Question 62 of 949
In a thermonuclear reaction, the total initial mass is 5.02 x 10−27kg and the total final mass is 5.01 x 10−27kg. The energy released in the process is
[c = 3.0 x 108 ms-1]
- A. 9.0 x 10-13J
- B. 9.0 x 10-12J
- C. 9.0 x 10-11J
- D. 9.0 x 10-10J
Correct Answer:
A
Explanation
To determine the energy released in a thermonuclear reaction based on the change in mass, we can use Einstein's famous equation:
\[
E = \Delta m c^2
\]
where:
- \(E\) is the energy released,
- \(\Delta m\) is the change in mass,
- \(c\) is the speed of light in a vacuum, approximately \(3.0 \times 10^8 \, \text{m/s}\).
### Step 1: Calculate the Change in Mass
First, we need to find the change in mass (\(\Delta m\)) during the reaction. The initial mass (\(m_i\)) and final mass (\(m_f\)) are given as follows:
- Initial mass, \(m_i = 5.02 \times 10^{-27} \, \text{kg}\)
- Final mass, \(m_f = 5.01 \times 10^{-27} \, \text{kg}\)
The change in mass is calculated as:
\[
\Delta m = m_i - m_f
\]
Substituting the values:
\[
\Delta m = 5.02 \times 10^{-27} \, \text{kg} - 5.01 \times 10^{-27} \, \text{kg} = 0.01 \times 10^{-27} \, \text{kg} = 1.0 \times 10^{-29} \, \text{kg}
\]
### Step 2: Calculate the Energy Released
Now, we can substitute \(\Delta m\) and \(c\) into the energy equation:
\[
E = \Delta m c^2
\]
Substituting the values:
\[
E = (1.0 \times 10^{-29} \, \text{kg}) \times (3.0 \times 10^8 \, \text{m/s})^2
\]
Calculating \(c^2\):
\[
c^2 = (3.0 \times 10^8)^2 = 9.0 \times 10^{16} \, \text{m}^2/\text{s}^2
\]
Now substituting back into the energy equation:
\[
E = 1.0 \times 10^{-29} \, \text{kg} \times 9.0 \times 10^{16} \, \text{m}^2/\text{s}^2
\]
Calculating the energy:
\[
E = 9.0 \times 10^{-13} \, \text{J}
\]
### Conclusion
The energy released in the thermonuclear reaction is:
\[
\boxed{9.0 \times 10^{-13} \, \text{J}}
\]
### Explanation of Options
- **Option A: \(9.0 \times 10^{-13} \, \text{J}\)** - This is the correct answer based on our calculations.
- **Option B: \(9.0 \times 10^{-12} \, \text{J}\)** - This is incorrect; it is an order of magnitude too high.
- **Option C: \(9.0 \times 10^{-11} \, \text{J}\)** - This is also incorrect; it is two orders of magnitude too high.
- **Option D: \(9.0 \times 10^{-10} \, \text{J}\)** - This is incorrect; it is three orders of magnitude too high.
### Common Pitfalls
1. **Miscalculating \(\Delta m\)**: Ensure that you subtract the final mass from the initial mass correctly.
2. **Forgetting to square \(c\)**: Remember that \(c\) must be squared in the energy equation.
3. **Unit conversion**: Ensure that all units are consistent, especially when dealing with mass in kilograms and energy in joules.
### Revision Summary
- Use \(E = \Delta m c^2\) to calculate energy from mass change.
- Calculate \(\Delta m\) by subtracting final mass from initial mass.
- Ensure \(c\) is squared in the energy calculation.
- Check your final answer against the options provided.