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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;








2












$begingroup$


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







share|cite|improve this question











$endgroup$


















    2












    $begingroup$


    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







    share|cite|improve this question











    $endgroup$














      2












      2








      2


      1



      $begingroup$


      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







      share|cite|improve this question











      $endgroup$




      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 5 hours ago









      Ben Crowell

      57.1k6172331




      57.1k6172331










      asked 8 hours ago









      João BravoJoão Bravo

      28718




      28718




















          3 Answers
          3






          active

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          4












          $begingroup$

          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.






          share|cite|improve this answer








          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.





          $endgroup$




















            2












            $begingroup$

            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.






            share|cite|improve this answer









            $endgroup$




















              0












              $begingroup$

              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.






              share|cite|improve this answer









              $endgroup$








              • 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











              Your Answer








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              3 Answers
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              $begingroup$

              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.






              share|cite|improve this answer








              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.





              $endgroup$

















                4












                $begingroup$

                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.






                share|cite|improve this answer








                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.





                $endgroup$















                  4












                  4








                  4





                  $begingroup$

                  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.






                  share|cite|improve this answer








                  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.







                  share|cite|improve this answer








                  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.








                  share|cite|improve this answer



                  share|cite|improve this answer






                  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




                  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.

























                      2












                      $begingroup$

                      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.






                      share|cite|improve this answer









                      $endgroup$

















                        2












                        $begingroup$

                        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.






                        share|cite|improve this answer









                        $endgroup$















                          2












                          2








                          2





                          $begingroup$

                          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.






                          share|cite|improve this answer









                          $endgroup$



                          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.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 5 hours ago









                          Ben CrowellBen Crowell

                          57.1k6172331




                          57.1k6172331





















                              0












                              $begingroup$

                              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.






                              share|cite|improve this answer









                              $endgroup$








                              • 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















                              0












                              $begingroup$

                              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.






                              share|cite|improve this answer









                              $endgroup$








                              • 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













                              0












                              0








                              0





                              $begingroup$

                              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.






                              share|cite|improve this answer









                              $endgroup$



                              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.







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              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












                              • 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

















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