A fusion reaction of the type given below
`._(1)^(2)D+._(1)^(2)D rarr ._(1)^(3)T+._(1)^(1)p+DeltaE`
is most promising for the production of power. Here D and T stand for deuterium and tritium, respectively. Calculate the mass of deuterium required per day for a power output of `10^(9) W`. Assume the efficiency of the process to be `50%`.
Given : `" "m(._(1)^(2)D)=2.01458 am u," "m(._(1)^(3)T)=3.01605 am u`
`m(._(1)^(1) p)=1.00728 am u` and `1 am u=930 MeV`.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the total energy output required per day
Given:
- Power output, \( P = 10^9 \) W
- Efficiency of the process, \( \eta = 50\% = 0.5 \)

The total energy output required per day can be calculated using the formula:
\[ \text{Energy} = \text{Power} \times \text{Time} \]
Convert 1 day into seconds:
\[ 1 \text{ day} = 24 \text{ hours} \times 3600 \text{ seconds/hour} = 86400 \text{ seconds} \]
Now, calculate the energy:
\[ \text{Energy} = 10^9 \text{ W} \times 86400 \text{ s} = 8.64 \times 10^{13} \text{ J} \]

### Step 2: Calculate the actual energy needed considering efficiency
Since the efficiency is 50%, the actual energy required from the fusion reaction is:
\[ \text{Actual Energy} = \frac{\text{Energy}}{\eta} = \frac{8.64 \times 10^{13} \text{ J}}{0.5} = 1.728 \times 10^{14} \text{ J} \]

### Step 3: Calculate the mass defect in the fusion reaction
The fusion reaction is:
\[ \text{D} + \text{D} \rightarrow \text{T} + \text{p} + \Delta E \]
Where:
- Mass of deuterium, \( m_D = 2.01458 \, \text{amu} \)
- Mass of tritium, \( m_T = 3.01605 \, \text{amu} \)
- Mass of proton, \( m_p = 1.00728 \, \text{amu} \)

Calculate the mass defect (\( \Delta m \)):
\[ \Delta m = 2m_D - (m_T + m_p) = 2(2.01458) - (3.01605 + 1.00728) \]
Calculating this gives:
\[ \Delta m = 4.02916 - 4.02333 = 0.00583 \, \text{amu} \]

### Step 4: Calculate the energy released per fusion reaction
Using the conversion factor \( 1 \, \text{amu} = 930 \, \text{MeV} \):
\[ \text{Energy released per fusion} = \Delta m \times 930 \, \text{MeV} = 0.00583 \times 930 \, \text{MeV} = 5.394 \, \text{MeV} \]
Convert MeV to Joules (1 MeV = \( 1.6 \times 10^{-13} \) J):
\[ \text{Energy released per fusion} = 5.394 \times 1.6 \times 10^{-13} \, \text{J} = 8.6304 \times 10^{-13} \, \text{J} \]

### Step 5: Calculate the number of fusion reactions required per second
\[ \text{Number of fusions per second} = \frac{\text{Actual Energy}}{\text{Energy released per fusion}} = \frac{1.728 \times 10^{14} \, \text{J}}{8.6304 \times 10^{-13} \, \text{J}} \approx 2.003 \times 10^{26} \text{ fusions/s} \]

### Step 6: Calculate the number of fusions required per day
\[ \text{Number of fusions per day} = 2.003 \times 10^{26} \text{ fusions/s} \times 86400 \text{ s} \approx 1.730 \times 10^{31} \text{ fusions/day} \]

### Step 7: Calculate the number of deuterium atoms required
Since each fusion reaction uses 2 deuterium atoms:
\[ \text{Number of deuterium atoms required} = 2 \times 1.730 \times 10^{31} \approx 3.460 \times 10^{31} \text{ deuterium atoms} \]

### Step 8: Calculate the mass of deuterium required
Using the mass of deuterium:
\[ \text{Mass of deuterium required} = \text{Number of deuterium atoms} \times \text{mass of one deuterium atom} \]
Convert deuterium mass from amu to kg:
\[ m_D = 2.01458 \, \text{amu} \times 1.66 \times 10^{-27} \, \text{kg/amu} \approx 3.344 \times 10^{-27} \, \text{kg} \]
Thus,
\[ \text{Mass of deuterium required} = 3.460 \times 10^{31} \times 3.344 \times 10^{-27} \approx 115.4 \, \text{kg} \]

### Final Answer
The mass of deuterium required per day for a power output of \( 10^9 \, \text{W} \) is approximately **115.4 kg**.