g Determine the magnitude of the electric field at the surface of a lead-208 nucleus, which contains 82 protons and 126 neutrons. Assume the lead nucleus has a volume 208 times that of one proton and consider a proton to be a sphere of radius 1.20 x 10-15 m.

Electric field at the surface of the the lead 208 = KQ / R²

where K = 8.99 * 10⁹ Nm² / C²

Q (total charge inside the nucleus) and e is the charge of a proton = Ne = 82 * 1.6 * 10⁻¹⁹ C = 1.312 * 10⁻¹⁷ C

V of the lead = 208 v of a proton assuming they both are sphere

4/3 πR³ = 208 (4/3 πr³) where R is the radius of the sphere and r is the radius of the proton

R³ = 208 r³

R = ∛ (208 r³) = 5.92r

replace r with 1.20 x 10-15 m

R = 5.92 * 1.20 x 10-15 m = 7.11 * 10⁻¹⁵ m

E = (8.99 * 10⁹ Nm² / C² * 1.312 * 10⁻¹⁷ C) / (7.11 * 10⁻¹⁵ m) ² = 0.233 * 10²² N/C = 2.33 * 10²¹ N/C