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QPages 121-124: Q11: HC Verma: Concepts of Physics - Volume 2 Electric Field Potential Pages 121-124: Q11 Solution: Consider a gold nucleus to be a sphere of radius 6.9 fermi in…

⚠️ Common Exam Traps & Conceptual Pitfalls

When calculating the Coulomb force between two protons at the farthest points inside a nucleus, the correct separation is the diameter, $2R$, not the radius $R$. Using $R$ would give a shorter distance and a larger force. Always convert fermi to meters before substituting into $F= rac{k e^2}{r^2}$.

Consider a gold nucleus to be a sphere of radius 6.9 fermi in which protons and neutrons are distributed. Find the force of repulsion between two protons situated at largest separation. Why do these protons not fly apart under this repulsion?

Step-by-Step Verified Solution

**Answer:** Force = 1.2 N. The protons do not fly apart because of the strong nuclear force. **Step-BY-STEP Explanation:** Largest separation between two protons in a nucleus of radius R = 6.9 fermi is 2R=13.82R = 13.8 fermi =13.8×10−15= 13.8 \times 10^{-15} m.
F=ke2r2=9×109×(1.6×10−19)2(13.8×10−15)2=1.2 NF = \frac{k e^2}{r^2} = \frac{9 \times 10^9 \times (1.6 \times 10^{-19})^2}{(13.8 \times 10^{-15})^2} = 1.2 \text{ N}
The strong nuclear force (short-range, attractive) overcomes the Coulomb repulsion inside the nucleus .

All Questions in this Chapter

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