Answer:
2.94 × 10²⁰ N/C
Explanation:
Given that:
The nucleus of "Lead-208 has 82 protons,
with a radius (r) 6.34×10-15 m, &
each electric charge has a value of 1.60218 × 10^-19 C
∴ The formula for calculating an electrical field at the surface of the nucleus is:
[tex]E=\frac{k*q}{r^2}[/tex]
Substituting our values into the equation above, we have;
E = [tex]\frac{8.98755*10^8*82(1.60218*10^{-19C)}}{(6.34*10^{-15}_m)^2}[/tex]
E = 2.93870499×10²⁰ N/C
E ≅ 2.94 × 10²⁰ N/C
