- #1
SUDOnym
- 90
- 1
Hello
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 particlee (so I know its mass) and also that I know its 10Mev particle?
Many Thanks!
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 particlee (so I know its mass) and also that I know its 10Mev particle?
Many Thanks!