Question regarding Rutherford's atomic model

AI Thread Summary
The discussion centers on understanding the derivation of the equation Ve = 2Vα in the context of Rutherford's atomic model. The original poster struggles with the assumption and arrives at Ve = 0, which contradicts known physics. A response clarifies that Ve = 2Vα is not merely an assumption but a conclusion derived from the equations of elastic collisions. It suggests setting up the equations for one-dimensional elastic collisions and deriving a general solution for the final velocities. The poster acknowledges their misunderstanding and realizes the importance of solving for Valpha' in relation to the other equations.
daselocution
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Homework Statement


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I am basically wondering how they got to the assumption that the Ve = 2Vα. I've tried it a few times and I keep on getting that based off of their other assumptions, Ve should =0, even though I know that this cannot possibly be the case.

Homework Equations



Conservation of momentum: Mava = Mav'a + meve

Conservation of energy: .5Mava2 = .5Mav'a2 + .5Meve2

The Attempt at a Solution



I really don't know what to do. I've tried starting under the assumption as given that Va≈Va', which leads to getting Ve=0. I tried to do a binomial expansion using the fact that Me<<Ma, but again I got zero. I'm really not sure how to approach this. Any help would be appreciated.
 
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daselocution said:
I am basically wondering how they got to the assumption that the Ve = 2Vα.
It's not an assumption, it's a conclusion.

Set up the equations for an elastic collision in one dimension, then get a general solution for the final velocities of both particles. Don't put in any numbers or the like, just get a general solution. Then assume that m_1 &gt;&gt; m_2, and see what you get for v_{2f}.
 
Thank you very much. That makes much more sense, I'm not really sure why I didn't just do that the first time. My mistake was not solving for Valpha' in terms of the other equations. Once I realized that I had to do that, it was much clearer.
 
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