Three `prop` -particles (`m_prop` = 4.0026) amu are combined to form a `C^12` nucleus. The energy released will be

Options

A. 4.02 Mev
B. 5.57 Mev
C. 3.39 Mev
D. 7.26 Mev

Answer

To solve the problem of finding the energy released when three alpha particles combine to form a \( C^{12} \) nucleus, we will follow these steps:

### Step 1: Determine the mass of three alpha particles
The mass of one alpha particle (\( m_{\text{alpha}} \)) is given as 4.0026 amu. Therefore, the mass of three alpha particles is calculated as follows:

\[ \text{Mass of 3 alpha particles} = 3 \times m_{\text{alpha}} = 3 \times 4.0026 \, \text{amu} = 12.0078 \, \text{amu} \]

### Step 2: Calculate the mass defect
The mass defect (\( \Delta m \)) is the difference between the total mass of the individual particles and the mass of the nucleus formed. The mass of the \( C^{12} \) nucleus is 12 amu. Thus, we can calculate the mass defect as follows:

\[ \Delta m = \text{Mass of 3 alpha particles} - \text{Mass of } C^{12} \]
\[ \Delta m = 12.0078 \, \text{amu} - 12 \, \text{amu} = 0.0078 \, \text{amu} \]

### Step 3: Convert the mass defect to energy
To find the energy released, we use the mass-energy equivalence principle, which states that the energy released (\( E \)) can be calculated using the formula:

\[ E = \Delta m \times c^2 \]

However, in nuclear physics, we use the conversion factor that 1 amu corresponds to approximately 931.5 MeV. Therefore, we can calculate the energy released as follows:

\[ E = \Delta m \times 931.5 \, \text{MeV/amu} \]
\[ E = 0.0078 \, \text{amu} \times 931.5 \, \text{MeV/amu} = 7.26 \, \text{MeV} \]

### Final Answer
The energy released when three alpha particles combine to form a \( C^{12} \) nucleus is approximately **7.26 MeV**.

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