Assuming that in a star, three alpha particle join in a single fusion reaction to form `._(6)C^(12)` nucleus. Calculate the energy released in this reaction. Given mass of `._(2)He^(4)` is 4.002604 a.m.u. and that of `._(6)C^(12)` is 12 a.m.u. Take 1a.m.u. =931MeV.

Explanation

To calculate the energy released in the fusion reaction where three alpha particles combine to form a carbon-12 nucleus, we can follow these steps:

### Step 1: Understand the reaction
In this reaction, three alpha particles (each being a helium-4 nucleus, denoted as \(_{2}^{4}\text{He}\)) combine to form a carbon-12 nucleus (\(_{6}^{12}\text{C}\)). The reaction can be represented as:
\[ 3 \, _{2}^{4}\text{He} \rightarrow \, _{6}^{12}\text{C} \]

### Step 2: Calculate the total mass of the reactants
The mass of one alpha particle (helium-4) is given as:
\[ \text{Mass of } _{2}^{4}\text{He} = 4.002604 \, \text{a.m.u.} \]
Thus, the total mass of three alpha particles is:
\[ \text{Total mass of reactants} = 3 \times 4.002604 \, \text{a.m.u.} = 12.007812 \, \text{a.m.u.} \]

### Step 3: Calculate the mass defect (Δm)
The mass defect (Δm) is the difference between the total mass of the reactants and the mass of the product (carbon-12):
\[ \Delta m = \text{Total mass of reactants} - \text{Mass of } _{6}^{12}\text{C} \]
Given that the mass of carbon-12 is:
\[ \text{Mass of } _{6}^{12}\text{C} = 12 \, \text{a.m.u.} \]
We can calculate Δm:
\[ \Delta m = 12.007812 \, \text{a.m.u.} - 12 \, \text{a.m.u.} = 0.007812 \, \text{a.m.u.} \]

### Step 4: Convert the mass defect to energy
To find the energy released in the reaction, we can use the formula:
\[ E = \Delta m \times 931 \, \text{MeV/a.m.u.} \]
Substituting the value of Δm:
\[ E = 0.007812 \, \text{a.m.u.} \times 931 \, \text{MeV/a.m.u.} \]

### Step 5: Calculate the energy
Now we perform the multiplication:
\[ E = 0.007812 \times 931 \approx 7.27 \, \text{MeV} \]

### Final Answer
The energy released in the fusion reaction is approximately:
\[ \boxed{7.27 \, \text{MeV}} \]

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