How Close Can Alpha Particles Get to a Gold Nucleus in Rutherford's Experiment?

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Homework Help Overview

The discussion revolves around a Rutherford gold experiment where alpha particles are directed at a gold foil. The original poster is attempting to determine the closest distance an alpha particle can approach a gold nucleus, given the charges and kinetic energy involved.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster uses a specific equation to calculate the closest distance but expresses frustration over the results. Other participants question the appropriateness of the formula and seek clarification on the principles involved, including the relationship between kinetic energy and distance.

Discussion Status

The discussion is ongoing, with participants raising questions about the original poster's approach and the underlying principles of energy conservation in the context of the experiment. There is no explicit consensus, but several lines of inquiry are being explored.

Contextual Notes

There is uncertainty regarding the distance of the alpha particle source from the gold sheet, which may impact the analysis. Additionally, the original poster's calculations and assumptions are under scrutiny.

fatboy12341
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in a rutherford gold experiment, alpha particles with sharge +2e and a kinetic energy of 7.7 Mev were beamed at gold foil. the nuclius of a gold atom contains 79 protons, giving it a charge of +79e. What is the closest distance that an alpha particle can get to a gold nucleus when it approaches head on ?


ok i used the equation En=k^2 x e^4 me . Z^2
2xh^2 h^2


i am getting n^2= -9.07x10^21


i know this is wrong and am getting quite frustrated
if some on could help me it would be great
 
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can some one tell me if i am eve using the right formula\
 
We don't know the distance of the alpha particle source with the gold sheet?
 
fatboy12341 said:
in a rutherford gold experiment, alpha particles with sharge +2e and a kinetic energy of 7.7 Mev were beamed at gold foil. the nuclius of a gold atom contains 79 protons, giving it a charge of +79e. What is the closest distance that an alpha particle can get to a gold nucleus when it approaches head on ?


ok i used the equation En=k^2 x e^4 me . Z^2
2xh^2 h^2


i am getting n^2= -9.07x10^21
I am not sure what you are doing here. Can you explain what the principle is here? How is the kinetic energy of the alpha particle related to the closest distance between the alpha and the gold nucleus? What is the kinetic energy of the alpha particle when it is at the closest distance? Where did the energy go?

AM
 
Adding to AM's questions, what is the electric potential energy of the alpha particle at its closest approach, and how is it related to the alpha particle's kinetic energy?
 

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