Assuming that about `20 M eV` of energy is released per fusion reaction `._(1)H^(2)+._(1)H^(3)rarr._(0)n^(1)+._(2)He^(4)`, the mass of `._(1)H^(2)` consumed per day in a future fusion reactor of powder `1 MW` would be approximately

Options

A. `0.001 g`
B. `0.2 g`
C. `10.0 g`
D. `2 g`

Answer

To solve the problem of determining the mass of \( _{1}H^{2} \) consumed per day in a future fusion reactor of power \( 1 \, \text{MW} \), we will follow these steps:

### Step 1: Calculate the total energy required per day
The power of the reactor is given as \( 1 \, \text{MW} \) (megawatt), which is equivalent to \( 10^6 \, \text{W} \). The energy required per day can be calculated using the formula:
\[ \text{Energy (E)} = \text{Power (P)} \times \text{Time (t)} \]
Where time in seconds for one day is:
\[ t = 24 \, \text{hours} \times 3600 \, \text{seconds/hour} = 86400 \, \text{seconds} \]
Thus, the energy required per day is:
\[ E = 10^6 \, \text{W} \times 86400 \, \text{s} = 8.64 \times 10^{10} \, \text{J} \]

### Step 2: Convert the energy released per fusion reaction
The energy released per fusion reaction is given as \( 20 \, \text{MeV} \). To convert this energy into joules, we use the conversion factor \( 1 \, \text{MeV} = 1.6 \times 10^{-13} \, \text{J} \):
\[ \text{Energy per reaction} = 20 \, \text{MeV} \times 1.6 \times 10^{-13} \, \text{J/MeV} = 3.2 \times 10^{-12} \, \text{J} \]

### Step 3: Calculate the number of fusion reactions required per day
To find the number of fusion reactions required to produce the total energy needed per day, we divide the total energy by the energy released per reaction:
\[ \text{Number of reactions} = \frac{E}{\text{Energy per reaction}} = \frac{8.64 \times 10^{10} \, \text{J}}{3.2 \times 10^{-12} \, \text{J}} \approx 2.7 \times 10^{22} \, \text{reactions} \]

### Step 4: Calculate the mass of \( _{1}H^{2} \) consumed
Each fusion reaction consumes one nucleus of \( _{1}H^{2} \). The molar mass of \( _{1}H^{2} \) (deuterium) is approximately \( 2 \, \text{g/mol} \). The mass of one molecule can be calculated using Avogadro's number \( N_A = 6.022 \times 10^{23} \):
\[ \text{Mass of one molecule} = \frac{2 \, \text{g/mol}}{N_A} = \frac{2 \, \text{g}}{6.022 \times 10^{23}} \approx 3.32 \times 10^{-24} \, \text{g} \]
Now, we can find the total mass of \( _{1}H^{2} \) consumed per day:
\[ \text{Total mass} = \text{Number of reactions} \times \text{Mass of one molecule} = 2.7 \times 10^{22} \times 3.32 \times 10^{-24} \, \text{g} \approx 0.0897 \, \text{g} \]

### Final Result
The mass of \( _{1}H^{2} \) consumed per day is approximately \( 0.1 \, \text{g} \).

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