The sun is believed to be getting its energy form the fusion of four protons to form a helium nucleus and a pair of positrons. Calculate the release of energy per fusion in MeV. Mass of proton=1.007825 a.m.u. , mass of positron =0.000549 a.m.u., mass of helium nucleus =4.002603 a.m.u. Take 1a.m.u. =931MeV.

Explanation

To calculate the release of energy per fusion of four protons into a helium nucleus and a pair of positrons, we will follow these steps:

### Step 1: Write the Fusion Reaction
The fusion reaction can be represented as:
\[ 4 \, \text{Protons} \rightarrow \text{Helium nucleus} + 2 \, \text{Positrons} \]
This can be written in terms of their symbols:
\[ 4 \, \text{H} \rightarrow \text{He} + 2 \, e^+ \]

### Step 2: Calculate the Mass Defect
The mass defect (\(\Delta m\)) is calculated using the masses of the reactants and products. The mass defect is given by:
\[ \Delta m = \text{(mass of reactants)} - \text{(mass of products)} \]
For our reaction:
- Mass of 4 protons = \(4 \times \text{mass of proton} = 4 \times 1.007825 \, \text{a.m.u.}\)
- Mass of helium nucleus = \(4.002603 \, \text{a.m.u.}\)
- Mass of 2 positrons = \(2 \times 0.000549 \, \text{a.m.u.}\)

Now substituting the values:
\[ \Delta m = (4 \times 1.007825) - (4.002603 + 2 \times 0.000549) \]

Calculating each term:
- Mass of 4 protons = \(4 \times 1.007825 = 4.0313 \, \text{a.m.u.}\)
- Mass of 2 positrons = \(2 \times 0.000549 = 0.001098 \, \text{a.m.u.}\)

Now substituting these values into the equation for mass defect:
\[ \Delta m = 4.0313 - (4.002603 + 0.001098) = 4.0313 - 4.003701 = 0.027599 \, \text{a.m.u.} \]

### Step 3: Calculate the Energy Released
The energy released per fusion can be calculated using the mass defect and the conversion factor \(1 \, \text{a.m.u.} = 931 \, \text{MeV}\):
\[ \text{Energy released} = \Delta m \times 931 \, \text{MeV} \]
Substituting the value of \(\Delta m\):
\[ \text{Energy released} = 0.027599 \times 931 \]

Calculating this:
\[ \text{Energy released} \approx 25.7 \, \text{MeV} \]

### Final Answer
The energy released per fusion of four protons to form a helium nucleus and a pair of positrons is approximately **25.7 MeV**.

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