# Can the decay rate of nuclear decay be proportional to the second/third exponent of number of nuclei?

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?

• Closely related: physics.stackexchange.com/questions/178233/… – Emilio Pisanty Jul 9 '17 at 13:05
• It's not the same @Emilo Pisanty – Mockingbird Jul 9 '17 at 13:07
• ... which is why I said it's related instead of a duplicate. Nevertheless, it does mean that you need to ask a sharper question: if your question is whether there are observations of decay that does not follow the simple exponential, then yes there are, as detailed in the link above. If you're asking whether there are observations of nuclear decay that specifically has the population derivative equal to a higher power of the population, then the answer is no, but you should make that clearer in the question. – Emilio Pisanty Jul 9 '17 at 13:13
• See my edit. Want to suggest something? – Mockingbird Jul 9 '17 at 13:55
• If you want to make an analogy with chemical reactions, then you should probably be considering nuclear reactions, rather than spontaneous nuclear decay. For example, you could consider the cross-section for neutron or proton transfer reactions as a function of the number of nucleons (mass number) in the target and projectile. – Ben Crowell Jul 9 '17 at 15:34