The atomic masses of deuteron, helium, neutron are 2.014 amu, 3.017 amu and 1.008 amu respectively. On fusion of 0.5 kg of deuterium,`""_(1)H^(2) + ""_(1)H^(2) to ""_(2)He^(3) + ""_(0)n^(1)`, the total energy released is

Options

A. `6.72 xx 10^(36)` J
B. `7.72 xx 10^(26)` J
C. `6.72 xx 10^(13)` J
D. `8.72 xx 10^(26)` J

Explanation

To solve the problem, we need to calculate the total energy released during the fusion of 0.5 kg of deuterium. The reaction given is:

\[ _{1}^{2}H + _{1}^{2}H \rightarrow _{2}^{3}He + _{0}^{1}n \]

### Step 1: Calculate the mass defect
First, we need to find the mass of the reactants and products.

**Mass of Reactants:**
- Mass of 2 deuterons:
\[ 2 \times 2.014 \, \text{amu} = 4.028 \, \text{amu} \]

**Mass of Products:**
- Mass of helium-3:
\[ 3.017 \, \text{amu} \]
- Mass of neutron:
\[ 1.008 \, \text{amu} \]
- Total mass of products:
\[ 3.017 + 1.008 = 4.025 \, \text{amu} \]

**Mass defect (Δm):**
\[ \Delta m = \text{Mass of Reactants} - \text{Mass of Products} \]
\[ \Delta m = 4.028 \, \text{amu} - 4.025 \, \text{amu} = 0.003 \, \text{amu} \]

### Step 2: Convert mass defect to energy
Using Einstein's equation \( E = \Delta m c^2 \), we can convert the mass defect into energy.

1 amu is equivalent to 931.5 MeV/c². Therefore, the energy released (in MeV) is:
\[ E = 0.003 \, \text{amu} \times 931.5 \, \text{MeV/amu} = 2.7945 \, \text{MeV} \]

### Step 3: Calculate the number of deuterium nuclei in 0.5 kg
To find the total energy released, we need to determine how many deuterium nuclei are in 0.5 kg of deuterium.

1. **Convert kg to grams:**
\[ 0.5 \, \text{kg} = 500 \, \text{g} \]

2. **Calculate the number of moles of deuterium:**
The molar mass of deuterium (D) is approximately 2.014 g/mol.
\[ \text{Number of moles} = \frac{500 \, \text{g}}{2.014 \, \text{g/mol}} \approx 248.8 \, \text{mol} \]

3. **Calculate the number of deuterium nuclei:**
Using Avogadro's number \( 6.022 \times 10^{23} \, \text{atoms/mol} \):
\[ \text{Number of deuterium nuclei} = 248.8 \, \text{mol} \times 6.022 \times 10^{23} \, \text{atoms/mol} \approx 1.496 \times 10^{26} \, \text{nuclei} \]

### Step 4: Calculate the number of reactions
Since each reaction requires 2 deuterium nuclei:
\[ \text{Number of reactions} = \frac{1.496 \times 10^{26}}{2} \approx 7.48 \times 10^{25} \]

### Step 5: Calculate total energy released
The total energy released can be calculated by multiplying the energy released per reaction by the number of reactions:
\[ \text{Total Energy} = \text{Energy per reaction} \times \text{Number of reactions} \]
\[ \text{Total Energy} = 2.7945 \, \text{MeV} \times 7.48 \times 10^{25} \]

1 MeV = \( 1.6 \times 10^{-13} \, \text{J} \):
\[ \text{Total Energy} = 2.7945 \times 7.48 \times 10^{25} \times 1.6 \times 10^{-13} \, \text{J} \]
\[ \text{Total Energy} \approx 3.365 \times 10^{13} \, \text{J} \]

### Final Answer
The total energy released during the fusion of 0.5 kg of deuterium is approximately:
\[ \boxed{3.365 \times 10^{13} \, \text{J}} \]