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Question 62 of 949

In a thermonuclear reaction, the total initial mass is 5.02 x 1027kg and the total final mass is 5.01 x 1027kg. 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.
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