The mass defect in a particular nuclear reaction is `0.3` grams. The amont of energy liberated in kilowatt hours is.
(Velocity of light `= 3 xx 10^8 m//s`).

Options

A. `1.5xx10^(6)`
B. `2.5xx10^(6)`
C. `3xx10^(6)`
D. `7.5xx10^(6)`

Explanation

To solve the problem of calculating the energy liberated from a mass defect of 0.3 grams in a nuclear reaction, we will follow these steps:

### Step 1: Convert mass defect from grams to kilograms
The mass defect is given as 0.3 grams. To convert this to kilograms, we use the conversion factor:
\[ \text{mass in kg} = \frac{\text{mass in grams}}{1000} \]
So,
\[ \Delta m = \frac{0.3 \text{ grams}}{1000} = 0.0003 \text{ kg} \]

### Step 2: Use Einstein's equation to calculate energy
According to Einstein's mass-energy equivalence principle, the energy (E) can be calculated using the formula:
\[ E = \Delta m \cdot c^2 \]
Where \(c\) is the speed of light, given as \(3 \times 10^8 \text{ m/s}\). Now substituting the values:
\[ E = 0.0003 \text{ kg} \cdot (3 \times 10^8 \text{ m/s})^2 \]

### Step 3: Calculate \(c^2\)
First, calculate \(c^2\):
\[ c^2 = (3 \times 10^8)^2 = 9 \times 10^{16} \text{ m}^2/\text{s}^2 \]

### Step 4: Calculate the energy in joules
Now substitute \(c^2\) back into the energy equation:
\[ E = 0.0003 \text{ kg} \cdot 9 \times 10^{16} \text{ m}^2/\text{s}^2 \]
\[ E = 2.7 \times 10^{13} \text{ joules} \]

### Step 5: Convert joules to kilowatt-hours
To convert joules to kilowatt-hours, we use the conversion factor:
\[ 1 \text{ kWh} = 3.6 \times 10^6 \text{ joules} \]
Thus, we convert the energy:
\[ E \text{ (in kWh)} = \frac{2.7 \times 10^{13} \text{ joules}}{3.6 \times 10^6 \text{ joules/kWh}} \]
Calculating this gives:
\[ E \text{ (in kWh)} = 7.5 \times 10^6 \text{ kWh} \]

### Final Answer
The amount of energy liberated in kilowatt-hours is:
\[ \boxed{7.5 \times 10^6 \text{ kWh}} \]

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