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Finding the kinetic energy of a proton in megaelectron volts?

  • Thread starter Violagirl
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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 = 106 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:

1/2mv2 = 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...
 

Answers and Replies

  • #2
Simon Bridge
Science Advisor
Homework Helper
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Approaching the nucleus charge +Ze, the proton charge +e is exchanging kinetic energy with potential energy.
It has to start out with at least as much kinetic energy as the difference in potential energy between it's start and finish radii. You know an equation for potential energy of a charge at a distance r from another charge.

Basically you seem to have confused eV with V.
1eV is the kinetic energy change in an electron moving through a potential difference of 1V
i.e. it is a unit of energy... so you can convert directly from Joules.
... but it is best practice to do your calculations with more convenient units - like:

##\small ke^2=1.4400\text{MeV.fm}##

... since you have radius in fm and you want energy in MeV.
http://www.phas.ubc.ca/~mcmillan/rqpdfs/3_particle_nature_of_matter.pdf

Similarly ##\small m_pc^2=938\text{MeV}## and ##\small E_K=(\gamma-1)m_pc^2 \approx \frac{1}{2}(m_pc^2)(v/c)^2##
... lets you get the speed in terms of the speed of light.

You'll find these numbers much much easier to remember and use than SI units ;)
 
Last edited:

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