Why is iron the peak of the binding energy curve?
$begingroup$
If Nickel-62 and Iron-58 have more binding energy per nucleon than Iron-56 does, then why is iron-56 shown as the peak of the binding energy curve? Also, does adding neutrons always make the atom more stable because it will increase the strong nuclear force but not add any more electrorepulsive force?
nuclear-physics binding-energy
$endgroup$
add a comment |
$begingroup$
If Nickel-62 and Iron-58 have more binding energy per nucleon than Iron-56 does, then why is iron-56 shown as the peak of the binding energy curve? Also, does adding neutrons always make the atom more stable because it will increase the strong nuclear force but not add any more electrorepulsive force?
nuclear-physics binding-energy
$endgroup$
$begingroup$
@PM2Ring may be "per nucleon"
$endgroup$
– Thomas Fritsch
4 hours ago
1
$begingroup$
Also, you should only have 1 question per question. But briefly, atoms with an excess of neutrons are unstable.
$endgroup$
– PM 2Ring
4 hours ago
add a comment |
$begingroup$
If Nickel-62 and Iron-58 have more binding energy per nucleon than Iron-56 does, then why is iron-56 shown as the peak of the binding energy curve? Also, does adding neutrons always make the atom more stable because it will increase the strong nuclear force but not add any more electrorepulsive force?
nuclear-physics binding-energy
$endgroup$
If Nickel-62 and Iron-58 have more binding energy per nucleon than Iron-56 does, then why is iron-56 shown as the peak of the binding energy curve? Also, does adding neutrons always make the atom more stable because it will increase the strong nuclear force but not add any more electrorepulsive force?
nuclear-physics binding-energy
nuclear-physics binding-energy
edited 1 hour ago
Student
asked 4 hours ago
StudentStudent
443
443
$begingroup$
@PM2Ring may be "per nucleon"
$endgroup$
– Thomas Fritsch
4 hours ago
1
$begingroup$
Also, you should only have 1 question per question. But briefly, atoms with an excess of neutrons are unstable.
$endgroup$
– PM 2Ring
4 hours ago
add a comment |
$begingroup$
@PM2Ring may be "per nucleon"
$endgroup$
– Thomas Fritsch
4 hours ago
1
$begingroup$
Also, you should only have 1 question per question. But briefly, atoms with an excess of neutrons are unstable.
$endgroup$
– PM 2Ring
4 hours ago
$begingroup$
@PM2Ring may be "per nucleon"
$endgroup$
– Thomas Fritsch
4 hours ago
$begingroup$
@PM2Ring may be "per nucleon"
$endgroup$
– Thomas Fritsch
4 hours ago
1
1
$begingroup$
Also, you should only have 1 question per question. But briefly, atoms with an excess of neutrons are unstable.
$endgroup$
– PM 2Ring
4 hours ago
$begingroup$
Also, you should only have 1 question per question. But briefly, atoms with an excess of neutrons are unstable.
$endgroup$
– PM 2Ring
4 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The "folk wisdom" that iron-56 has the highest binding energy per nucleon is in fact incorrect. I can't do much better than citing an article on the subject:
M. P. Fewell, "The atomic nuclide with the highest mean binding energy". Am. J. Phys. 63, 653–658 (1995).
The author of that work traces this misconception back to texts on stellar nucleosynthesis in the '50s and '60s. Stellar nucleosynthesis does favor the production of iron over nickel, and the author postulates that this fact may have been conflated with the peak of the binding energy curve.
We can roughly model nuclei as having a set of "proton energy levels" and "neutron energy levels". Since both protons and neutrons are spin-$frac12$ fermions, this means that one can have at most two neutrons per energy level in the nucleus. Adding more neutrons to the nucleus will thus result in the neutrons being piled into higher-energy states.
However, neutrons can undergo beta-decay into protons: $n to p^+ + e^- + bar{nu}$. Suppose a neutron is in a relatively high energy level in the nucleus, and there is a vacant proton energy level below it. It can be energetically favorable for this neutron to turn into a proton and drop into this lower energy level. Thus, nuclei with too many neutrons will tend to undergo beta decay. (The same argument shows why nuclei with too many protons will tend to undergo inverse beta decay.)
$endgroup$
$begingroup$
Inverse beta decay isn't so common, compared to normal beta decay. And when it does happen, it's often in the form of electron capture, rather than positron emission. Nuclei with a large proton excess tend to spit out protons, borderline heavy nuclei can reduce their need for neutrons by emitting alphas.
$endgroup$
– PM 2Ring
3 hours ago
$begingroup$
So if nickel is more stable/tightly bound than iron, why do stars favor iron production over nickel?
$endgroup$
– Student
1 hour ago
add a comment |
$begingroup$
From Wikipedia:
Iron-56 (56Fe) is the most efficiently bound nucleus meaning that it has the least average mass per nucleon. However, nickel-62 is the most tightly bound nucleus in terms of energy of binding per nucleon. (Nickel-62's higher energy of binding does not translate to a larger mean mass loss than Fe-56, because Ni-62 has a slightly higher ratio of neutrons/protons than does iron-56, and the presence of the heavier neutrons increases nickel-62's average mass per nucleon).
In any case, iron-56 is not necessarily usually shown (anymore) as the highest binding energy per nucleon, mostly because it's not usually possible to show the tiny difference in binding energy between iron-56 and nickel-62. The difference is so tiny that accounting for the fact that the neutron is 0.1% heavier than the proton, instead of treating all nucleons equally, changes which is bound more tightly. Because of this tiny difference, it's unlikely that you'll see one depicted as higher than the other on any plot that actually shows both of them, like this one (source: https://www.asc.ohio-state.edu/kagan.1/phy367/Lectures/P367_lec_14.html):
Can you tell which of the two has higher binding energy? I certainly can't.
In addition, adding neutrons does not always make the nucleus more stable. In the nuclear shell model, protons and neutrons, being distinguishable from each other, occupy distinct sets of energy levels in the nucleus. (As a side note, the neutron energy levels are shifted lower than the proton energy levels because there is no Coulomb repulsion among neutrons. This is why the heavier stable nuclei tend to have more neutrons - you can fill more of the neutron energy levels for the same valence nucleon energy.) Since they're both fermions, you have to add new neutrons to higher and higher energy levels. Adding too many neutrons, enough to put the highest filled neutron energy level significantly above the highest proton energy level, will cause the highest-energy neutrons to spontaneously convert to protons via beta decay so that they can occupy a lower (proton) energy level.
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "151"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f460457%2fwhy-is-iron-the-peak-of-the-binding-energy-curve%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The "folk wisdom" that iron-56 has the highest binding energy per nucleon is in fact incorrect. I can't do much better than citing an article on the subject:
M. P. Fewell, "The atomic nuclide with the highest mean binding energy". Am. J. Phys. 63, 653–658 (1995).
The author of that work traces this misconception back to texts on stellar nucleosynthesis in the '50s and '60s. Stellar nucleosynthesis does favor the production of iron over nickel, and the author postulates that this fact may have been conflated with the peak of the binding energy curve.
We can roughly model nuclei as having a set of "proton energy levels" and "neutron energy levels". Since both protons and neutrons are spin-$frac12$ fermions, this means that one can have at most two neutrons per energy level in the nucleus. Adding more neutrons to the nucleus will thus result in the neutrons being piled into higher-energy states.
However, neutrons can undergo beta-decay into protons: $n to p^+ + e^- + bar{nu}$. Suppose a neutron is in a relatively high energy level in the nucleus, and there is a vacant proton energy level below it. It can be energetically favorable for this neutron to turn into a proton and drop into this lower energy level. Thus, nuclei with too many neutrons will tend to undergo beta decay. (The same argument shows why nuclei with too many protons will tend to undergo inverse beta decay.)
$endgroup$
$begingroup$
Inverse beta decay isn't so common, compared to normal beta decay. And when it does happen, it's often in the form of electron capture, rather than positron emission. Nuclei with a large proton excess tend to spit out protons, borderline heavy nuclei can reduce their need for neutrons by emitting alphas.
$endgroup$
– PM 2Ring
3 hours ago
$begingroup$
So if nickel is more stable/tightly bound than iron, why do stars favor iron production over nickel?
$endgroup$
– Student
1 hour ago
add a comment |
$begingroup$
The "folk wisdom" that iron-56 has the highest binding energy per nucleon is in fact incorrect. I can't do much better than citing an article on the subject:
M. P. Fewell, "The atomic nuclide with the highest mean binding energy". Am. J. Phys. 63, 653–658 (1995).
The author of that work traces this misconception back to texts on stellar nucleosynthesis in the '50s and '60s. Stellar nucleosynthesis does favor the production of iron over nickel, and the author postulates that this fact may have been conflated with the peak of the binding energy curve.
We can roughly model nuclei as having a set of "proton energy levels" and "neutron energy levels". Since both protons and neutrons are spin-$frac12$ fermions, this means that one can have at most two neutrons per energy level in the nucleus. Adding more neutrons to the nucleus will thus result in the neutrons being piled into higher-energy states.
However, neutrons can undergo beta-decay into protons: $n to p^+ + e^- + bar{nu}$. Suppose a neutron is in a relatively high energy level in the nucleus, and there is a vacant proton energy level below it. It can be energetically favorable for this neutron to turn into a proton and drop into this lower energy level. Thus, nuclei with too many neutrons will tend to undergo beta decay. (The same argument shows why nuclei with too many protons will tend to undergo inverse beta decay.)
$endgroup$
$begingroup$
Inverse beta decay isn't so common, compared to normal beta decay. And when it does happen, it's often in the form of electron capture, rather than positron emission. Nuclei with a large proton excess tend to spit out protons, borderline heavy nuclei can reduce their need for neutrons by emitting alphas.
$endgroup$
– PM 2Ring
3 hours ago
$begingroup$
So if nickel is more stable/tightly bound than iron, why do stars favor iron production over nickel?
$endgroup$
– Student
1 hour ago
add a comment |
$begingroup$
The "folk wisdom" that iron-56 has the highest binding energy per nucleon is in fact incorrect. I can't do much better than citing an article on the subject:
M. P. Fewell, "The atomic nuclide with the highest mean binding energy". Am. J. Phys. 63, 653–658 (1995).
The author of that work traces this misconception back to texts on stellar nucleosynthesis in the '50s and '60s. Stellar nucleosynthesis does favor the production of iron over nickel, and the author postulates that this fact may have been conflated with the peak of the binding energy curve.
We can roughly model nuclei as having a set of "proton energy levels" and "neutron energy levels". Since both protons and neutrons are spin-$frac12$ fermions, this means that one can have at most two neutrons per energy level in the nucleus. Adding more neutrons to the nucleus will thus result in the neutrons being piled into higher-energy states.
However, neutrons can undergo beta-decay into protons: $n to p^+ + e^- + bar{nu}$. Suppose a neutron is in a relatively high energy level in the nucleus, and there is a vacant proton energy level below it. It can be energetically favorable for this neutron to turn into a proton and drop into this lower energy level. Thus, nuclei with too many neutrons will tend to undergo beta decay. (The same argument shows why nuclei with too many protons will tend to undergo inverse beta decay.)
$endgroup$
The "folk wisdom" that iron-56 has the highest binding energy per nucleon is in fact incorrect. I can't do much better than citing an article on the subject:
M. P. Fewell, "The atomic nuclide with the highest mean binding energy". Am. J. Phys. 63, 653–658 (1995).
The author of that work traces this misconception back to texts on stellar nucleosynthesis in the '50s and '60s. Stellar nucleosynthesis does favor the production of iron over nickel, and the author postulates that this fact may have been conflated with the peak of the binding energy curve.
We can roughly model nuclei as having a set of "proton energy levels" and "neutron energy levels". Since both protons and neutrons are spin-$frac12$ fermions, this means that one can have at most two neutrons per energy level in the nucleus. Adding more neutrons to the nucleus will thus result in the neutrons being piled into higher-energy states.
However, neutrons can undergo beta-decay into protons: $n to p^+ + e^- + bar{nu}$. Suppose a neutron is in a relatively high energy level in the nucleus, and there is a vacant proton energy level below it. It can be energetically favorable for this neutron to turn into a proton and drop into this lower energy level. Thus, nuclei with too many neutrons will tend to undergo beta decay. (The same argument shows why nuclei with too many protons will tend to undergo inverse beta decay.)
answered 3 hours ago
Michael SeifertMichael Seifert
15k22852
15k22852
$begingroup$
Inverse beta decay isn't so common, compared to normal beta decay. And when it does happen, it's often in the form of electron capture, rather than positron emission. Nuclei with a large proton excess tend to spit out protons, borderline heavy nuclei can reduce their need for neutrons by emitting alphas.
$endgroup$
– PM 2Ring
3 hours ago
$begingroup$
So if nickel is more stable/tightly bound than iron, why do stars favor iron production over nickel?
$endgroup$
– Student
1 hour ago
add a comment |
$begingroup$
Inverse beta decay isn't so common, compared to normal beta decay. And when it does happen, it's often in the form of electron capture, rather than positron emission. Nuclei with a large proton excess tend to spit out protons, borderline heavy nuclei can reduce their need for neutrons by emitting alphas.
$endgroup$
– PM 2Ring
3 hours ago
$begingroup$
So if nickel is more stable/tightly bound than iron, why do stars favor iron production over nickel?
$endgroup$
– Student
1 hour ago
$begingroup$
Inverse beta decay isn't so common, compared to normal beta decay. And when it does happen, it's often in the form of electron capture, rather than positron emission. Nuclei with a large proton excess tend to spit out protons, borderline heavy nuclei can reduce their need for neutrons by emitting alphas.
$endgroup$
– PM 2Ring
3 hours ago
$begingroup$
Inverse beta decay isn't so common, compared to normal beta decay. And when it does happen, it's often in the form of electron capture, rather than positron emission. Nuclei with a large proton excess tend to spit out protons, borderline heavy nuclei can reduce their need for neutrons by emitting alphas.
$endgroup$
– PM 2Ring
3 hours ago
$begingroup$
So if nickel is more stable/tightly bound than iron, why do stars favor iron production over nickel?
$endgroup$
– Student
1 hour ago
$begingroup$
So if nickel is more stable/tightly bound than iron, why do stars favor iron production over nickel?
$endgroup$
– Student
1 hour ago
add a comment |
$begingroup$
From Wikipedia:
Iron-56 (56Fe) is the most efficiently bound nucleus meaning that it has the least average mass per nucleon. However, nickel-62 is the most tightly bound nucleus in terms of energy of binding per nucleon. (Nickel-62's higher energy of binding does not translate to a larger mean mass loss than Fe-56, because Ni-62 has a slightly higher ratio of neutrons/protons than does iron-56, and the presence of the heavier neutrons increases nickel-62's average mass per nucleon).
In any case, iron-56 is not necessarily usually shown (anymore) as the highest binding energy per nucleon, mostly because it's not usually possible to show the tiny difference in binding energy between iron-56 and nickel-62. The difference is so tiny that accounting for the fact that the neutron is 0.1% heavier than the proton, instead of treating all nucleons equally, changes which is bound more tightly. Because of this tiny difference, it's unlikely that you'll see one depicted as higher than the other on any plot that actually shows both of them, like this one (source: https://www.asc.ohio-state.edu/kagan.1/phy367/Lectures/P367_lec_14.html):
Can you tell which of the two has higher binding energy? I certainly can't.
In addition, adding neutrons does not always make the nucleus more stable. In the nuclear shell model, protons and neutrons, being distinguishable from each other, occupy distinct sets of energy levels in the nucleus. (As a side note, the neutron energy levels are shifted lower than the proton energy levels because there is no Coulomb repulsion among neutrons. This is why the heavier stable nuclei tend to have more neutrons - you can fill more of the neutron energy levels for the same valence nucleon energy.) Since they're both fermions, you have to add new neutrons to higher and higher energy levels. Adding too many neutrons, enough to put the highest filled neutron energy level significantly above the highest proton energy level, will cause the highest-energy neutrons to spontaneously convert to protons via beta decay so that they can occupy a lower (proton) energy level.
$endgroup$
add a comment |
$begingroup$
From Wikipedia:
Iron-56 (56Fe) is the most efficiently bound nucleus meaning that it has the least average mass per nucleon. However, nickel-62 is the most tightly bound nucleus in terms of energy of binding per nucleon. (Nickel-62's higher energy of binding does not translate to a larger mean mass loss than Fe-56, because Ni-62 has a slightly higher ratio of neutrons/protons than does iron-56, and the presence of the heavier neutrons increases nickel-62's average mass per nucleon).
In any case, iron-56 is not necessarily usually shown (anymore) as the highest binding energy per nucleon, mostly because it's not usually possible to show the tiny difference in binding energy between iron-56 and nickel-62. The difference is so tiny that accounting for the fact that the neutron is 0.1% heavier than the proton, instead of treating all nucleons equally, changes which is bound more tightly. Because of this tiny difference, it's unlikely that you'll see one depicted as higher than the other on any plot that actually shows both of them, like this one (source: https://www.asc.ohio-state.edu/kagan.1/phy367/Lectures/P367_lec_14.html):
Can you tell which of the two has higher binding energy? I certainly can't.
In addition, adding neutrons does not always make the nucleus more stable. In the nuclear shell model, protons and neutrons, being distinguishable from each other, occupy distinct sets of energy levels in the nucleus. (As a side note, the neutron energy levels are shifted lower than the proton energy levels because there is no Coulomb repulsion among neutrons. This is why the heavier stable nuclei tend to have more neutrons - you can fill more of the neutron energy levels for the same valence nucleon energy.) Since they're both fermions, you have to add new neutrons to higher and higher energy levels. Adding too many neutrons, enough to put the highest filled neutron energy level significantly above the highest proton energy level, will cause the highest-energy neutrons to spontaneously convert to protons via beta decay so that they can occupy a lower (proton) energy level.
$endgroup$
add a comment |
$begingroup$
From Wikipedia:
Iron-56 (56Fe) is the most efficiently bound nucleus meaning that it has the least average mass per nucleon. However, nickel-62 is the most tightly bound nucleus in terms of energy of binding per nucleon. (Nickel-62's higher energy of binding does not translate to a larger mean mass loss than Fe-56, because Ni-62 has a slightly higher ratio of neutrons/protons than does iron-56, and the presence of the heavier neutrons increases nickel-62's average mass per nucleon).
In any case, iron-56 is not necessarily usually shown (anymore) as the highest binding energy per nucleon, mostly because it's not usually possible to show the tiny difference in binding energy between iron-56 and nickel-62. The difference is so tiny that accounting for the fact that the neutron is 0.1% heavier than the proton, instead of treating all nucleons equally, changes which is bound more tightly. Because of this tiny difference, it's unlikely that you'll see one depicted as higher than the other on any plot that actually shows both of them, like this one (source: https://www.asc.ohio-state.edu/kagan.1/phy367/Lectures/P367_lec_14.html):
Can you tell which of the two has higher binding energy? I certainly can't.
In addition, adding neutrons does not always make the nucleus more stable. In the nuclear shell model, protons and neutrons, being distinguishable from each other, occupy distinct sets of energy levels in the nucleus. (As a side note, the neutron energy levels are shifted lower than the proton energy levels because there is no Coulomb repulsion among neutrons. This is why the heavier stable nuclei tend to have more neutrons - you can fill more of the neutron energy levels for the same valence nucleon energy.) Since they're both fermions, you have to add new neutrons to higher and higher energy levels. Adding too many neutrons, enough to put the highest filled neutron energy level significantly above the highest proton energy level, will cause the highest-energy neutrons to spontaneously convert to protons via beta decay so that they can occupy a lower (proton) energy level.
$endgroup$
From Wikipedia:
Iron-56 (56Fe) is the most efficiently bound nucleus meaning that it has the least average mass per nucleon. However, nickel-62 is the most tightly bound nucleus in terms of energy of binding per nucleon. (Nickel-62's higher energy of binding does not translate to a larger mean mass loss than Fe-56, because Ni-62 has a slightly higher ratio of neutrons/protons than does iron-56, and the presence of the heavier neutrons increases nickel-62's average mass per nucleon).
In any case, iron-56 is not necessarily usually shown (anymore) as the highest binding energy per nucleon, mostly because it's not usually possible to show the tiny difference in binding energy between iron-56 and nickel-62. The difference is so tiny that accounting for the fact that the neutron is 0.1% heavier than the proton, instead of treating all nucleons equally, changes which is bound more tightly. Because of this tiny difference, it's unlikely that you'll see one depicted as higher than the other on any plot that actually shows both of them, like this one (source: https://www.asc.ohio-state.edu/kagan.1/phy367/Lectures/P367_lec_14.html):
Can you tell which of the two has higher binding energy? I certainly can't.
In addition, adding neutrons does not always make the nucleus more stable. In the nuclear shell model, protons and neutrons, being distinguishable from each other, occupy distinct sets of energy levels in the nucleus. (As a side note, the neutron energy levels are shifted lower than the proton energy levels because there is no Coulomb repulsion among neutrons. This is why the heavier stable nuclei tend to have more neutrons - you can fill more of the neutron energy levels for the same valence nucleon energy.) Since they're both fermions, you have to add new neutrons to higher and higher energy levels. Adding too many neutrons, enough to put the highest filled neutron energy level significantly above the highest proton energy level, will cause the highest-energy neutrons to spontaneously convert to protons via beta decay so that they can occupy a lower (proton) energy level.
answered 3 hours ago
probably_someoneprobably_someone
17.5k12857
17.5k12857
add a comment |
add a comment |
Thanks for contributing an answer to Physics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f460457%2fwhy-is-iron-the-peak-of-the-binding-energy-curve%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
@PM2Ring may be "per nucleon"
$endgroup$
– Thomas Fritsch
4 hours ago
1
$begingroup$
Also, you should only have 1 question per question. But briefly, atoms with an excess of neutrons are unstable.
$endgroup$
– PM 2Ring
4 hours ago