Answer
(a) $^{207}Pb$ has 82 protons, 125 neutrons, and 82 electrons.
(b) $E = 2.34\times 10^{21}~N/C$
Work Step by Step
(a) $^{207}Pb$
A lead atom has 82 protons.
Since this atom has 207 nucleons, the atom must have 125 neutrons.
Since this atom is neutral, it must have 82 electrons.
(b) We can assume that the nucleus is a sphere.
We can find the electric field strength at the surface of the nucleus:
$E = \frac{q}{4\pi \epsilon_0~r^2}$
$E = \frac{(82)(1.6\times 10^{-19}~C)}{(4\pi)(8.854\times 10^{-12}~F/m)~(7.1\times 10^{-15}~m)^2}$
$E = 2.34\times 10^{21}~N/C$
