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- Thread starter sodaboy7
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Khashishi

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Of course, the solution is already available for 1/r^2 forces (Rutherford scattering).

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Ok. I got the first part. But can you please elaborate "effective potential" stuff a little more.

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Khashishi

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dx/dt = dy/dt = dz/dt = 0, because there are no forces because these particles don't collide. But if you look at dr/dt, dr/dt > 0. There is an effective potential that pushes the particle away from the center of mass. see http://en.wikipedia.org/wiki/Effective_potential

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http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/rutsca2.html

Plugging values of nuclear Z, the scattered angle, and the incident alpha particle energy into the empty boxes is sufficient to determine the

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Thanks

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