If you add neutrons to a nucleus, each is more weakly bound than the last. Eventually, the last neutron added is unbound, so it comes right back out. Usually, this happens within a time comparable to $1\times10^{-23}$ sec. For each proton number, Z, there is a maximum number of neutrons, call it Nd, which can be in a nucleus with Z protons. The set of nuclides $(Z,Nd)$ is a curve on a Z,N plane known as the neutron dripline. The neutron dripline defines the maximum size a nucleus with a given number of protons may have.

If a nucleus with Z protons has too few neutrons, one of two things will happen. It may eject a proton or it may fission. Large nuclei will almost invariably fission, though, so that's the important criterion. The simplest workable model of an atomic nucleus is the "liquid drop model". Since its charges are trying to push it apart, though, thinking of a nucleus as a tiny, highly stressed balloon gives a better idea of the forces in play. Electric repulsion varies as $(Z^2 / r_{eff})$ where reff is distance between equivalent point charges. What pulls the nucleus together is what amounts to surface tension - unbalanced nuclear cohesion - and the total "surface energy" stored varies as $(r^2)$, where r is nuclear radius. The ratio between Coulomb and surface energies is defined by $(Z^2 / r_{eff})*(1/ r^2) = K$. Set $r_{eff} = r$. Nuclear volume is proportional to the total number of particles, $A = Z+N$, in a collection. That means r varies as $A^{1/3}$, so $(Z^2 / r^3) = K = (Z^2)/A$. K is called a "fissility parameter". A given value of K defines a set of nuclei which have similar liquid-drop-model barriers against spontaneous fission. For specified value of K, $N(Z) = (1/K)*(Z^2) - Z$ defines a curve of constant fission barrier height on the $(Z,N)$ plane. One particular curve defines the line dividing sets of nucleons for which a fission barrier exists and sets of nucleons which do not. In other words, it defines the minimum number of neutrons which a nucleus of given Z may have.

At least one nuclear model includes nuclei with up to $330$ neutrons and $175$ protons(1). An equation for the neutron dripline as a function of Z can be extracted from their dripline. A second equation for $N/Z$ as $f(Z)$ can be used to construct an alternate dripline curve. KUTY's neutron dripline does not show any dramatic changes below $N=330$. Still, when extrapolating into the unknown, it seems prudent to consider the upper limit to neutron count in a nucleus to be $1/4$ order of magnitude ($1.77$) times larger.

Liquid-drop theory predicts immediate fission for $K>50$; however, the liquid drop model does not account for the additional binding produced by nuclear structure. The maximum K value for any nucleus in the KUTY model can be used as a guide to how large K must be to overcome these corrections. Taking that value as the geometric mean between $K=50$ and the value of K to be used gives $K=102$. (This was the highest of three techniques tried).

For large Z, the fission curve rises faster than the dripline curve. The point at which they meet is the largest possible nucleus. Anything larger will immediately decay by neutron emission or fission. Nominally, the largest nucleus $Z=592$, $N=2846$ - but that is way too much precision for this sort of calculation. It's reasonable to say that the largest possible nucleus has $Z <600$ and $N < 3000$.

If you add neutrons to a nucleus, each is more weakly bound than the last. Eventually, the last neutron added is unbound, so it comes right back out. Usually, this happens within a time comparable to $1\times10^{-23}$ sec. For each proton number, Z, there is a maximum number of neutrons, call it Nd, which can be in a nucleus with Z protons. The set of nuclides $(Z,Nd)$ is a curve on a Z,N plane known as the neutron dripline. The neutron dripline defines the maximum size a nucleus with a given number of protons may have.

If a nucleus with Z protons has too few neutrons, one of two things will happen: It may eject a proton or it may fission. Large nuclei will almost invariably fission, though, so that's the important criterion. The simplest workable model of an atomic nucleus is the "liquid drop model". Since its charges are trying to push it apart, though, thinking of a nucleus as a tiny, highly stressed balloon gives a better idea of the forces in play. Electric repulsion varies as $(Z^2 / r_{eff})$ where $r_{eff}$ is distance between equivalent point charges. What pulls the nucleus together is what amounts to surface tension – unbalanced nuclear cohesion – and the total "surface energy" stored varies as $(r^2)$, where r is nuclear radius. The ratio between Coulomb and surface energies is defined by $(Z^2 / r_{eff})*(1/ r^2) = K$. Set $r_{eff} = r$. Nuclear volume is proportional to the total number of particles, $A = Z+N$, in a collection. That means r varies as $A^{1/3}$, so $(Z^2 / r^3) = K = (Z^2)/A$. K is called a "fissility parameter." A given value of K defines a set of nuclei which have similar liquid-drop-model barriers against spontaneous fission. For specified value of K, $N(Z) = (1/K)*(Z^2) - Z$ defines a curve of constant fission barrier height on the $(Z,N)$ plane. One particular curve defines the line dividing sets of nucleons for which a fission barrier exists and sets of nucleons which do not. In other words, it defines the minimum number of neutrons which a nucleus of given Z may have.

At least one nuclear model includes nuclei with up to $330$ neutrons and $175$ protons⁽¹⁾. An equation for the neutron dripline as a function of Z can be extracted from their dripline. A second equation for $N/Z$ as $f(Z)$ can be used to construct an alternate dripline curve. KUTY's neutron dripline does not show any dramatic changes below $N=330$. Still, when extrapolating into the unknown, it seems prudent to consider the upper limit to neutron count in a nucleus to be $1/4$ order of magnitude ($1.77$) times larger.

Liquid-drop theory predicts immediate fission for $K>50$; however, the liquid drop model does not account for the additional binding produced by nuclear structure. The maximum K value for any nucleus in the KUTY model can be used as a guide to how large K must be to overcome these corrections. Taking that value as the geometric mean between $K=50$ and the value of K to be used gives $K=102$. (This was the highest of three techniques tried.)

For large Z, the fission curve rises faster than the dripline curve. The point at which they meet is the largest possible nucleus. Anything larger will immediately decay by neutron emission or fission. Nominally, the largest nucleus $Z=592$, $N=2846$ - but that is way too much precision for this sort of calculation. It's reasonable to say that the largest possible nucleus has $Z <600$ and $N < 3000$.

Defining an "atomic nucleus" as a set of protons and neutrons, in a common nuclear potential well, whose mean life is large with respect to the time it took the set to form. (A nuclear interaction takes place over a span of time on the order of 1E-23 sec.)

If you add neutrons to a nucleus, each is more weakly bound than the last. Eventually, the last neutron added is unbound, so it comes right back out. Usually, this happens within a time comparable to 1E-23 sec. For each proton number, Z, there is a maximum number of neutrons, call it Nd, which can be in a nucleus with Z protons. The set of nuclides (Z,Nd) is a curve on a Z,N plane known as the neutron dripline. The neutron dripline defines the maximum size a nucleus with a given number of protons may have.

If a nucleus with Z protons has too few neutrons, one of two things will happen. It may eject a proton or it may fission. Large nuclei will almost invariably fission, though, so that's the important criterion. The simplest workable model of an atomic nucleus is the "liquid drop model". Since its charges are trying to push it apart, though, thinking of a nucleus as a tiny, highly stressed balloon gives a better idea of the forces in play. Electric repulsion varies as (Z^2 / reff) where reff is distance between equivalent point charges. What pulls the nucleus together is what amounts to surface tension - unbalanced nuclear cohesion - and the total "surface energy" stored varies as (r^2), where r is nuclear radius. The ratio between Coulomb and surface energies is defined by (Z^2 / reff)(1/ r^2) = K. Set reff = r. Nuclear volume is proportional to the total number of particles, A = Z+N, in a collection. That means r varies as A^(1/3), so (Z^2 / r^3) = K = (Z^2)/A. K is called a "fissility parameter". A given value of K defines a set of nuclei which have similar liquid-drop-model barriers against spontaneous fission. For specified value of K, N(Z) = (1/K)(Z^2) - Z defines a curve of constant fission barrier height on the (Z,N) plane. One particular curve defines the line dividing sets of nucleons for which a fission barrier exists and sets of nucleons which do not. In other words, it defines the minimum number of neutrons which a nucleus of given Z may have.

At least one nuclear model includes nuclei with up to 330 neutrons and 175 protons(1). An equation for the neutron dripline as a function of Z can be extracted from their dripline. A second equation for N/Z as f(Z) can be used to construct an alternate dripline curve. KUTY's neutron dripline does not show any dramatic changes below N=330. Still, when extrapolating into the unknown, it seems prudent to consider the upper limit to neutron count in a nucleus to be 1/4 order of magnitude (1.77) times larger.

Liquid-drop theory predicts immediate fission for K>50; however, the liquid drop model does not account for the additional binding produced by nuclear structure. The maximum K value for any nucleus in the KUTY model can be used as a guide to how large K must be to overcome these corrections. Taking that value as the geometric mean between K=50 and the value of K to be used gives K=102. (This was the highest of three techniques tried).

For large Z, the fission curve rises faster than the dripline curve. The point at which they meet is the largest possible nucleus. Anything larger will immediately decay by neutron emission or fission. Nominally, the largest nucleus Z=592, N=2846 - but that is way too much precision for this sort of calculation. It's reasonable to say that the largest possible nucleus has Z <600 and N < 3000.

Defining an "atomic nucleus" as a set of protons and neutrons, in a common nuclear potential well, whose mean life is large with respect to the time it took the set to form. (A nuclear interaction takes place over a span of time on the order of $1\times10^{-23}$ sec.)

If you add neutrons to a nucleus, each is more weakly bound than the last. Eventually, the last neutron added is unbound, so it comes right back out. Usually, this happens within a time comparable to $1\times10^{-23}$ sec. For each proton number, Z, there is a maximum number of neutrons, call it Nd, which can be in a nucleus with Z protons. The set of nuclides $(Z,Nd)$ is a curve on a Z,N plane known as the neutron dripline. The neutron dripline defines the maximum size a nucleus with a given number of protons may have.

If a nucleus with Z protons has too few neutrons, one of two things will happen. It may eject a proton or it may fission. Large nuclei will almost invariably fission, though, so that's the important criterion. The simplest workable model of an atomic nucleus is the "liquid drop model". Since its charges are trying to push it apart, though, thinking of a nucleus as a tiny, highly stressed balloon gives a better idea of the forces in play. Electric repulsion varies as $(Z^2 / r_{eff})$ where reff is distance between equivalent point charges. What pulls the nucleus together is what amounts to surface tension - unbalanced nuclear cohesion - and the total "surface energy" stored varies as $(r^2)$, where r is nuclear radius. The ratio between Coulomb and surface energies is defined by $(Z^2 / r_{eff})*(1/ r^2) = K$. Set $r_{eff} = r$. Nuclear volume is proportional to the total number of particles, $A = Z+N$, in a collection. That means r varies as $A^{1/3}$, so $(Z^2 / r^3) = K = (Z^2)/A$. K is called a "fissility parameter". A given value of K defines a set of nuclei which have similar liquid-drop-model barriers against spontaneous fission. For specified value of K, $N(Z) = (1/K)*(Z^2) - Z$ defines a curve of constant fission barrier height on the $(Z,N)$ plane. One particular curve defines the line dividing sets of nucleons for which a fission barrier exists and sets of nucleons which do not. In other words, it defines the minimum number of neutrons which a nucleus of given Z may have.

At least one nuclear model includes nuclei with up to $330$ neutrons and $175$ protons(1). An equation for the neutron dripline as a function of Z can be extracted from their dripline. A second equation for $N/Z$ as $f(Z)$ can be used to construct an alternate dripline curve. KUTY's neutron dripline does not show any dramatic changes below $N=330$. Still, when extrapolating into the unknown, it seems prudent to consider the upper limit to neutron count in a nucleus to be $1/4$ order of magnitude ($1.77$) times larger.

Liquid-drop theory predicts immediate fission for $K>50$; however, the liquid drop model does not account for the additional binding produced by nuclear structure. The maximum K value for any nucleus in the KUTY model can be used as a guide to how large K must be to overcome these corrections. Taking that value as the geometric mean between $K=50$ and the value of K to be used gives $K=102$. (This was the highest of three techniques tried).

For large Z, the fission curve rises faster than the dripline curve. The point at which they meet is the largest possible nucleus. Anything larger will immediately decay by neutron emission or fission. Nominally, the largest nucleus $Z=592$, $N=2846$ - but that is way too much precision for this sort of calculation. It's reasonable to say that the largest possible nucleus has $Z <600$ and $N < 3000$.