Rutherford's 1911 Experiment: Electric Field & Deflection Formula

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Homework Help Overview

The discussion revolves around Rutherford's 1911 experiment concerning the scattering of alpha particles and the associated electric field and deflection formulas. The original poster seeks clarification on deriving the deflection formula from the given electric field expression.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss integrating the vertical component of the force to derive the deflection angle. Questions arise about the trajectory of the alpha particle and the relationship between various parameters such as the radius of the atom and the perpendicular distance from the center.

Discussion Status

Some participants have offered suggestions on how to approach the problem, including integrating forces and changing variables. There is an acknowledgment of the need for clarity in the problem statement and the importance of adhering to forum guidelines.

Contextual Notes

There is mention of specific variables and constants in the formulas, as well as a note on the importance of using the correct terminology regarding the nucleus and the center of the atom. Participants also highlight the need for proper formatting in the original post to facilitate assistance.

D.Hilbert
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Hi,

My question is about the famous paper by E. Rutherford on the scattering of alpha particles (1911). The paper is easy to find on internet. Rutherford gives the formula for the electric field inside an atom, at a distance r from the nucleus (here reduced to a point):

X = N e (1/r^2 - r/R^3)

Here N e in the electric charge in the nucleus and R is the radius of the atom. After he says

It is not difficult to show that the deflection (supposed small) of an electrified particle due to this field is given by

theta = b/p (1 - p^2/R^2)^(3/2)

where p is the perpendicular from the center on the path.

I can obtain X but I don't see where the formula for theta comes from.

Any suggestion?

Thanks DH
 
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If I were old Ernest, I would integrate the vertical component of the force over the section of the trajectory through the nucleus. If theta is small enough, pretending it is straight may be OK, and it's easier.
 
Hi,
Thanks for the answer.
Let the trajectory of the alpha particle be on the x-axis.

Are you suggesting to calculate

\int_{t1}^{t2} Fy dt

where Fy is the y-coordinate of the force?

DH
 
I found Rutherford-1911 where your formulas feature. There also is an explanation of what b stands for, which you would have included in your problem statement if you would have used the template. Please use it from now on. PF has a simple rule: no template, no assistance. It would have saved me some time that I could have used for others. Now I have to do some errands, so I am short on time.

But yes, (read: atom in my post, the nucleus is considered pointlike, somewhat incongruent in this context: it still had to be discovered. Better to speak of the center of the atom, but never mind). t1 and t2 can be related to R and p and the speed of the ##\alpha##. So you change from dt to dx. Some ##\beta## comes in with ## \sin\beta = {x\over R}##; perhaps you go from dx to d##\beta##. All constants go into b and there you are!
 
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