Recent content by SuperPokenerd
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Rutherford Scattering cross section
I thought so, since the problem says that theta goes between pi/2 and pi i.e. large angle scattering is from 90-180 degrees. I don't see how I'm supposed to get it to be 4pi instead of 8/3.- SuperPokenerd
- Post #3
- Forum: Advanced Physics Homework Help
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Rutherford Scattering cross section
Homework Statement Integrate the rutherford cross section over the backward hemisphere to get 4pi(sigma0(E)) Homework Equations Rutherford cross section is sigma0(E)/sin^4(theta/2) The Attempt at a Solution When I integrate this with the limits pi/2 to pi i get sigma0(E)*(8/3) i...- SuperPokenerd
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- Cross Cross section Rutherford Rutherford scattering Scattering Scattering cross section Section
- Replies: 3
- Forum: Advanced Physics Homework Help
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Abundancy of Uranium 235 when the earth was formed.
Got it wrong so now I got an 80 on my HW. Thanks for nothing.- SuperPokenerd
- Post #5
- Forum: Advanced Physics Homework Help
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Abundancy of Uranium 235 when the earth was formed.
ok, so I've worked it up to the point where I have the new abundancy = .0065e^(lambda238-lambda235)*4.5billion years but it's wrong. How can i fix this?- SuperPokenerd
- Post #4
- Forum: Advanced Physics Homework Help
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Abundancy of Uranium 235 when the earth was formed.
Well, N/N0 would be the probability that a nucleus has decayed in the given period of time which in this case would be -4.5billion years if we take t=0 to be the present. And can't N be larger than N0 if we are going back in time?- SuperPokenerd
- Post #3
- Forum: Advanced Physics Homework Help
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Abundancy of Uranium 235 when the earth was formed.
Homework Statement The Earth is about 4.5 billion years old. If 235U is 0.65% abundant today, how abundant was it when the Earth formed? Note, in this case abundancy is defined as the ratio of Uranium 235 to Uranium 238Homework Equations R=N(lambda) N=N0e^-lambda(t) Half Life = ln(2)/lambda...- SuperPokenerd
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- Earth Uranium
- Replies: 4
- Forum: Advanced Physics Homework Help