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I actually wanted to posted this in the "Homework" section but it is currently working for me..

The problem is:

Calculate the cross-section for the scattering of a 10 MeV alpha particle by a gold nucleus [tex]_{79}^{197}Au[/tex] through an angle greater than (a) 10 degrees (b) 20 degrees c 30 degrees.

My answer:

I know the relevant equation is:

[tex]\sigma=\pi(\frac{Zze^{2}}{4\pi\epsilon_{0}mv_{0}^{2}}\cot\frac{\theta}{2})^{2}[/tex]

so its really just a plug and chug kind of problem... only thing I can't figure out is how to get the value for v_0.... i know that is hidden somewhere in the fact that it is a 10 Mev particle and I assume E=mc^2 plays a roll too.... but trying the following does not provide me with a useful value of v_o:

[tex]E=mc^{2}+\frac{1}{2}mv_{0}^{2}[/tex]

neither does:

[tex]E=\frac{1}{2}mv_{0}^{2}[/tex]

so my question is, how do I find the value of v_0 given that I know its an alpha particale (so I know its mass) and also that I know its 10Mev particle?

Many Thanks!

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# Scattering of Mev particles

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