Charged particles such as protons are accelerated to high velocities and are allowed to collide with atomic nuclei to probe their internal structure. The electric potential outside of a point charge Ze. A) A lead nucleus (Z - 82) is approximately described as a sphere of radius 7 x 10-15 m. How much kinetic energy in megaelectron volts (1Me V = 106 electron volts) must a proton have to overcome the electrical repulsion and reach its surface? B) What is the corresponding initial velocity?
1/2mv2 = e Δ V
V = U/q
U = W = F * D
F = q * E
E = kq/r2
The Attempt at a Solution
I posted all key equations above that I used to try to solve for part a. So I started out by solving for E:
E = kQ/r2 = (9.0 x 109 N m2/r2 (1.6 x 10-19 C)/(7 x 10-15 m)2 = 2.93 x 1019 N/C
From there, I used E to find F:
F = q * E = (1.6 x 10-19 C) (2.93 x 1019 N/C) = 4.688 N
From there, I found U:
U = (F) ( D) = (4.688 N) (7 x 10-15 m) = 3.2816 x 10-14 J
Finally, solved for V:
V = U/q = (3.2861 x 10-14 J)/(1.6 x 10-19 C = 2.05 x 105 V
Solving for megavolts:
2.05 x 105 (1 meV/106 V) = .205 MeV
However, this turned out to be wrong. The answer in my solutions manual says that it should be 16.8 MeV. What did I do wrong in this problem?
Otherwise I figured for part b, I would use the equation:
1/2mv2 = eV to find v, the initial velocity. But can't quite yet without MeV for part a...