For the `D-T` fusion reaction, find the rate at which deuterium & trithium are consumed to produce `1 MW`. The `Q`-value of `D-T` reactions is `17.6 MeV` & assume all the energy from the fusion rection is available.

Explanation

To solve the problem of finding the rate at which deuterium and tritium are consumed to produce 1 MW from the D-T fusion reaction, we can follow these steps:

### Step 1: Understand the Energy Requirement
We need to produce 1 MW of power.
1 MW = \(10^6\) watts.

### Step 2: Convert the Q-value to Joules
The Q-value for the D-T fusion reaction is given as 17.6 MeV. We need to convert this energy into joules for our calculations.
1 MeV = \(1.6 \times 10^{-13}\) joules.

So,
\[ Q = 17.6 \, \text{MeV} = 17.6 \times 1.6 \times 10^{-13} \, \text{J} \]
Calculating this gives:
\[ Q = 28.16 \times 10^{-13} \, \text{J} \]

### Step 3: Calculate the Number of Reactions per Second
To find the number of fusion reactions needed per second to produce 1 MW, we use the formula:
\[ \text{Number of reactions per second} = \frac{\text{Power}}{\text{Energy per reaction}} \]
Substituting the values we have:
\[ \text{Number of reactions per second} = \frac{10^6 \, \text{W}}{28.16 \times 10^{-13} \, \text{J}} \]

Calculating this gives:
\[ \text{Number of reactions per second} = \frac{10^6}{28.16 \times 10^{-13}} \approx 3.55 \times 10^{16} \, \text{reactions/second} \]

### Step 4: Determine the Rate of Consumption of Deuterium and Tritium
In the D-T fusion reaction, one deuterium nucleus and one tritium nucleus are consumed per reaction. Therefore, the rate at which deuterium and tritium are consumed is equal to the number of reactions per second.

Thus, the rate of consumption of deuterium and tritium is:
\[ \text{Rate of consumption of D and T} = 3.55 \times 10^{16} \, \text{nuclei/second} \]

### Final Answer
The rate at which deuterium and tritium are consumed to produce 1 MW of power is approximately \(3.55 \times 10^{16}\) nuclei per second for each of D and T.

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