The atomic masses of the hydrogen isotopes are
Hydrogen `m_1H^1=1.007825` amu
Deuterium `m_1H^2=2.014102` amu
Tritium `m_1H^3=3.016049` amu
The energy released in the reaction,
`_1H^2+_1H^2rarr_1H^3+_1H^1` is nearly

Options

A. (a) 1MeV
B. (b) 2MeV
C. (c) 4MeV
D. (d) 8 MeV

Explanation

To find the energy released in the reaction \( _1H^2 + _1H^2 \rightarrow _1H^3 + _1H^1 \), we will follow these steps:

### Step 1: Identify the masses of the reactants and products
- Mass of Deuterium (\( _1H^2 \)): \( m_1H^2 = 2.014102 \) amu
- Mass of Tritium (\( _1H^3 \)): \( m_1H^3 = 3.016049 \) amu
- Mass of Hydrogen (\( _1H^1 \)): \( m_1H^1 = 1.007825 \) amu

### Step 2: Calculate the total mass of the reactants
Since there are two deuterium nuclei in the reactants:
\[ \text{Total mass of reactants} = 2 \times m_1H^2 = 2 \times 2.014102 \, \text{amu} = 4.028204 \, \text{amu} \]

### Step 3: Calculate the total mass of the products
The products consist of one tritium nucleus and one hydrogen nucleus:
\[ \text{Total mass of products} = m_1H^3 + m_1H^1 = 3.016049 \, \text{amu} + 1.007825 \, \text{amu} = 4.023874 \, \text{amu} \]

### Step 4: Calculate the change in mass (\( \Delta m \))
\[ \Delta m = \text{mass of reactants} - \text{mass of products} = 4.028204 \, \text{amu} - 4.023874 \, \text{amu} = 0.004330 \, \text{amu} \]

### Step 5: Convert the change in mass to energy
Using the conversion factor \( 1 \, \text{amu} = 931.5 \, \text{MeV/c}^2 \):
\[ \text{Energy released} (Q) = \Delta m \times 931.5 \, \text{MeV} = 0.004330 \, \text{amu} \times 931.5 \, \text{MeV} \approx 4.037 \, \text{MeV} \]

### Step 6: Round the energy to a suitable value
The energy released is approximately \( 4 \, \text{MeV} \).

### Final Answer
The energy released in the reaction is nearly \( 4 \, \text{MeV} \).

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