Assuming that four hydrogen atom combine to form a helium atom and two positrons, each of mass `0.00549u`, calculate the energy released. Given `m(._(1)H^(1))=1.007825u, m(._(2)He^(4))=4.002604u`.

Options

A. 25.71 MeV
B. 20.71 MeV
C. 22.75 MeV
D. 23.50 MeV

Answer

To calculate the energy released when four hydrogen atoms combine to form a helium atom and two positrons, we can follow these steps:

### Step 1: Write the Reaction
The reaction can be represented as:
\[ 4 \, ^1H \rightarrow \, ^4He + 2 \, e^+ \]
Where:
- \( ^1H \) is a hydrogen atom.
- \( ^4He \) is a helium atom.
- \( e^+ \) is a positron.

### Step 2: Calculate the Masses
We are given the following masses:
- Mass of one hydrogen atom, \( m(._{1}H^{1}) = 1.007825 \, u \)
- Mass of one helium atom, \( m(._{2}He^{4}) = 4.002604 \, u \)
- Mass of one positron, \( m(e^+) = 0.00549 \, u \)

### Step 3: Calculate the Total Mass of Reactants
The total mass of the reactants (4 hydrogen atoms) is:
\[ \text{Total mass of reactants} = 4 \times m(._{1}H^{1}) = 4 \times 1.007825 \, u = 4.0313 \, u \]

### Step 4: Calculate the Total Mass of Products
The total mass of the products (1 helium atom and 2 positrons) is:
\[ \text{Total mass of products} = m(._{2}He^{4}) + 2 \times m(e^+) \]
\[ = 4.002604 \, u + 2 \times 0.00549 \, u = 4.002604 \, u + 0.01098 \, u = 4.013584 \, u \]

### Step 5: Calculate the Change in Mass (Δm)
The change in mass (Δm) is given by:
\[ \Delta m = \text{Total mass of reactants} - \text{Total mass of products} \]
\[ = 4.0313 \, u - 4.013584 \, u = 0.017716 \, u \]

### Step 6: Convert Mass Change to Energy
Using Einstein's equation \( E = \Delta m \cdot c^2 \), and knowing that \( 1 \, u \) corresponds to \( 931 \, MeV \):
\[ Q = \Delta m \cdot 931 \, MeV/u \]
\[ Q = 0.017716 \, u \cdot 931 \, MeV/u = 16.5 \, MeV \]

### Step 7: Conclusion
The energy released during the reaction is approximately:
\[ Q \approx 16.5 \, MeV \]