It is proposed to use the nuclear fusion reaction,
`_1^2H+_1^2Hrarr_2^4He`
in a nuclear reactor 200 MW rating. If the energy from the above reaction is used with a 25 per cent efficiency in the reactor, how many grams of deuterium fuel will be needed per day?(The masses of `_1^2H` and `_2^4He are 2.0141 atommic mass units and 4.0026 atomic mass units respectively.)

Explanation

To solve the problem, we will follow a series of steps to find out how many grams of deuterium fuel are needed per day for a nuclear reactor with a 200 MW rating, using the given fusion reaction.

### Step 1: Calculate the mass defect (Δm)
The mass defect is calculated as the difference between the mass of the reactants and the mass of the products.

\[ \Delta m = \text{mass of reactants} - \text{mass of products} \]

Given:
- Mass of deuterium (\(_1^2H\)) = 2.0141 amu
- Mass of helium (\(_2^4He\)) = 4.0026 amu

Since two deuterium nuclei are fused:
\[ \Delta m = (2 \times 2.0141) - 4.0026 = 4.0282 - 4.0026 = 0.0256 \text{ amu} \]

### Step 2: Convert mass defect to energy released (ΔE)
Using Einstein's equation \(E = mc^2\), we can convert the mass defect to energy. First, we convert the mass defect from amu to joules.

1 amu = 931.48 MeV, and 1 MeV = \(1.6 \times 10^{-13}\) Joules.

\[ \Delta E = \Delta m \times 931.48 \text{ MeV} = 0.0256 \times 931.48 \text{ MeV} = 23.85 \text{ MeV} \]

Now convert this to joules:
\[ \Delta E = 23.85 \times 1.6 \times 10^{-13} \text{ J} = 3.816 \times 10^{-12} \text{ J} \]

### Step 3: Calculate the energy available to the reactor
Given that the reactor operates at 25% efficiency, we calculate the energy available to the reactor:

\[ E_{\text{available}} = 0.25 \times \Delta E = 0.25 \times 3.816 \times 10^{-12} \text{ J} = 9.54 \times 10^{-13} \text{ J} \]

### Step 4: Calculate the energy required per day
The total energy required per day can be calculated using the formula:

\[ E_{\text{required}} = \text{Power} \times \text{Time} \]

Given:
- Power = 200 MW = \(200 \times 10^6 \text{ W}\)
- Time = 1 day = \(24 \times 3600 \text{ s}\)

Calculating the energy required:
\[ E_{\text{required}} = 200 \times 10^6 \times (24 \times 3600) = 200 \times 10^6 \times 86400 = 1.728 \times 10^{13} \text{ J} \]

### Step 5: Calculate the number of deuterons needed
The total number of deuterons needed can be calculated by dividing the total energy required by the energy released per deuteron.

Since two deuterons are involved in the reaction, the energy released per deuteron is:
\[ E_{\text{per deuteron}} = \frac{\Delta E}{2} = \frac{3.816 \times 10^{-12}}{2} = 1.908 \times 10^{-12} \text{ J} \]

Now, calculate the number of deuterons needed:
\[ n = \frac{E_{\text{required}}}{E_{\text{per deuteron}}} = \frac{1.728 \times 10^{13}}{1.908 \times 10^{-12}} \approx 9.05 \times 10^{24} \]

### Step 6: Calculate the moles of deuterium
Using Avogadro's number (\(N_A = 6.022 \times 10^{23}\)):
\[ \text{moles of deuterium} = \frac{n}{N_A} = \frac{9.05 \times 10^{24}}{6.022 \times 10^{23}} \approx 15.03 \text{ moles} \]

### Step 7: Calculate the mass of deuterium needed
The molar mass of deuterium is approximately 2 g/mol. Thus, the mass of deuterium needed is:
\[ \text{mass} = \text{moles} \times \text{molar mass} = 15.03 \times 2 \approx 30.06 \text{ grams} \]

### Final Answer
The mass of deuterium fuel needed per day is approximately **30.06 grams**.

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