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conantee
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« on: January 02, 2010, 01:41:38 PM »

42. A proton beam is incident on a scatterer 0.1 centimeter thick. The scatterer contains 10^{20} target nuclei per cubic centimeter. In passing through the scatterer, one proton per incident million is scattered, The scattering cross section is
(A) 10^{-29} \mbox{ cm^2}
(B) 10^{-27} \mbox{ cm^2}
(C) 10^{-25} \mbox{ cm^2}
(D) 10^{-23} \mbox{ cm^2}
(E) 10^{-21} \mbox{ cm^2}
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« Reply #1 on: January 02, 2010, 01:56:32 PM »

First, check the choices, you don't have do accurate numerical calculation here. So, try to approximate and simplify your calculation.

Now, if you don't remember how cross section works. Just recognize that, the thicker scatterer, the less protons pass the way through it. You should also expect exponential behavior here (read about this topic if you don't notice that). You should also know that higher density of nuclei causes more scattering in the scatterer, so it decreases the number of protons passing the way through it. Combining this knowledge together and dimensional analysis. You have
r = e^{- n \sigma x}, where r is the ratio of number of protons passing through the scatterer per number of incoming protons, n is the number density of nucleis in the scatterer, \sigma is the cross-section area, and x is thickness.

Now, one proton per incident million is scattered. so, r = 1-1/100000 = 1-10^{-6}
Also, we can approximate e^{-n \sigma x} \approx 1 - n \sigma x since the cross-section is very small (you can see from choices).

Therefore, n \sigma x = 10^{-6} and \sigma = \frac{10^{-6}}{nx} = \frac{10^{-6}}{(10^{20})(0.1)} = 10^{25} \mbox{ cm^2}

The correct answer is (C). 38 of 100 people answer it correctly.
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