Calculate the energy released during the combination of four hydrogen atoms and forming a helium atom along with two positrons. Use the atomic masses given as follows:
`m(""_(1)H^(1)) = 1.007824 u, m(""_(2)He^(4)) = 4.002604`u and mass of each positrons = 0.000548 u.

Answer

To calculate the energy released during the combination of four hydrogen atoms to form one helium atom and two positrons, we can follow these steps:

### Step 1: Calculate the mass of the reactants
We have 4 hydrogen atoms, and the mass of each hydrogen atom is given as:
\[ m(H) = 1.007824 \, \text{u} \]

Thus, the total mass of 4 hydrogen atoms is:
\[ m_{\text{reactants}} = 4 \times m(H) = 4 \times 1.007824 \, \text{u} = 4.031296 \, \text{u} \]

### Step 2: Calculate the mass of the products
The products of the reaction are 1 helium atom and 2 positrons. The mass of the helium atom is given as:
\[ m(He) = 4.002604 \, \text{u} \]
And the mass of each positron is:
\[ m(e^+) = 0.000548 \, \text{u} \]

Thus, the total mass of the products is:
\[ m_{\text{products}} = m(He) + 2 \times m(e^+) = 4.002604 \, \text{u} + 2 \times 0.000548 \, \text{u} = 4.002604 \, \text{u} + 0.001096 \, \text{u} = 4.003700 \, \text{u} \]

### Step 3: Calculate the mass defect
The mass defect (\( \Delta m \)) is the difference between the mass of the reactants and the mass of the products:
\[ \Delta m = m_{\text{reactants}} - m_{\text{products}} = 4.031296 \, \text{u} - 4.003700 \, \text{u} = 0.027596 \, \text{u} \]

### Step 4: Convert the mass defect to energy
To find the energy released, we use the formula:
\[ E = \Delta m \times c^2 \]
Where \( c^2 \) in terms of energy is given by:
\[ 1 \, \text{u} = 931.5 \, \text{MeV/c}^2 \]

Thus, the energy released is:
\[ E = 0.027596 \, \text{u} \times 931.5 \, \text{MeV/u} = 25.69 \, \text{MeV} \]

### Final Answer
The energy released during the combination of four hydrogen atoms to form one helium atom and two positrons is approximately \( 25.69 \, \text{MeV} \).

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