I understand that greater binding energy per nucleon implies a more stable atom and atoms undergo nuclear fusion and fission to attain higher binding energy per nucleon. The binding energy per nucleon curve (https://en.wikipedia.org/wiki/File:Binding_energy_curve_-_common_isotopes.svg) shows that radioactive isotopes of hydrogen (deuterium and tritium) are more stable than hydrogen itself. Is this not incorrect? Is my assumption of greater binding energy per nucleon $=>$ more stable atom incorrect?
1 Answer
$^1_1H$ is a special case as the proton does not have any other nucleons to bind onto. I suppose that you could call the binding energy per nucleon zero which means that you require no energy to split up the nucleus of $^1_1H$ into its constituent parts?
Note also that the binding energy per nucleon is not necessarily the full measure of whether a nucleus is stable or not.
$^2_1H$ has a binding energy per nucleon of 1.11 MeV and is a stable isotope of hydrogen whereas $^3_1H$ has a binding energy per nucleon of 2.83 MeV and is an unstable isotope of hydrogen with a half life of 12.32 years.
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$\begingroup$ It would be very kind of you if you could elaborate on "the binding energy per nucleon is not necessarily the full measure of whether a nucleus is stable or not." Could you please explain what the values of binding energy per nucleon actually imply? $\endgroup$ Commented Apr 17, 2016 at 9:08
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$\begingroup$ @Saprativ Ray: "[...] elaborate on "the binding energy per nucleon is not necessarily the full measure of whether a nucleus is stable or not."" -- For example, tritium decays by a weak process. "what the values of binding energy per nucleon actually imply?" -- Comparison to all individual nucleons being separated from each other. (Btw.: yet another/more relevant notion of "stability" is: Which nucleus is the most resilient against gamma-induced fission?.) $\endgroup$ Commented Apr 17, 2016 at 15:59