How many kilowatt hours of energy are released from 25 g deuterium `""_(1)^(2)H` fuel in the fusion reaction, `""_(1)^(2)H+""_(1)^(2)Hto""_(2)^(4)He+gamma` where the masses are `""_(1)^(2)H=2.014102u` and `""_(2)^(4)He=4.002603u`

Options

A. `1xx10^(6)kWh`
B. `3xx10^(6)kWh`
C. `2xx10^(6)kWh`
D. `4xx10^(6)kWh`

Answer

To solve the problem of how many kilowatt hours of energy are released from 25 g of deuterium fuel in the fusion reaction, we will follow these steps:

### Step 1: Calculate the mass defect (Δm)
The mass defect (Δm) in the fusion reaction can be calculated using the masses of the reactants and products. The reaction is:

\[ \text{D} + \text{D} \rightarrow \text{He} + \gamma \]

Where:
- Mass of deuterium (\( \text{D} \)) = \( 2.014102 \, u \)
- Mass of helium (\( \text{He} \)) = \( 4.002603 \, u \)

The mass defect is given by:

\[ \Delta m = 2 \times m_{\text{D}} - m_{\text{He}} \]

Substituting the values:

\[ \Delta m = 2 \times 2.014102 \, u - 4.002603 \, u = 4.028204 \, u - 4.002603 \, u = 0.025601 \, u \]

### Step 2: Convert the mass defect to energy
Using Einstein's equation \( E = \Delta m c^2 \), we convert the mass defect to energy. The energy equivalent of 1 atomic mass unit (u) is approximately \( 931.5 \, \text{MeV} \).

Thus, the energy released in one reaction is:

\[ E = \Delta m \times 931.5 \, \text{MeV} = 0.025601 \, u \times 931.5 \, \text{MeV/u} \approx 23.8 \, \text{MeV} \]

### Step 3: Calculate the number of moles of deuterium
Next, we calculate the number of moles of deuterium in 25 g:

\[ \text{Molar mass of deuterium} = 2 \, \text{g/mol} \]

\[ \text{Number of moles} = \frac{25 \, \text{g}}{2 \, \text{g/mol}} = 12.5 \, \text{moles} \]

### Step 4: Calculate the number of reactions
Since each reaction consumes 2 moles of deuterium, the number of reactions that can occur is:

\[ \text{Number of reactions} = \frac{12.5 \, \text{moles}}{2 \, \text{moles/reaction}} = 6.25 \, \text{reactions} \]

### Step 5: Calculate the total energy released
The total energy released from all reactions is:

\[ E_{\text{total}} = \text{Number of reactions} \times E_{\text{per reaction}} = 6.25 \times 23.8 \, \text{MeV} \approx 148.75 \, \text{MeV} \]

### Step 6: Convert energy from MeV to Joules
To convert MeV to Joules, we use the conversion factor \( 1 \, \text{MeV} = 1.6 \times 10^{-13} \, \text{J} \):

\[ E_{\text{total}} = 148.75 \, \text{MeV} \times 1.6 \times 10^{-13} \, \text{J/MeV} \approx 2.38 \times 10^{-11} \, \text{J} \]

### Step 7: Convert energy from Joules to kilowatt hours
Finally, we convert Joules to kilowatt hours using the conversion \( 1 \, \text{kWh} = 3.6 \times 10^6 \, \text{J} \):

\[ E_{\text{total}} \, (\text{kWh}) = \frac{2.38 \times 10^{-11} \, \text{J}}{3.6 \times 10^6 \, \text{J/kWh}} \approx 6.61 \times 10^{-18} \, \text{kWh} \]

### Conclusion
The total energy released from 25 g of deuterium in the fusion reaction is approximately \( 6.61 \times 10^{-18} \, \text{kWh} \).

---