To solve the problem step by step, we will follow the outlined process for calculating the total energy released in the fusion reaction of deuterium.
### Step 1: Identify the reaction and the masses involved
The fusion reaction is:
\[ _1^2H + _1^2H \rightarrow _2^3He + _0^1n \]
The masses given in atomic mass units (amu) are:
- Mass of deuteron (H) = 2.015 amu
- Mass of helium (He) = 3.017 amu
- Mass of neutron (n) = 1.009 amu
### Step 2: Calculate the mass defect (ΔM)
The mass defect is calculated as:
\[ \Delta M = \text{Mass of reactants} - \text{Mass of products} \]
The mass of the reactants (2 deuterons):
\[ \text{Mass of reactants} = 2 \times 2.015 \, \text{amu} = 4.030 \, \text{amu} \]
The mass of the products (1 helium and 1 neutron):
\[ \text{Mass of products} = 3.017 \, \text{amu} + 1.009 \, \text{amu} = 4.026 \, \text{amu} \]
Now, calculate the mass defect:
\[ \Delta M = 4.030 \, \text{amu} - 4.026 \, \text{amu} = 0.004 \, \text{amu} \]
### Step 3: Calculate the number of deuterium nuclei in 1 kg
To find the number of nuclei in 1 kg of deuterium:
1 kg = 1000 grams
Using Avogadro's number (approximately \(6.022 \times 10^{23}\) nuclei/mol):
The number of moles of deuterium in 1 kg:
\[ \text{Number of moles} = \frac{1000 \, \text{g}}{2 \, \text{g/mol}} = 500 \, \text{mol} \]
(Deuterium has a molar mass of approximately 2 g/mol)
Now, calculate the number of nuclei:
\[ \text{Number of nuclei} = 500 \, \text{mol} \times 6.022 \times 10^{23} \, \text{nuclei/mol} = 3.011 \times 10^{26} \, \text{nuclei} \]
### Step 4: Calculate the energy released per reaction
The energy released per reaction can be calculated using the mass defect:
\[ E = \Delta M \times c^2 \]
Where \(c^2\) can be converted from amu to MeV:
\[ 1 \, \text{amu} = 931.5 \, \text{MeV}/c^2 \]
Thus,
\[ E = 0.004 \, \text{amu} \times 931.5 \, \text{MeV/amu} = 3.726 \, \text{MeV} \]
### Step 5: Calculate the total energy released for 1 kg of deuterium
The total energy released for all the reactions in 1 kg of deuterium:
\[ \text{Total Energy} = \text{Number of reactions} \times E \]
Since each reaction involves 2 deuterons:
\[ \text{Number of reactions} = \frac{3.011 \times 10^{26}}{2} = 1.5055 \times 10^{26} \]
Now calculate the total energy:
\[ \text{Total Energy} = 1.5055 \times 10^{26} \times 3.726 \, \text{MeV} \]
\[ \text{Total Energy} = 5.6 \times 10^{26} \, \text{MeV} \]
### Step 6: Convert MeV to Joules
To convert MeV to Joules:
\[ 1 \, \text{MeV} = 1.6 \times 10^{-13} \, \text{J} \]
Thus,
\[ \text{Total Energy in Joules} = 5.6 \times 10^{26} \, \text{MeV} \times 1.6 \times 10^{-13} \, \text{J/MeV} \]
\[ \text{Total Energy in Joules} = 8.96 \times 10^{13} \, \text{J} \]
### Final Answer
The total energy released when 1 kg of deuterium undergoes complete fusion is approximately:
\[ \text{Total Energy} \approx 9 \times 10^{13} \, \text{J} \]
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