Determine the magnitude of the electric field at the surface of a lead-196 nucleus, which contains 82 protons and 114 neutrons. Assume the lead nucleus has a volume 196 times that of one proton and consider a proton to be a sphere of radius 1.20 ✕ 10⁻¹⁵ m.

2.4 * 10 ²¹ N/C

Explanation:

The lead-196 nucleus has 82 proton

Q (charge on the nucleus) = 82e where e = 1.602 * 10⁻¹⁹ C

Q = 82 * 1.602 * 10⁻¹⁹ C = 1.314 * 10⁻¹⁷ C

vp (volume of proton) = 4/3π r³

V, volume of lead nucleus = 4/3πR³

4/3πR³ = 196 * 4/3π r³

R = ∛ (196r³) = 5.81 r = 5.81 * 1.20 ✕ 10⁻¹⁵ m = 6.97 * 10⁻¹⁵ m

magnitude of the electric field = KQ/R² = 8.99 * 10⁹ * 1.314 * 10⁻¹⁷ C / (6.97 * 10⁻¹⁵ m) ² = 2.4 * 10 ²¹ N/C