For a Xenon-125 nucleus (radius 6.0 fm, Z = 54) with a proton located 2.0 fm from the surface, what is the Coulomb force on the proton? Use k ≈ 8.99×10^9 N·m^2/C^2 and e = 1.602×10^-19 C.

• A. 498 N
• B. 49.8 N
• C. 4.98 N
• D. 4.98×10^3 N

Answer: (C)

Coulomb’s law is the key idea: the force between charges falls off as 1/r^2, and outside a uniformly charged sphere the field looks like that of a point charge at the center. The nucleus has total charge Q = Z e, so the force on a proton (charge e) at distance r from the center is F = k Z e^2 / r^2.

Here the nuclear radius is 6.0 fm. The proton sits 2.0 fm from the surface, so the distance from the center is r = 6.0 + 2.0 = 8.0 fm = 8.0 × 10^-15 m. With Z = 54, e = 1.602 × 10^-19 C, and k ≈ 8.99 × 10^9 N·m^2/C^2:

• e^2 ≈ (1.602 × 10^-19)^2 ≈ 2.57 × 10^-38 C^2

• Z e^2 ≈ 54 × 2.57 × 10^-38 ≈ 1.39 × 10^-36 C^2

• r^2 ≈ (8.0 × 10^-15)^2 ≈ 6.4 × 10^-29 m^2

F ≈ (8.99 × 10^9) × (1.39 × 10^-36) / (6.4 × 10^-29) ≈ 1.95 × 10^2 N

So the Coulomb force is about 1.95 × 10^2 N, directed radially outward from the nucleus.