Rutherford's atom experiment problem

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Homework Help Overview

The problem involves Rutherford's experiment, specifically the interaction between a helium nucleus and a gold nucleus. The original poster seeks to determine how close the helium nucleus will come to the gold nucleus before reversing direction, given specific parameters such as mass, charge, and initial velocity.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the use of energy concepts and electric field equations. There is a focus on the original poster's attempt to apply a specific formula, which some participants question for clarity and dimensional consistency.

Discussion Status

Guidance has been offered regarding the need to think about energy in the context of the problem. Participants are exploring the relationships between the variables involved, and there is an emphasis on understanding the underlying principles rather than simply applying formulas.

Contextual Notes

There is a suggestion that the original poster's approach may lack some necessary components for dimensional consistency. Participants are encouraged to clarify their equations and assumptions, indicating a potential gap in understanding the problem setup.

Dana Fishel
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Homework Statement


In 1911, Ernest Rutherford discovered the nucleus of the atom by observing the scattering of helium nuclei from gold nuclei. If a helium nucleus with a mass of 6.68
multiply.gif
10^-27 kg, a charge of +2e, and an initial velocity of 1.30
multiply.gif
10^7 m/s is projected head-on toward a gold nucleus with a charge of +79e, how close will the helium atom come to the gold nucleus before it stops and turns around? (Assume the gold nucleus is held in place by other gold atoms and does not move.)

Homework Equations


electric field equations involving k constant, charge, mass, and velocity

The Attempt at a Solution


I was told by a classmate that this should work:

d = (9e9)4(79)(1.602e-19)^2 / (6.68e-27)(1.3e7)
= 8.4e-7 meters
(where e means x10^#)

But it's wrong, and I'm not sure which equation this solution came from in the first place.
 
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Dana Fishel said:
d = (9e9)4(79)(1.602e-19)^2 / (6.68e-27)(1.3e7)
It's not reasonable to expect others to decode that by reverse engineering the numbers.
Please post it as an equation using symbols, preferably defining them.
 
Think energy.
 
Haru is absolutely right. But you yourself can easily do the reverse engineering, isn't it ?

d = (9e9)4(79)(1.602e-19)^2 / (6.68e-27)(1.3e7)

I see a kcoulomb, q1, q2, m, and a v. In short: all you need.
But dimensionally it doesn't fit. Andrev's hint should be enough to discover the little thingy that's still missing from your friendly classmate's spoiler !

[warning]: make sure you learn from this. Getting the answer from a mate means that in fact you miss the exercise. You can't have someone else train for the marathon in your place and still expect to be able to run it yourself!
 
It's very similar to throwing a ball upwards, except in this case the force increases drastically "upwards".
 

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