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I was studying and came across this quote from Giancoli's Physics For Scientists & Engineers with Modern Physics. in Chapter 40. "From the uncertainty principle with $\Delta$x larger, we see that $\Delta$p and the minimum momentum can be less than when they are separate. That is, the molecule has less energy than the two separate atoms, and so is more stable."

Where $\Delta$x is larger because electrons can move around both the nuclei. I get the feeling that the way it was formulated is incorrect. Since the uncertainty principle only says something about the uncertainty and not about the value? Have I missed anything? And if not, what is a more correct way to explain the lower energy level? Sorry if this question was asked already, I couldn't find a question that solves this problem.

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  • $\begingroup$ Indeed, this quote does not make much sense. $\endgroup$
    – my2cts
    Dec 20 '19 at 20:32
  • $\begingroup$ Can . Too glib to the point of meaninglessness. Two atoms may or may not bind to a molecule. It is a delicate balance between kinetic and potential energies; it is not guaranteed that the two atoms can lower their total energy to a lower molecule energy just because they are apart! Look at the prototype dihydrogen cation. Consider this. The chemistry devil is in the details. $\endgroup$ Dec 20 '19 at 21:12
  • $\begingroup$ Sure it is handwaving but maybe it makes more sense if you also consider the preceding sentence of your quote: "Why a bond is formed can also be understood from the energy point of view. When the two H atoms approach close to one another, if the spins of their electrons are opposite, the electrons can occupy the same pace, as discussed above. This means that each electron can now move about in the space of two atoms instead of in the volume of only one. Because each electron now occupies more space, it is less well localized", and then followed by your quote. $\endgroup$
    – hyportnex
    Dec 20 '19 at 22:47
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The uncertainty principle is also related to the kinetic energy of an electron. The hydrogen atom confines the electron to a small sphere with an average radius of $0.5$ angstroms. If you restrict the accessible space of the electron to a smaller average distance from the nucleus, say $0.3$ angstroms, then the kinetic energy will increase dramatically, as occurs in the particle in a box model. Thus, the smaller the space that the electron can occupy, the smaller the fluctuation of its position, $\Delta x$, and the fluctuation of the momentum, $\Delta p$, increase by a a quantity approximately given by the Heisenberg uncertainty principle.

If you add a proton to the system, creating the H$_2^+$ ion, then the electron has another proton to move around and the accessible space for the electron is expanded in the ground state. Now, the fluctuation in the position is larger than the H atom and that makes $\Delta p$ smaller. Result: the kinetic energy, and also the total energy, is lower than the case of an isolated H. The argument for the H$_2$ molecule would be similar.

It is important to note that this is only a qualitative explanation. A better argument for H$_2$ require the Schrödinger equation.

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