Consider the following nuclear reaction
`._(1)^(2)D+._(1)^(2)_Dto._(1)^(3)D+._(1)^(1)P`
Given that `m(_(1)^(3)T)=3.01605` amu,
`m(_(1)^(3)p)=1.00728` amu
The mass of deuterium `(._(1)^(2)D)` required per day in order to produce a power output of `10^(9)W` (with `50%` efficiency) is `n.3` kg where `n` is a digit. Find `n`

Explanation

To solve the problem, we need to analyze the given nuclear reaction and calculate the mass of deuterium required to produce a specific power output. Let's break down the steps:

### Step 1: Understand the Nuclear Reaction
The nuclear reaction given is:
\[ _{1}^{2}D + _{1}^{2}D \rightarrow _{1}^{3}D + _{1}^{1}P \]
This indicates that two deuterium nuclei combine to form one tritium nucleus and one proton.

### Step 2: Calculate the Mass Defect
To find the energy released in the reaction, we first need to calculate the mass defect. The mass defect is the difference between the mass of the reactants and the mass of the products.

1. **Mass of Reactants:**
- Mass of 2 deuterium nuclei:
\[ m_{reactants} = 2 \times m(_{1}^{2}D) \]

2. **Mass of Products:**
- Mass of products (1 tritium and 1 proton):
\[ m_{products} = m(_{1}^{3}D) + m(_{1}^{1}P) \]
- Given:
- \( m(_{1}^{3}D) = 3.01605 \) amu
- \( m(_{1}^{1}P) = 1.00728 \) amu

### Step 3: Substitute Values
Let’s denote the mass of deuterium as \( m(_{1}^{2}D) \). We can find the mass defect as follows:
\[ \Delta m = m_{reactants} - m_{products} \]
\[ \Delta m = 2 \times m(_{1}^{2}D) - (3.01605 + 1.00728) \]

### Step 4: Calculate Energy Released
The energy released in the reaction can be calculated using Einstein's mass-energy equivalence:
\[ E = \Delta m \cdot c^2 \]
Where \( c \) is the speed of light (\( c \approx 3 \times 10^8 \) m/s). To convert energy to joules, we need to convert amu to kg:
\[ 1 \text{ amu} = 1.660539 \times 10^{-27} \text{ kg} \]

### Step 5: Calculate Power Output
Given that the efficiency of the process is 50%, the actual power output from the reaction is:
\[ P_{actual} = \frac{P_{desired}}{\text{Efficiency}} = \frac{10^9 \text{ W}}{0.5} = 2 \times 10^9 \text{ W} \]

### Step 6: Calculate Mass of Deuterium Required
Using the energy released per reaction, we can find how many reactions are needed to produce the desired power output in one day (86400 seconds):
\[ \text{Total energy required in one day} = P_{actual} \times \text{time} \]

Finally, we can find the mass of deuterium required by dividing the total energy by the energy released per reaction and multiplying by the mass of deuterium.

### Final Calculation
1. Calculate \( \Delta m \) using the mass of deuterium.
2. Calculate the energy released per reaction.
3. Find the total energy required for one day.
4. Calculate the number of reactions needed and the corresponding mass of deuterium.