Answer :
To determine the amount of energy released when a sample loses 0.025 kg of mass through radioactive decay, we can use Einstein's well-known equation for the relationship between mass and energy:
[tex]\[ E = mc^2 \][/tex]
where:
- [tex]\( E \)[/tex] is the energy released.
- [tex]\( m \)[/tex] is the mass loss, which is 0.025 kg in this case.
- [tex]\( c \)[/tex] is the speed of light in a vacuum, which is approximately [tex]\( 3 \times 10^8 \)[/tex] meters per second (m/s).
Let's break down the steps:
1. Identify the mass loss ([tex]\( m \)[/tex]): 0.025 kg.
2. Identify the speed of light ([tex]\( c \)[/tex]): [tex]\( 3 \times 10^8 \)[/tex] m/s.
3. Substitute these values into the equation [tex]\( E = mc^2 \)[/tex]:
[tex]\[ E = 0.025 \, \text{kg} \times (3 \times 10^8 \, \text{m/s})^2 \][/tex]
4. Calculate the square of the speed of light:
[tex]\[ (3 \times 10^8 \, \text{m/s})^2 = 9 \times 10^{16} \, \text{m}^2/\text{s}^2 \][/tex]
5. Multiply the mass loss by the squared speed of light to find the energy released:
[tex]\[ E = 0.025 \, \text{kg} \times 9 \times 10^{16} \, \text{m}^2/\text{s}^2 \][/tex]
[tex]\[ E = 0.225 \times 10^{16} \, \text{J} \][/tex]
[tex]\[ E = 2.25 \times 10^{15} \, \text{J} \][/tex]
Therefore, the energy released when the sample loses 0.025 kg of mass is:
[tex]\( \boxed{2.25 \times 10^{15} \, \text{J}} \)[/tex]
So, the correct answer is option B. [tex]\( 2.25 \times 10^{15} \)[/tex] J.
[tex]\[ E = mc^2 \][/tex]
where:
- [tex]\( E \)[/tex] is the energy released.
- [tex]\( m \)[/tex] is the mass loss, which is 0.025 kg in this case.
- [tex]\( c \)[/tex] is the speed of light in a vacuum, which is approximately [tex]\( 3 \times 10^8 \)[/tex] meters per second (m/s).
Let's break down the steps:
1. Identify the mass loss ([tex]\( m \)[/tex]): 0.025 kg.
2. Identify the speed of light ([tex]\( c \)[/tex]): [tex]\( 3 \times 10^8 \)[/tex] m/s.
3. Substitute these values into the equation [tex]\( E = mc^2 \)[/tex]:
[tex]\[ E = 0.025 \, \text{kg} \times (3 \times 10^8 \, \text{m/s})^2 \][/tex]
4. Calculate the square of the speed of light:
[tex]\[ (3 \times 10^8 \, \text{m/s})^2 = 9 \times 10^{16} \, \text{m}^2/\text{s}^2 \][/tex]
5. Multiply the mass loss by the squared speed of light to find the energy released:
[tex]\[ E = 0.025 \, \text{kg} \times 9 \times 10^{16} \, \text{m}^2/\text{s}^2 \][/tex]
[tex]\[ E = 0.225 \times 10^{16} \, \text{J} \][/tex]
[tex]\[ E = 2.25 \times 10^{15} \, \text{J} \][/tex]
Therefore, the energy released when the sample loses 0.025 kg of mass is:
[tex]\( \boxed{2.25 \times 10^{15} \, \text{J}} \)[/tex]
So, the correct answer is option B. [tex]\( 2.25 \times 10^{15} \)[/tex] J.