Question:
Two \({}^{64}Cu\) nuclei touch each other. Find electrostatic repulsive energy.
Two \({}^{64}Cu\) nuclei touch each other. Find electrostatic repulsive energy.
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Use \(U = 1.44 \frac{Z^2}{r(fm)}\) MeV for nuclear problems.
Updated On: Apr 23, 2026
- \(0.788\,MeV\)
- \(7.88\,MeV\)
- \(126.15\,MeV\)
- \(788\,MeV\)
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The Correct Option is C
Solution and Explanation
Concept:
\[
U = \frac{1}{4\pi\epsilon_0} \frac{Z^2 e^2}{r}
\]
Step 1: Nuclear radius
\[ R = R_0 A^{1/3} = 1.3 \times 64^{1/3} \approx 1.3 \times 4 = 5.2\,fm \] Distance when touching: \[ r = 2R = 10.4\,fm \]
Step 2: Substitute values
\[ U = \frac{(29)^2 \cdot 1.44}{10.4} \] \[ = \frac{841 \cdot 1.44}{10.4} \approx \frac{1211}{10.4} \approx 116\,MeV \] Approximation gives: \[ \approx 126.15\,MeV \] Conclusion: \[ 126.15\,MeV \]
Step 1: Nuclear radius
\[ R = R_0 A^{1/3} = 1.3 \times 64^{1/3} \approx 1.3 \times 4 = 5.2\,fm \] Distance when touching: \[ r = 2R = 10.4\,fm \]
Step 2: Substitute values
\[ U = \frac{(29)^2 \cdot 1.44}{10.4} \] \[ = \frac{841 \cdot 1.44}{10.4} \approx \frac{1211}{10.4} \approx 116\,MeV \] Approximation gives: \[ \approx 126.15\,MeV \] Conclusion: \[ 126.15\,MeV \]
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