One milligram of matter is converted into energy. The energy released will be

Options

A. `9xx10^6` J
B. `9xx10^8` J
C. `9xx10^10` J
D. `9xx10^12` J

Explanation

To solve the problem of finding the energy released when one milligram of matter is converted into energy, we will use Einstein's equation of mass-energy equivalence, which is given by:

\[ E = mc^2 \]

where:
- \( E \) is the energy,
- \( m \) is the mass,
- \( c \) is the speed of light in a vacuum (approximately \( 3 \times 10^8 \) m/s).

### Step-by-Step Solution:

**Step 1: Convert the mass from milligrams to kilograms.**
- Given mass = 1 mg
- We know that \( 1 \text{ mg} = 10^{-3} \text{ g} \) and \( 1 \text{ g} = 10^{-3} \text{ kg} \).
- Therefore, \( 1 \text{ mg} = 10^{-3} \text{ g} = 10^{-6} \text{ kg} \).

**Step 2: Use Einstein's equation to calculate the energy.**
- We have \( m = 10^{-6} \text{ kg} \) and \( c = 3 \times 10^8 \text{ m/s} \).
- Plugging these values into the equation \( E = mc^2 \):

\[ E = (10^{-6} \text{ kg}) \times (3 \times 10^8 \text{ m/s})^2 \]

**Step 3: Calculate \( c^2 \).**
- First, calculate \( (3 \times 10^8)^2 \):

\[ (3 \times 10^8)^2 = 9 \times 10^{16} \text{ m}^2/\text{s}^2 \]

**Step 4: Substitute \( c^2 \) back into the energy equation.**
- Now substitute \( c^2 \) into the energy equation:

\[ E = (10^{-6} \text{ kg}) \times (9 \times 10^{16} \text{ m}^2/\text{s}^2) \]

**Step 5: Perform the multiplication.**
- Multiply the values:

\[ E = 9 \times 10^{10} \text{ kg m}^2/\text{s}^2 = 9 \times 10^{10} \text{ joules} \]

### Final Answer:
The energy released when one milligram of matter is converted into energy is:

\[ E = 9 \times 10^{10} \text{ joules} \]