Question

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.

Answer 1

Answer:

Explanation:

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