Neutron–proton ratio and isotope stability

Question

I have studied that most of the isotopes (not all of them!) with a neutron–proton ratio of $\ge 1.5$ are unstable; but it is obvious that this is not true in some cases like carbon-14 or technetium-99.

Is it possible for an isotope to have a neutron/proton ratio of more than 1.5 and still be stable? If yes, could you please give an example?

Answer 1

In context of this answer, stable means not undergoing a radioactive decay, regardless of the value of measured half-life.(+)

• For light nuclei till about 40 nucleons, the neutron/proton ratio of stable isobars(the same nucleon count) is generally around 1.0. ( $\ce{^3He, T}$ are quite exception. )
• Nuclei too aside from this ratio ( See Valley of stability ) are unstable, undergoing beta decay ( electron or positron emission, or electron capture ) toward the stability valley.
• Extremely proton/neutron rich nuclides may emit a respective nucleon.
• For nuclei with more than 40 nucleons, the "stability valley ratios" steadily climb from 1.0 for calcium toward about 1.5 for lead. It is because the short distance strong nuclear force gets harder time at balancing the long distance electrostatic repulsion of protons.
• The most stable nuclei are $\ce{^{56}_{26}Fe}$ ( the lowest mass per nucleon ) and $\ce{^{62}_{28}Ni}$ ( the highest binding energy per nucleon ).
• The element with the highest number of stable nuclides (10) is $\ce{_{50}Sn}$ - see magic numbers below.
• The heaviest stable yet nuclei reach the neutron/proton ratio near 1.5 ( 1.53 for $\ce{^{208}_{82}Pb}$), but they are already on the edge of cohesive ability of the strong nuclear force.
• There are nuclides that are observationally stable, but theoretically they could undergo a decay to more stable nuclides.
• All known nuclides with Z > 82 ( bismuth and above ) are unstable, adding emission of alpha particle, spontaneous or stimulated fission to the decay repertoir.

There are few other semi-empirical rules for nuclide stability :

• If the element has the odd number of protons, it has maximally 2 stable isotopes. Technetium and promethium have bad-luck here, everything taken by even-proton neighbors.
• If there are 2 isobars ( nuclei of the same nucleon number) of 2 neighbor elements, (like $\ce{^{40}K, ^{40}Ca}$), maximally one of them is stable ( the one with lower mass ). The unstable one usually decays by some of 3 modes of beta decay. ( $\ce{^{40}K}$ by all of them ).
• Generally, nuclei with lowest energy/mass and the highest stability are those with even number of protons as well as of neutrons ( even-even ). Then come odd-even(either way) ones and the least stable are generally odd-odd nuclei. This is related to stabilization of nucleus by nucleon spin coupling.
• There are semi-empirically determined so-called nucleon magic numbers
  2, 8, 20, 28, 50, 82, 126 ,
  related to nucleus quantum structure, predicting relative extra stability of a nucleus. The first 3 magic numbers apply to 3 "double-magic nuclei $\ce{^4_2He, ^{16}_8O, ^{40}_{20}Ca}$. the higher cases for the same numbers are not stable, as they miss the stability ratio, like $\ce{^{56}_{28}Ni}$.(+++)
• Some nuclei, often "even-even" ones, like $\ce{^{128}Te,^{82}Se}$ are conditionally stable (not to be confused with meta-stable nuclear isomers), being at the local energy minimum among their isobars. Their "odd-odd" neighbors (like $\ce{^{128}I}$) are even less stable with higher energy, so the regular beta decay is not applicable. But they can undergo very slow double beta decay, simultaneously converting 2 neutrons to protons (or vice versa ): $\ce{^{128}_{52}Te -> ^{128}_{54}Xe + 2 e- + 2 \bar{\nu_e}}$. This can be considered as a kind of quantum tunneling through the state with higher energy than the original nucleus has.
• Some of these "change Z by 2" events may be achieved by alpha emission even for medium mass nuclei. The lightest known alpha emitter is accidentally again tellurium: $\ce{^{104}_{52}Te -> ^{100}_{50}Sn + ^{4}_{2}He}$(++), unless we count the trivial case $\ce{^8_4Be -> 2 ^4_2He}$.

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(+) Very strictly speaking, elements before iron and behind nickel are not thermodynamically stable, but for most "conventionally stable" nuclei no spontaneous decay/fusion/fission is observed, directly nor indirectly.
(++) This is possibly related to the fact $\ce{^{100}_{50}Sn}$ is "double-magic" nucleus. But in spite of that, it decays anyway by emission of positron or proton, as the ratio 1.0 is too low for tin.
(+++) OTOH, the non-nequal double-magic 20/28 $\ce{^{48}_{20}Ca}$ with neutron/proton ratio 1.4 is unstable, but with exceptionally long half-time $\pu{6.4E19 \mathrm{year}}$, being the lightest nuclide undergoing the double beta decay.

Answer 2

The $N/Z$ numbers you see in general chemistry textbooks are just guiding principles. First of all, ask the question, what is the meaning of a stable isotope? Don't take these numbers and guiding principles too literally. In the comments you have been given so many examples. Let me give another example of $\ce{^{128}Te}$ ($Z=52$, $N/Z= 1.46$). This must be unstable as per the rigid general chemistry rules.

$\ce{^{128}Te}$ undergoes a very rare type of decay and its half life is on the order of $10^{24}$ years. Is this an unstable nucleus? The age of the universe is on the order of $10^{10}$ years. Compare this magnitude. You will have to wait for an infinite time to see it decay. This is stable by all standards. There are so many exceptions.

$\ce{^{124}Xe}$ ($Z=54$, $N/Z =1.29$), also "stable" on an indefinite time-scale. Wait for about $10^{22}$ years to watch it decay.

The examples in the comments deal with U and Pb nuclei with N/Z > 1.5. All isotopes of U are unstable, but $\ce{^{208}Pb}$, is stable.