Energy released in fusion of `1 kg` of deuterium nuclei.

Options

A. `8xx10^(13) J`
B. `3xx10^(27) J`
C. `2xx10^(7) "kWH"`
D. `8xx10^(23) "MeV"`

Answer

To find the energy released in the fusion of 1 kg of deuterium nuclei, we can follow these steps:

### Step 1: Understand the Fusion Reaction
The fusion reaction of deuterium nuclei can be represented as:
\[ 2 \, ^2H \rightarrow \, ^3He + n + 3.27 \, \text{MeV} \]
This means that when two deuterium nuclei fuse, they produce helium-3 and a neutron, releasing 3.27 MeV of energy.

### Step 2: Calculate the Number of Deuterium Nuclei in 1 kg
First, we need to determine how many deuterium nuclei are present in 1 kg of deuterium. The molar mass of deuterium (which is essentially 2 hydrogen nuclei) is approximately 2 g/mol.

1. Convert 1 kg to grams:
\[ 1 \, \text{kg} = 1000 \, \text{g} \]

2. Calculate the number of moles of deuterium:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{1000 \, \text{g}}{2 \, \text{g/mol}} = 500 \, \text{mol} \]

3. Use Avogadro's number (\(6.022 \times 10^{23}\) nuclei/mol) to find the total number of deuterium nuclei:
\[ \text{Number of nuclei} = \text{Number of moles} \times \text{Avogadro's number} = 500 \, \text{mol} \times 6.022 \times 10^{23} \, \text{nuclei/mol} \]
\[ = 3.011 \times 10^{26} \, \text{nuclei} \]

### Step 3: Calculate the Total Energy Released
Now, we can calculate the total energy released from the fusion of all these deuterium nuclei.

1. Convert the energy released per fusion from MeV to Joules:
\[ 3.27 \, \text{MeV} = 3.27 \times 10^6 \, \text{eV} \]
Using the conversion \(1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}\):
\[ 3.27 \times 10^6 \, \text{eV} = 3.27 \times 10^6 \times 1.6 \times 10^{-19} \, \text{J} \]
\[ = 5.232 \times 10^{-13} \, \text{J} \]

2. Calculate the total energy released:
\[ \text{Total energy} = \text{Energy per nucleus} \times \text{Number of nuclei} \]
\[ = 5.232 \times 10^{-13} \, \text{J} \times 3.011 \times 10^{26} \, \text{nuclei} \]
\[ = 1.577 \times 10^{14} \, \text{J} \]

### Step 4: Final Result
The total energy released in the fusion of 1 kg of deuterium nuclei is approximately:
\[ \approx 1.58 \times 10^{14} \, \text{J} \text{ or } 1.58 \times 10^{14} \, \text{J} \approx 1.6 \times 10^{14} \, \text{J} \]

### Summary
The energy released in the fusion of 1 kg of deuterium nuclei is approximately \(1.58 \times 10^{14} \, \text{J}\).

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