Impact parameter of alpha particles

Click For Summary
SUMMARY

The discussion focuses on calculating the impact parameter for alpha particles with an energy of 7.7 MeV deflecting by 2 degrees when interacting with a thin gold foil. The impact parameter equation used is b = (k * qα * Q) / (mα * v²) * cot(θ/2). Participants clarified the values for k, qα, and Q, confirming that k = 8.988 x 10^9 N m²/C², qα = 2(1.6 x 10^-19 C), and Q = 79(1.6 x 10^-19 C). The final calculation yielded an impact parameter of b = 2.70 x 10^-31 m, prompting further discussion on the necessity of converting MeV to eV and the correct application of charge values.

PREREQUISITES
  • Understanding of classical mechanics and energy conservation principles
  • Familiarity with electrostatic force equations, specifically Coulomb's law
  • Knowledge of particle physics, particularly alpha particle interactions
  • Ability to perform unit conversions, specifically MeV to eV
NEXT STEPS
  • Study the derivation and application of the impact parameter equation in nuclear physics
  • Learn about the conversion of MeV to eV and its significance in particle interactions
  • Explore the role of charge quantization in particle physics, focusing on elementary charge (e)
  • Investigate the effects of different materials on alpha particle deflection and scattering
USEFUL FOR

Students and educators in physics, particularly those focusing on nuclear physics and particle interactions, as well as researchers interested in alpha particle behavior in various materials.

stripes
Messages
262
Reaction score
0

Homework Statement



For alpha particles of 7.7 MeV, what impact parameter will result in a deflection of 2 degrees for a thin gold foil?

Homework Equations



impact parameter equation, [tex]b=\frac{kq_{\alpha}Q}{m_{\alpha}v^{2}}cot\frac{\theta}{2}[/tex]

The Attempt at a Solution



Pretty easy, just plug and play into[tex]b=\frac{kq_{\alpha}Q}{m_{\alpha}v^{2}}cot\frac{\theta}{2}[/tex], but my textbook and instructor aren't very specific with what numbers we put in exactly. We know that mv^2 = 2KE, but as for the k, q and Q, do I just put k = 8.988x10^9, q(alpha) = 2(1.6x10^-19) and Q = 79(1.6x10^-19)?

I plugged in all of these numbers, along with theta = 2 degrees, and I get b = 2.70x10^-31 m. That seems kind of small for a deflection of 2 degrees. Did I do something wrong? Am I plugging in the wrong numbers?

Thank you all in advance.
 
Physics news on Phys.org
After taking a closer look, I think my problems are:

1. I need to convert MeV to eV,
2. For q(alpha), use q = 2,
3. For Q, use Q =79,
4. Multiply k*q(alpha)*Q*e for e = 1.6x10^-19.

I did this for another question for which I had the answer, and I got the correct answer. Is this the problem here? Do I always need to use the above 4 solutions to my problem?

The thing is, i was under the assumption that q = 2e, and then Q = 79e, which means you need to have e^2 in the numerator...but that is not the case...I don't get that...
 

Similar threads

Replies
9
Views
1K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 3 ·
Replies
3
Views
5K
Replies
9
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 10 ·
Replies
10
Views
8K
Replies
1
Views
21K
  • · Replies 5 ·
Replies
5
Views
16K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K