the mass defect (Δm) when four hydrogen nuclei combine to form one helium nucleus is 0.0265 atomic mass units (a.m.u).

2. **Use Einstein's Mass-Energy Equivalence**:
According to Einstein's equation, the energy (E) released due to a mass defect is given by the formula:
\[ E = \Delta m \cdot c^2 \]
where:
- \(E\) is the energy in mega electron volts (MeV),
- \(\Delta m\) is the mass defect in a.m.u,
- \(c\) is the speed of light.

3. **Convert Mass Defect to Energy**:
We know that \(c^2\) can be expressed in terms of energy per atomic mass unit:
\[ c^2 = 931.5 \text{ MeV/a.m.u} \]
Therefore, we can rewrite the energy equation as:
\[ E = \Delta m \cdot 931.5 \text{ MeV/a.m.u} \]

4. **Substitute the Values**:
Now, substituting the given mass defect into the equation:
\[ E = 0.0265 \text{ a.m.u} \cdot 931.5 \text{ MeV/a.m.u} \]

5. **Calculate the Energy**:
Performing the multiplication:
\[ E = 0.0265 \cdot 931.5 \approx 24.68 \text{ MeV} \]

6. **Final Answer**:
Thus, the energy released when four hydrogen nuclei combine to form a helium nucleus is approximately:
\[ E \approx 24.7 \text{ MeV} \]