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 number of fusion reactions required to generate 1kWh is nearly

Options

A. (a) `10^8`
B. (b) `10^18`
C. (c) `10^28`
D. (d) `10^38`

Explanation

To find the number of fusion reactions required to generate 1 kWh of energy, we will follow these steps:

### Step 1: Calculate the energy released per fusion reaction.
We will consider the fusion reaction of deuterium (D) and tritium (T):
\[ ^2H + ^2H \rightarrow ^3H + ^1H \]
The atomic masses are given as:
- Deuterium \( m_{^2H} = 2.014102 \) amu
- Tritium \( m_{^3H} = 3.016049 \) amu
- Hydrogen \( m_{^1H} = 1.007825 \) amu

Using the mass-energy equivalence, we can calculate the energy released:
\[ \Delta m = 2 \times m_{^2H} - (m_{^3H} + m_{^1H}) \]
Substituting the values:
\[ \Delta m = 2 \times 2.014102 - (3.016049 + 1.007825) \]
Calculating:
\[ \Delta m = 4.028204 - 4.023874 = 0.004330 \text{ amu} \]

### Step 2: Convert the mass defect to energy.
Using the conversion factor \( 1 \text{ amu} = 931.48 \text{ MeV} \):
\[ E = \Delta m \times 931.48 \text{ MeV} \]
Calculating:
\[ E = 0.004330 \times 931.48 \approx 4.03 \text{ MeV} \]

### Step 3: Convert energy from MeV to Joules.
Using the conversion \( 1 \text{ MeV} = 1.6 \times 10^{-13} \text{ Joules} \):
\[ E = 4.03 \text{ MeV} \times 1.6 \times 10^{-13} \text{ J/MeV} \approx 6.448 \times 10^{-13} \text{ J} \]

### Step 4: Calculate the total energy required for 1 kWh.
1 kWh is equivalent to:
\[ 1 \text{ kWh} = 1000 \text{ W} \times 3600 \text{ s} = 3.6 \times 10^6 \text{ J} \]

### Step 5: Calculate the number of fusion reactions needed.
Let \( n \) be the number of fusion reactions required:
\[ n \times E = 3.6 \times 10^6 \text{ J} \]
Substituting for \( E \):
\[ n \times 6.448 \times 10^{-13} = 3.6 \times 10^6 \]
Solving for \( n \):
\[ n = \frac{3.6 \times 10^6}{6.448 \times 10^{-13}} \approx 5.58 \times 10^{18} \]

### Step 6: Conclusion.
Thus, the number of fusion reactions required to generate 1 kWh is approximately:
\[ \boxed{10^{18}} \]