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Imagine you have two 1 kg samples of a radioactive material. one sample is molded into...

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eyehawk | (Level 1) Honors

Posted September 27, 2013 at 3:19 PM via web

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Imagine you have two 1 kg samples of a radioactive material. one sample is molded into a sphere; as a result, it has a relatively small surface area. the other sample is spread out in a thin sheet, with a very large surface area. which of these samples is more likely to experience a chain reaction? why?

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valentin68 | College Teacher | (Level 3) Associate Educator

Posted September 27, 2013 at 5:19 PM (Answer #1)

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Usually a nuclear reaction (including nuclear chain reactions) takes place as a result of a nucleus absorbing an incoming nuclear particle (like a neutron in the case of uranium fission). The rate of reaction depends on (is proportional to) the total flux of the incoming particles and on the probability that one of the incoming particle hits the target nucleus and is absorbed into it. This probability for a certain particle to hit and be absorbed by a nucleus  is called nuclear cross section. Thinking on atomic distances, this cross section should be the same for all identical atoms regardless if they are placed together in a small sphere space or if they are spread on to a very large area. This is called the atomic nuclear cross section. However on larger scale, the macroscopic cross section depends on the density of nuclei in the target. Therefore the macroscopic cross section for the nuclei in the sphere will be bigger than the macroscopic cross section for the atoms in the thin layer (spread over a larger distance).

You can think at this somewhat different. For the same flux of incoming particles that triggers the reaction there will be more nuclei to capture them if the nuclei are packed in a sphere than if the nuclei are spread in a thin layer.

Thus the sample molded into the shape of a sphere will have a much higher probability to experience a chain reaction than the sample spread over a large surface area.

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eyehawk | (Level 1) Honors

Posted September 28, 2013 at 2:22 AM (Reply #1)

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Thanks for your help

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