Show that the energy released when three alpha particles fuse to form `""^(12)C` is 7.27 MeV. The atomic mass of `""^(4)He` is 4.0026u and that of `""^(12)C` is 12.0000u.

Answer

To show that the energy released when three alpha particles fuse to form \(^{12}\text{C}\) is 7.27 MeV, we can follow these steps:

### Step 1: Understand the Reaction
We know that three alpha particles (which are \(^{4}\text{He}\) nuclei) fuse to form one \(^{12}\text{C}\) nucleus. The reaction can be represented as:
\[ 3 \, ^4\text{He} \rightarrow \, ^{12}\text{C} + \text{Energy} \]

### Step 2: Calculate the Mass of Reactants
The mass of three alpha particles is:
\[ \text{Mass of reactants} = 3 \times \text{mass of } ^4\text{He} = 3 \times 4.0026 \, u = 12.0078 \, u \]

### Step 3: Calculate the Mass of Products
The mass of the product (the \(^{12}\text{C}\) nucleus) is:
\[ \text{Mass of products} = \text{mass of } ^{12}\text{C} = 12.0000 \, u \]

### Step 4: Calculate the Mass Defect
The mass defect (\(\Delta m\)) is the difference between the total mass of the reactants and the total mass of the products:
\[ \Delta m = \text{Mass of reactants} - \text{Mass of products} = 12.0078 \, u - 12.0000 \, u = 0.0078 \, u \]

### Step 5: Convert Mass Defect to Energy
We can convert the mass defect into energy using Einstein's equation \(E = \Delta m c^2\). The conversion factor from atomic mass units (u) to MeV is approximately 931.5 MeV/u:
\[ E = \Delta m \times 931.5 \, \text{MeV/u} = 0.0078 \, u \times 931.5 \, \text{MeV/u} \]

### Step 6: Calculate the Energy Released
Now, we perform the calculation:
\[ E = 0.0078 \times 931.5 \approx 7.27 \, \text{MeV} \]

### Conclusion
Thus, the energy released when three alpha particles fuse to form \(^{12}\text{C}\) is approximately 7.27 MeV.

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