- #1

Violagirl

- 114

- 0

## Homework Statement

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 = 10

^{6}electron volts) must a proton have to overcome the electrical repulsion and reach its surface? B) What is the corresponding initial velocity?

## Homework Equations

Kinetic energy:

1/2mv

^{2}= e Δ V

V = U/q

U = W = F * D

F = q * E

E = kq/r

^{2}

## 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/r

^{2}= (9.0 x 10

^{9}N m

^{2}/r

^{2}(1.6 x 10

^{-19}C)/(7 x 10

^{-15}m)

^{2}= 2.93 x 10

^{19}N/C

From there, I used E to find F:

F = q * E = (1.6 x 10

^{-19}C) (2.93 x 10

^{19}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 10

^{5}V

Solving for megavolts:

2.05 x 10

^{5}(1 meV/10

^{6}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:

1/2mv

^{2}= eV to find v, the initial velocity. But can't quite yet without MeV for part a...