To find the nuclear binding energy using the given mass defect and Einstein’s equation \(E = mc^2\), we can substitute the mass defect and the speed of light into the equation.
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Given:
- Mass defect, \(m = 3.2008 \times 10^{-27}\) kg
- Speed of light, \(c = 3 \times 10^8\) m/s
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Applying the equation \(E = mc^2\):
\[ E = (3.2008 \times 10^{-27} \text{ kg}) \times (3 \times 10^{8} \text{ m/s})^2 \]
- Calculate \(c^2\):
\[ (3 \times 10^{8} \text{ m/s})^2 = 9 \times 10^{16} \text{ m}^2/\text{s}^2 \]
- Now substitute \(c^2\) back into the energy equation:
\[ E = 3.2008 \times 10^{-27} \text{ kg} \times 9 \times 10^{16} \text{ m}^2/\text{s}^2 \]
- Calculate \(E\):
\[ E = 3.2008 \times 9 \times 10^{-27 + 16} \] \[ E = 28.8072 \times 10^{-11} \text{ joules} \] \[ E = 2.88072 \times 10^{-10} \text{ joules} \]
To match significant figures, we can round it to \(2.8807 \times 10^{-10}\) joules.
Thus, the correct answer is:
C. \(2.8807 \times 10^{-10}\) joules.