So I was reading about the stability of elements based on Nuclear Binding Energy, and I saw that the 'Iron group' of elements were most tightly bound and hence most stable, and that is why the graph peaks there. Why do elements that come after Iron, which are less stable even exist? And if they do, why do they not constantly strive to achieve Iron-like stability?
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2$\begingroup$ By your argument, all the elements lighter than Fe should be striving to be iron as well. Those elements that are stable, are stable against the usual (alpha, beta, gamma, fission) processes that can occur to an isolated atom. $\endgroup$– Jon CusterCommented Jun 24, 2015 at 23:04
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$\begingroup$ Yeah but once the fission causes iron to form, why go further than that when it's the most stable? Anything after that would inherently be less stable than what it was before. How can it be a spontaneous process? $\endgroup$– Sreekar VoletiCommented Jun 25, 2015 at 0:12
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$\begingroup$ It's not a spontaneous process but a non-equilibrium process in supernovae. These nuclei are being generated faster in the initial conditions of the explosion than they are being destroyed during the later phases. You are correct that there would only be microscopic amounts of these nuclei, if the only processes that would create them were in thermodynamic equilibrium. $\endgroup$– CuriousOneCommented Jun 25, 2015 at 5:01
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There are a couple of related questions:
- What elements can be created in the fusion process of different types of stars?
- What is the heaviest element possible produced in a supernova?
though surprisingly I can't find an exact duplicate (which probably just means I didn't look hard enough).
Iron is the most stable nucleus so in principle all other nuclei should fuse or fission to form iron, but the reaction is extremely slow because large kinetic barriers exist. If heavy nuclei are formed faster than they can decay then we end up with a signficiant concentration of the heavy nuclei.
In supernovae and stars heavy nuclei can be formed by the r-process and the s-process respectively. In normal stars the temperature is not high enough for the heavy nuclei to decay to iron at any significant rate (though they can be destroyed in other nuclear reactions). The temperature in supernovae may be high enough, but the high temperature lasts for too short a time. In either case the end result is a significant concentration of the heavy nuclei.
answered Jun 25, 2015 at 9:46
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$\begingroup$ There's a great answer by Rob Jeffries about how heavy nuclei are formed: physics.stackexchange.com/a/141180/123208 But I can't find other answers (apart from this one) about why they're formed. $\endgroup$– PM 2RingCommented Aug 12, 2020 at 8:09