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A equation we all come across in high-school physics:

$$\frac{-dN}{dt}= kN$$ where N is number nuclei left

Is this always true for spontaneous nuclear decays? In chemistry, we find second, third order reactions. Similarly, has anyone found spontaneous decay where the decay rate is proportional not to the first exponent of remaining nuclei, rather to second or third exponent?

Bill N
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Mockingbird
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1 Answers1

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The nuclear forces responsible for radioactive decay are short ranged and so isolated from other forces (such as the much weaker E$M forces) that this results in the exponential decay law. I believe there have been experiments that demonstrate that these decays can be slightly affected by exposure to very strong E&M fields, but this is a special circumstance that normally does not occur. I would have to Google to find references to those experiments. I believe the experiments were conducted on nuclear isomers.

Edit: The experiments that I was remembering took place between 1998 and 2007, but a Google search reveals that those experiments have now been discredited. You may read about this episode here. The search terms that I used were: nuclear, isomer, decay, stimulated. If you follow the links resulting from this search, you may find the original sources.

Lewis Miller
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