In nuclear fusion, One gram hydrogen is converted into `0.993 gm`. If the efficiency of the generator be `5%`, then energy obtained in `KWH` is

Options

A. `8.75xx10^(3)`
B. `4.75xx10^(3)`
C. `5.75xx10^(3)`
D. `3.73xx10^(3)`

Explanation

To solve the problem step-by-step, we can follow this approach:

### Step 1: Calculate the change in mass
The initial mass of hydrogen is 1 gram, and the final mass after fusion is 0.993 grams.

\[ \Delta m = m_{\text{initial}} - m_{\text{final}} = 1 \, \text{g} - 0.993 \, \text{g} = 0.007 \, \text{g} \]

### Step 2: Convert the change in mass to kilograms
Since we need the mass in kilograms for our calculations, we convert grams to kilograms:

\[ \Delta m = 0.007 \, \text{g} = 0.007 \times 10^{-3} \, \text{kg} = 7 \times 10^{-6} \, \text{kg} \]

### Step 3: Use Einstein's mass-energy equivalence to find energy
Using the formula \(E = \Delta mc^2\), where \(c\) (the speed of light) is approximately \(3 \times 10^8 \, \text{m/s}\):

\[ E = (7 \times 10^{-6} \, \text{kg}) \times (3 \times 10^8 \, \text{m/s})^2 \]
\[ E = 7 \times 10^{-6} \times 9 \times 10^{16} = 63 \times 10^{10} \, \text{J} = 6.3 \times 10^{11} \, \text{J} \]

### Step 4: Convert energy from joules to kilowatt-hours
To convert joules to kilowatt-hours, we use the conversion factor \(1 \, \text{kWh} = 3.6 \times 10^6 \, \text{J}\):

\[ E_{\text{kWh}} = \frac{6.3 \times 10^{11} \, \text{J}}{3.6 \times 10^6 \, \text{J/kWh}} \approx 1.75 \times 10^{5} \, \text{kWh} \]

### Step 5: Calculate the energy obtained considering the efficiency
Given that the efficiency of the generator is 5%, we find the actual energy output:

\[ E_{\text{output}} = E_{\text{kWh}} \times \frac{5}{100} = 1.75 \times 10^{5} \, \text{kWh} \times 0.05 = 8.75 \times 10^{3} \, \text{kWh} \]

### Final Answer
The energy obtained in kilowatt-hours is approximately:

\[ \boxed{8750 \, \text{kWh}} \]