Answer :
To determine how much energy is released when 1 kg of mass is lost through radioactive decay, we should use Albert Einstein's famous equation [tex]\( E = mc^2 \)[/tex].
Here are the steps to solve this problem:
1. Identify the given values:
- Mass lost ([tex]\( m \)[/tex]): 1 kg
- Speed of light ([tex]\( c \)[/tex]): [tex]\( 3 \times 10^8 \)[/tex] meters per second
2. Recall the formula to calculate energy:
[tex]\( E = mc^2 \)[/tex]
Where:
- [tex]\( E \)[/tex] is the energy released
- [tex]\( m \)[/tex] is the mass lost
- [tex]\( c \)[/tex] is the speed of light
3. Substitute the given values into the formula:
[tex]\[ E = (1 \, \text{kg}) \times (3 \times 10^8 \, \text{m/s})^2 \][/tex]
4. Calculate the value within the parentheses first:
[tex]\[ (3 \times 10^8 \, \text{m/s})^2 = 9 \times 10^{16} \, \text{m}^2/\text{s}^2 \][/tex]
5. Multiply the mass by the squared speed of light:
[tex]\[ E = 1 \times 9 \times 10^{16} \, \text{kg} \cdot \text{m}^2/\text{s}^2 = 9 \times 10^{16} \, \text{J} \][/tex]
6. So, the energy released is [tex]\( 9 \times 10^{16} \)[/tex] Joules.
Therefore, the correct answer is:
A. [tex]\( 9 \times 10^{16} \, \text{J} \)[/tex]
Here are the steps to solve this problem:
1. Identify the given values:
- Mass lost ([tex]\( m \)[/tex]): 1 kg
- Speed of light ([tex]\( c \)[/tex]): [tex]\( 3 \times 10^8 \)[/tex] meters per second
2. Recall the formula to calculate energy:
[tex]\( E = mc^2 \)[/tex]
Where:
- [tex]\( E \)[/tex] is the energy released
- [tex]\( m \)[/tex] is the mass lost
- [tex]\( c \)[/tex] is the speed of light
3. Substitute the given values into the formula:
[tex]\[ E = (1 \, \text{kg}) \times (3 \times 10^8 \, \text{m/s})^2 \][/tex]
4. Calculate the value within the parentheses first:
[tex]\[ (3 \times 10^8 \, \text{m/s})^2 = 9 \times 10^{16} \, \text{m}^2/\text{s}^2 \][/tex]
5. Multiply the mass by the squared speed of light:
[tex]\[ E = 1 \times 9 \times 10^{16} \, \text{kg} \cdot \text{m}^2/\text{s}^2 = 9 \times 10^{16} \, \text{J} \][/tex]
6. So, the energy released is [tex]\( 9 \times 10^{16} \)[/tex] Joules.
Therefore, the correct answer is:
A. [tex]\( 9 \times 10^{16} \, \text{J} \)[/tex]