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 mass of deuterium, `_1H^2` that would be needed to generate 1 kWh. (energy released by 1 fusion is 4MeV)

Options

A. (a) `3.7kg`
B. (b) `3.7g`
C. (c) `3.7xx10^-5kg`
D. (d) `3.7xx10^-8kg`

Explanation

To solve the problem of finding the mass of deuterium needed to generate 1 kWh of energy, we will follow these steps:

### Step 1: Convert 1 kWh to Joules
1 kWh (kilowatt-hour) is equivalent to \( 1 \text{ kW} \times 3600 \text{ s} \).
\[ 1 \text{ kWh} = 1000 \text{ W} \times 3600 \text{ s} = 3.6 \times 10^6 \text{ J} \]

### Step 2: Convert the energy released per fusion from MeV to Joules
The energy released by one fusion reaction is given as 4 MeV. We need to convert this to Joules using the conversion factor \( 1 \text{ MeV} = 1.6 \times 10^{-13} \text{ J} \).
\[ E = 4 \text{ MeV} \times 1.6 \times 10^{-13} \text{ J/MeV} = 6.4 \times 10^{-13} \text{ J} \]

### Step 3: Calculate the number of fusion reactions required
Using the power equation \( P = n \cdot E \), where \( n \) is the number of fusion reactions, we can rearrange to find \( n \):
\[ n = \frac{P}{E} \]
Substituting the values:
\[ n = \frac{3.6 \times 10^6 \text{ J}}{6.4 \times 10^{-13} \text{ J}} \approx 5.625 \times 10^{18} \]

### Step 4: Determine the number of deuterium nuclei needed
Since each fusion reaction requires 2 deuterium nuclei, the total number of deuterium nuclei required is:
\[ n' = 2n = 2 \times 5.625 \times 10^{18} \approx 1.125 \times 10^{19} \]

### Step 5: Calculate the mass of deuterium needed
The mass of deuterium can be calculated using the formula:
\[ \text{mass} = \frac{n' \cdot m_d}{N_A} \]
where \( m_d \) is the mass of deuterium (2.014102 amu) and \( N_A \) is Avogadro's number (\( 6.022 \times 10^{23} \) nuclei/mol). First, convert the mass of deuterium from amu to kg:
\[ m_d = 2.014102 \text{ amu} \times 1.66 \times 10^{-27} \text{ kg/amu} \approx 3.34 \times 10^{-27} \text{ kg} \]
Now substituting the values:
\[ \text{mass} = \frac{1.125 \times 10^{19} \cdot 3.34 \times 10^{-27} \text{ kg}}{6.022 \times 10^{23}} \approx 0.3735 \times 10^{-4} \text{ kg} \]

### Step 6: Convert the mass to kg
Finally, we convert the mass into a more manageable form:
\[ \text{mass} \approx 3.735 \times 10^{-8} \text{ kg} \]

### Final Answer
The mass of deuterium needed to generate 1 kWh of energy is approximately:
\[ \text{mass} \approx 3.73 \times 10^{-8} \text{ kg} \]

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