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I have two questions :

  1. In a simple hydrogen atom $(^1_1H)$, there is no neutron. So, if there is no nuclear force to hold the nucleus together then what is the thing that is holding the atom together? Why doesn't the proton get out of the atom ?

  2. If there is no neutron, does this mean that the binding energy of Hydrogen atom is $0$ ? I came across a question, in which binding energy of helium atom was to be calculated and in the solution the atomic mass of helium atom was subtracted from the sum of 2 Hydrogen atoms $(2\space\times 1.0078u)$ and 2 neutrons $(2\space\times1.0087u$), so wouldn't this imply that the binding energy of hydrogen is $0$?

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    $\begingroup$ For both 1 and 2, it seems you think of the nucleus as a bound system - but it's a single proton. It's not clear to me why you would need nuclear force to hold a single proton together or what the "binding energy" of a single proton is supposed to be. $\endgroup$ – ACuriousMind Feb 24 '17 at 12:22
  • $\begingroup$ 1.What's holding the atom together..? 2.And is the binding energy 0 or not..? $\endgroup$ – Mitchell Feb 24 '17 at 12:30
  • $\begingroup$ This question (v2) seems to conflate nuclear physics and atomic physics. $\endgroup$ – Qmechanic Feb 24 '17 at 14:28
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    $\begingroup$ @BhavyaSharma I think the heart of your confusion is that you believe there's a difference between "hydrogen nuclei" and "proton," but there is not. The phenomena of "binding energy" is manifest for a single proton (in the form of bound quarks and also for the proton-electron pair), but this is somewhat different than what is normally meant by "binding energy" in the nuclear context, which is the difference in weight between the nucleus and its constituent hadrons. There is no difference in weight between a proton and a hydrogen nucleus because they are the same thing. $\endgroup$ – Bobak Hashemi Feb 26 '17 at 11:27
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The nucleus of hydrogen is just a proton, protons are stable. There are bound valence quarks in the nucleus however, and they are held together by the strong force. The proton-electron version of hydrogen does have the same feature you mentioned though, that it weighs slightly less than the mass of a free proton plus the mass of a free electron due to binding energy of the pair. But the binding energy there is much smaller than what you get from the strong force.

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The hydrogen atom does not need a neutron. Neutrons exist in the atom to hold the positively charged protons together, otherwise they would repel each other. With only one proton, there is no need for a neutron. Also, the electron would stay in a stable orbit because since the neutron has no charge the electron is not attracted to it, and therefore, nothing is really too different.

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  • $\begingroup$ Not related to the question but I curious about why the atom is neutral from outside and by this the protons charge has exactly to cancel out the electrons charge BUT in helium (with two protons canceling out the electric field of two electrons) the two protons repeal each other? My conclusion, or there has to be a net electric field from the electrons (in the case of repealing force between the two protons) or there isn't net field of the two electrons and three isn't any repealing force between the protons. $\endgroup$ – HolgerFiedler Feb 24 '17 at 17:29
  • $\begingroup$ Well, in a helium atom, there are 2 protons, 2 neutrons, and 2 electrons, the charges of all of these particles would cancel out and make it neutral. $\endgroup$ – TECTEC3 Studios Feb 24 '17 at 17:33
  • $\begingroup$ So there is no repulsion between the protons? $\endgroup$ – HolgerFiedler Feb 24 '17 at 17:35
  • $\begingroup$ There is repulsion, known as proton degeneracy, but the neutrons use the strong force to hold all of the protons together in a helium atom. $\endgroup$ – TECTEC3 Studios Feb 24 '17 at 17:38
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    $\begingroup$ Don't think of degeneracy in terms of contact—it's a quantum phenomena and follows quantum rules. In that picture electrons in an atom are not little balls and they don't have positions. They occupy stationary states with well defined energy and angular momentum but without well defined position or linear momentum. The electrons in a Helium atom (in its ground state) are degenerate in the sense that they completely fill the $n=1$ shell and any added electrons would necessarily need to have higher energy. $\endgroup$ – dmckee Feb 24 '17 at 18:09

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