Does a proton have a binding energy?The “binding energy” of bonded particles adds mass?Tritium decay is spontaneous even if the binding energy of tritium is higher than the binding energy of 3He. Why?Negative or positive binding energy?Calculating binding energy of a moleculeWhy is there a difference in measured binding energy vs formula for binding energy?Binding energy of the daughter nucleusDefect mass and Binding energyIf a nucleus is radioactive, will another nucleus with lesser binding energy be necessarily radioactive?For which kind of nuclear reactions is the nuclear binding energy per nucleon appropriate?$E = mc^2$ and Binding Energy Confusion
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Does a proton have a binding energy?
The “binding energy” of bonded particles adds mass?Tritium decay is spontaneous even if the binding energy of tritium is higher than the binding energy of 3He. Why?Negative or positive binding energy?Calculating binding energy of a moleculeWhy is there a difference in measured binding energy vs formula for binding energy?Binding energy of the daughter nucleusDefect mass and Binding energyIf a nucleus is radioactive, will another nucleus with lesser binding energy be necessarily radioactive?For which kind of nuclear reactions is the nuclear binding energy per nucleon appropriate?$E = mc^2$ and Binding Energy Confusion
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
When calculating the $Q$-value, $Q = Delta M cdot c^2$, of this reaction:
$$ ^6Li (alpha, p) ^9Be quad iff quad alpha + ^6Li longrightarrow ^9Be + p $$
The $Q$-value can also be written in terms of binding energies $BE$.
Should I consider the binding energy of the proton $p$ $ $ ~$1ucdot c^2$ or $0$?
nuclear-physics physical-chemistry nuclear-engineering protons binding-energy
edited 5 hours ago
Ben Crowell
57.1k6172331
asked 8 hours ago
João BravoJoão Bravo
28718
$endgroup$
add a comment |
$begingroup$
When calculating the $Q$-value, $Q = Delta M cdot c^2$, of this reaction:
$$ ^6Li (alpha, p) ^9Be quad iff quad alpha + ^6Li longrightarrow ^9Be + p $$
The $Q$-value can also be written in terms of binding energies $BE$.
Should I consider the binding energy of the proton $p$ $ $ ~$1ucdot c^2$ or $0$?
nuclear-physics physical-chemistry nuclear-engineering protons binding-energy
edited 5 hours ago
Ben Crowell
57.1k6172331
asked 8 hours ago
João BravoJoão Bravo
28718
$endgroup$
add a comment |
$begingroup$
When calculating the $Q$-value, $Q = Delta M cdot c^2$, of this reaction:
$$ ^6Li (alpha, p) ^9Be quad iff quad alpha + ^6Li longrightarrow ^9Be + p $$
The $Q$-value can also be written in terms of binding energies $BE$.
Should I consider the binding energy of the proton $p$ $ $ ~$1ucdot c^2$ or $0$?
nuclear-physics physical-chemistry nuclear-engineering protons binding-energy
edited 5 hours ago
Ben Crowell
57.1k6172331
asked 8 hours ago
João BravoJoão Bravo
28718
$endgroup$
When calculating the $Q$-value, $Q = Delta M cdot c^2$, of this reaction:
$$ ^6Li (alpha, p) ^9Be quad iff quad alpha + ^6Li longrightarrow ^9Be + p $$
The $Q$-value can also be written in terms of binding energies $BE$.
Should I consider the binding energy of the proton $p$ $ $ ~$1ucdot c^2$ or $0$?
nuclear-physics physical-chemistry nuclear-engineering protons binding-energy
nuclear-physics physical-chemistry nuclear-engineering protons binding-energy
edited 5 hours ago
Ben Crowell
57.1k6172331
asked 8 hours ago
João BravoJoão Bravo
28718
edited 5 hours ago
Ben Crowell
57.1k6172331
asked 8 hours ago
João BravoJoão Bravo
28718
edited 5 hours ago
Ben Crowell
57.1k6172331
edited 5 hours ago
Ben Crowell
57.1k6172331
edited 5 hours ago
Ben Crowell
57.1k6172331
57.1k6172331
asked 8 hours ago
João BravoJoão Bravo
28718
asked 8 hours ago
João BravoJoão Bravo
28718
asked 8 hours ago
João BravoJoão Bravo
28718
28718
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Note that binding energy is relative, like the gravitational potential energy in certain respects. That said, the binding energy of the proton could be considered non-zero if its sub-particles, i.e. quarks, are studied. However, since in the energy region of your mentioned reaction these sub-particles play no direct role (except for perhaps the underlying physics), it is more reasonable if you consider the binding energy of the nucleons, i.e. protons and neutrons, as zero. Note that if you consider the binding energy to be non-zero for both sides of the reaction, similar final results would be obtained.
answered 8 hours ago
Seyed Mohsen AyyoubzadehSeyed Mohsen Ayyoubzadeh
1511
New contributor
Seyed Mohsen Ayyoubzadeh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
No, a proton doesn't have a well-defined binding energy. That would be the energy required in order to separate it into three quarks, but free quarks don't exist.
answered 5 hours ago
Ben CrowellBen Crowell
57.1k6172331
$endgroup$
add a comment |
$begingroup$
Yes, protons have binding energy when they combine with neutrons to form a nucleus. In the fusion process where nucleons combine to form helium, they require less mass within the nucleus than they did as free particles. This surplus mass is emitted as binding energy, which is where the sun gets its power. When a nucleon leaves the nucleus,it follows that this binding energy must be re-supplied to it before it can become a free particle.
answered 6 hours ago
Michael WalsbyMichael Walsby
46216
$endgroup$
1
$begingroup$
I understand that, but I was just asking from the point of view of the formation of a proton
$endgroup$
– João Bravo
6 hours ago
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Note that binding energy is relative, like the gravitational potential energy in certain respects. That said, the binding energy of the proton could be considered non-zero if its sub-particles, i.e. quarks, are studied. However, since in the energy region of your mentioned reaction these sub-particles play no direct role (except for perhaps the underlying physics), it is more reasonable if you consider the binding energy of the nucleons, i.e. protons and neutrons, as zero. Note that if you consider the binding energy to be non-zero for both sides of the reaction, similar final results would be obtained.
answered 8 hours ago
Seyed Mohsen AyyoubzadehSeyed Mohsen Ayyoubzadeh
1511
New contributor
Seyed Mohsen Ayyoubzadeh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Note that binding energy is relative, like the gravitational potential energy in certain respects. That said, the binding energy of the proton could be considered non-zero if its sub-particles, i.e. quarks, are studied. However, since in the energy region of your mentioned reaction these sub-particles play no direct role (except for perhaps the underlying physics), it is more reasonable if you consider the binding energy of the nucleons, i.e. protons and neutrons, as zero. Note that if you consider the binding energy to be non-zero for both sides of the reaction, similar final results would be obtained.
answered 8 hours ago
Seyed Mohsen AyyoubzadehSeyed Mohsen Ayyoubzadeh
1511
New contributor
Seyed Mohsen Ayyoubzadeh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Note that binding energy is relative, like the gravitational potential energy in certain respects. That said, the binding energy of the proton could be considered non-zero if its sub-particles, i.e. quarks, are studied. However, since in the energy region of your mentioned reaction these sub-particles play no direct role (except for perhaps the underlying physics), it is more reasonable if you consider the binding energy of the nucleons, i.e. protons and neutrons, as zero. Note that if you consider the binding energy to be non-zero for both sides of the reaction, similar final results would be obtained.
answered 8 hours ago
Seyed Mohsen AyyoubzadehSeyed Mohsen Ayyoubzadeh
1511
New contributor
Seyed Mohsen Ayyoubzadeh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Note that binding energy is relative, like the gravitational potential energy in certain respects. That said, the binding energy of the proton could be considered non-zero if its sub-particles, i.e. quarks, are studied. However, since in the energy region of your mentioned reaction these sub-particles play no direct role (except for perhaps the underlying physics), it is more reasonable if you consider the binding energy of the nucleons, i.e. protons and neutrons, as zero. Note that if you consider the binding energy to be non-zero for both sides of the reaction, similar final results would be obtained.
answered 8 hours ago
Seyed Mohsen AyyoubzadehSeyed Mohsen Ayyoubzadeh
1511
New contributor
Seyed Mohsen Ayyoubzadeh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 8 hours ago
Seyed Mohsen AyyoubzadehSeyed Mohsen Ayyoubzadeh
1511
New contributor
Seyed Mohsen Ayyoubzadeh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 8 hours ago
Seyed Mohsen AyyoubzadehSeyed Mohsen Ayyoubzadeh
1511
answered 8 hours ago
Seyed Mohsen AyyoubzadehSeyed Mohsen Ayyoubzadeh
1511
1511
New contributor
Seyed Mohsen Ayyoubzadeh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Seyed Mohsen Ayyoubzadeh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
$begingroup$
No, a proton doesn't have a well-defined binding energy. That would be the energy required in order to separate it into three quarks, but free quarks don't exist.
answered 5 hours ago
Ben CrowellBen Crowell
57.1k6172331
$endgroup$
add a comment |
$begingroup$
No, a proton doesn't have a well-defined binding energy. That would be the energy required in order to separate it into three quarks, but free quarks don't exist.
answered 5 hours ago
Ben CrowellBen Crowell
57.1k6172331
$endgroup$
add a comment |
$begingroup$
No, a proton doesn't have a well-defined binding energy. That would be the energy required in order to separate it into three quarks, but free quarks don't exist.
answered 5 hours ago
Ben CrowellBen Crowell
57.1k6172331
$endgroup$
No, a proton doesn't have a well-defined binding energy. That would be the energy required in order to separate it into three quarks, but free quarks don't exist.
answered 5 hours ago
Ben CrowellBen Crowell
57.1k6172331
answered 5 hours ago
Ben CrowellBen Crowell
57.1k6172331
answered 5 hours ago
Ben CrowellBen Crowell
57.1k6172331
answered 5 hours ago
Ben CrowellBen Crowell
57.1k6172331
57.1k6172331
add a comment |
add a comment |
$begingroup$
Yes, protons have binding energy when they combine with neutrons to form a nucleus. In the fusion process where nucleons combine to form helium, they require less mass within the nucleus than they did as free particles. This surplus mass is emitted as binding energy, which is where the sun gets its power. When a nucleon leaves the nucleus,it follows that this binding energy must be re-supplied to it before it can become a free particle.
answered 6 hours ago
Michael WalsbyMichael Walsby
46216
$endgroup$
1
$begingroup$
I understand that, but I was just asking from the point of view of the formation of a proton
$endgroup$
– João Bravo
6 hours ago
add a comment |
$begingroup$
Yes, protons have binding energy when they combine with neutrons to form a nucleus. In the fusion process where nucleons combine to form helium, they require less mass within the nucleus than they did as free particles. This surplus mass is emitted as binding energy, which is where the sun gets its power. When a nucleon leaves the nucleus,it follows that this binding energy must be re-supplied to it before it can become a free particle.
answered 6 hours ago
Michael WalsbyMichael Walsby
46216
$endgroup$
1
$begingroup$
I understand that, but I was just asking from the point of view of the formation of a proton
$endgroup$
– João Bravo
6 hours ago
add a comment |
$begingroup$
Yes, protons have binding energy when they combine with neutrons to form a nucleus. In the fusion process where nucleons combine to form helium, they require less mass within the nucleus than they did as free particles. This surplus mass is emitted as binding energy, which is where the sun gets its power. When a nucleon leaves the nucleus,it follows that this binding energy must be re-supplied to it before it can become a free particle.
answered 6 hours ago
Michael WalsbyMichael Walsby
46216
$endgroup$
Yes, protons have binding energy when they combine with neutrons to form a nucleus. In the fusion process where nucleons combine to form helium, they require less mass within the nucleus than they did as free particles. This surplus mass is emitted as binding energy, which is where the sun gets its power. When a nucleon leaves the nucleus,it follows that this binding energy must be re-supplied to it before it can become a free particle.
answered 6 hours ago
Michael WalsbyMichael Walsby
46216
answered 6 hours ago
Michael WalsbyMichael Walsby
46216
answered 6 hours ago
Michael WalsbyMichael Walsby
46216
answered 6 hours ago
Michael WalsbyMichael Walsby
46216
46216
1
$begingroup$
I understand that, but I was just asking from the point of view of the formation of a proton
$endgroup$
– João Bravo
6 hours ago
add a comment |
1
$begingroup$
I understand that, but I was just asking from the point of view of the formation of a proton
$endgroup$
– João Bravo
6 hours ago
1
1
$begingroup$
I understand that, but I was just asking from the point of view of the formation of a proton
$endgroup$
– João Bravo
6 hours ago
$begingroup$
I understand that, but I was just asking from the point of view of the formation of a proton
$endgroup$
– João Bravo
6 hours ago
add a comment |
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