Calculating Distance of Closest Approach in Rutherford's Scattering Experiment

In summary, in Rutherford's scattering experiments, an alpha particle with an initial kinetic energy of 3.6 MeV is fired towards a gold atom. By using the equation for electrical potential energy, the distance of closest approach between the alpha particle and the gold nucleus can be found by setting the kinetic energy equal to the potential energy and solving for r. The correct answer is 1.01008 x 10^-32 meters.
  • #1
deenuh20
50
0

Homework Statement



In Rutherford's scattering experiments, alpha particles (charge = +2e) were fired at a gold foil. Consider an alpha particle, very far from the gold foil, with an initial kinetic energy of 3.6 MeV heading directly for a gold atom (charge +79e). The alpha particle will come to rest when all its initial kinetic energy has been converted to electrical potential energy. Find the distance of closest approach between the alpha particle and the gold nucleus.

Homework Equations



Force:
mv^2=kZe^2/r

k=8.99x10^9 N^2/C^2
Z= # protons
e=1.6x10^-19 C

The Attempt at a Solution



From the question, I'm assuming that the gold and the alpha particle never collide. Thus, I figured that by Newton's third law, I can deduce the distance (r) between them. First, I did 3.6MeV*2 to find mv^2 (since KE=1/2mv^2 and I know that all KE is converted to PE). I got 7200 eV. Then, I plugged in 79 for Z on the other side of the equation and tried solving for r. I came up with 2.525E-30 m. However this answer is not correct. Had I missed an important step? Thanks in advance for your help!
 
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  • #2
Hi,

Conceptually you've got this question: (assume that all of the particle's initial kinetic energy is converted into electrical potential energy). All of your problems stem from silly mistakes here:

deenuh20 said:
Force:
mv^2=kZe^2/r

1. As you correctly stated in the body of the solution, this equation has nothing to do with forces, but rather is an equation of energies.

2. So, the left hand side has mv^2 = 2*KE. That's all well and good, but how did you get rid of the 1/2? Answer: by multiplying BOTH sides of the equation by two. I don't see that on the right hand side.

3. You appear to have forgotten that your particle is an alpha particle with charge +2e. Your equation for the Coulomb potential energy is correct, but it applies to a particle of charge e in the vicinity of a nucleus of charge Ze. Remember the most general form that applies to the potential energy of a system of any two charges:

[tex] U = \frac{kq_1q_2}{r} [/tex]

Thus, for this problem you would have kZe(2e)/r = 2kZe^2/r

Let me know if these corrections do the trick.
 
  • #3
deenuh20 said:
First, I did 3.6MeV*2 to find mv^2 (since KE=1/2mv^2 and I know that all KE is converted to PE). I got 7200 eV.

Ok, here's a question for you: how the hell does 3.6 MILLION electron volts times two equal 7200 eV?

Also, did you convert everything to SI units when doing your calculation?
 
  • #4
cepheid said:
Ok, here's a question for you: how the hell does 3.6 MILLION electron volts times two equal 7200 eV?

Also, did you convert everything to SI units when doing your calculation?

Oops! For some reason, I thought 3.6MeV was 3.6 x 10^3 eV, but its 10^6!


However, I tried what you proposed in your first post, and I understand the reasoning...but I'm still not arriving at the correct answer. Here's what I'm doing:

Total E= 1/2mv^2 - k(q1)(q2)/r

mv^2=2k(q1)(q2)/r

KE = 1/2mv^2
3.6MeV * 2 = 7.2 x10^6 eV

r= 2k(q1)(q2)/7.2 x10^6 eV
= [(2)(8.99x10^9)(79)(2)(1.6x10^-19)^2]/[7.2 x10^6 eV]
= 1.01008 x 10^-32 m

which is Incorrect. :confused:
 
  • #5
*bump for viewing*
 
  • #6
Convert the kinetic energy from eV to Joules.
 
  • #7
Thank you! I got the answer :)
 

1. What was the purpose of the Rutherford Experiment?

The purpose of the Rutherford Experiment was to investigate the structure of the atom and determine the location of its positive charge.

2. Who conducted the Rutherford Experiment?

The Rutherford Experiment was conducted by New Zealand physicist Ernest Rutherford in 1911.

3. What were the main findings of the Rutherford Experiment?

The main findings of the Rutherford Experiment were that atoms have a small, dense, positively charged nucleus at the center and that most of the atom's mass is concentrated in this nucleus.

4. How was the Rutherford Experiment conducted?

The Rutherford Experiment involved firing alpha particles at a thin gold foil and observing their deflection. This was done using a specially designed apparatus called the Geiger-Marsden experiment.

5. How did the Rutherford Experiment contribute to our understanding of the atom?

The Rutherford Experiment revolutionized our understanding of the atom by disproving the previously accepted "plum pudding" model and providing evidence for the existence of a small, dense nucleus at the center of the atom. It also laid the foundation for the development of the modern atomic model.

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