Energy of electrons described by hydrogen wavefunctions

In a previous post, we showed the Bohr atom as represented at the Georgia State University Physics Department’s website in the following way:

Screen shot 2013-03-21 at 8.03.10 PM

Is it possible to derive the total energy of electrons in the Bohr model from Schrodinger’s wavefunctions?

In general, we can calculate the total energy, E, of the electron as the sum of its Coulombic potential, U_C and kinetic energy, E_k: E = U_C + E_k

With, E, we can calculate the average total energy, E_{avg}, using: E_{avg}=\int\int\int{r^2\sin{\phi}E\psi^2}drd{\phi}d{\theta}

So what is the form of E?

The average potential energy, U_{C,avg}, is calculated using Coulomb’s law: U_{C,avg}=\int_r^\infty-\frac{ke^2}{r^2}dr

The average kinetic energy, E_{k,avg}, can be determined if the electrons travel around the nucleus in a circular fashion.*

From Newton’s second law: F_{net}=ma and a=m\frac{v^2}{r} for circular motion
while F_C=\frac{ke^2}{r^2}. Therefore, m\frac{v^2}{r}=\frac{ke^2}{r^2}, so E_k = 1/2m{v^2}=\frac{1}{2}\frac{ke^2}{r} = -\frac{1}{2}U_{C,avg}

This means that we have our average total energy for the circulating electron: E_{avg} = U_{C,avg} + E_{k,avg} = U_{C,avg} - \frac{1}{2}U_{C,avg} = 1/2U_{C,avg}, and E_{avg}=\int\int\int{r^2\sin{\phi}\left(-\frac{1}{2}\frac{ke^2}{r}\right)\psi^2}drd{\phi}d{\theta}

From here, it is a simple matter of applying the integral we learned earlier: \int_0^\infty{t^ne^{-at}}dt = n!\left(\frac{1}{a}\right)^{n+1} to the \psi_{1s}^2 and \psi_{2s}^2 to derive the total energy of the electrons in these Bohr orbits:

The Energy of the 1s electron
E_{1s,avg} = \int_0^{2\pi}\int_0^{\pi}\int_0^\infty{r^2\cdot{\sin{\phi}}\cdot\left(-\frac{1}{2}\frac{ke^2}{r}\right)\cdot{\left(\frac{1}{\sqrt{\pi}a_0^{3/2}}e^{-\frac{r}{a_0}}\right)^2}}drd{\phi}d{\theta}
E_{1s,avg} = \frac{-4\pi{ke^2}}{2\pi{a_0}^3}\int_0^\infty{r\cdot{\left(e^{-\frac{2r}{a_0}}\right)}}dr
E_{1s,avg} = \frac{-4\pi{ke^2}}{2\pi{a_0}^3}{1!\cdot{\left(\frac{a_0}{2}\right)^2}}

E_{1s,avg} = \frac{-{ke^2}}{{2a_0}} = -13.60571 eV

The Energy of the 2s electron
E_{2s,avg} = \int_0^{2\pi}\int_0^{\pi}\int_0^\infty{r^2\cdot{\sin{\phi}}\cdot\left(-\frac{1}{2}\frac{ke^2}{r}\right)\cdot{\left(\frac{1}{4\sqrt{2\pi}a_0^{3/2}}(2-\frac{r}{a_0})e^{-\frac{r}{2a_0}}\right)^2}}drd{\phi}d{\theta}
E_{2s,avg} = \frac{-4\pi{ke^2}}{64\pi{a_0^{3}}}\int_0^\infty{{\left(4r-4\frac{r^2}{a_0}+\frac{r^3}{a_0^2}\right)e^{-\frac{r}{a_0}}}}dr
E_{2s,avg} = \frac{-{ke^2}}{16{a_0^{3}}}\left(4\cdot1!\cdot{a_0}^2-\frac{4\cdot2!a_0^3}{a_0}+\frac{3!a_0^4}{a_0^2}\right)
E_{2s,avg} = -\frac{{ke^2}}{{8a_0}} = -3.401427 eV

Looking at the energy differences between these two energy levels, the \Delta{E} = h\nu = 10.20428 eV can be calculated and compared to the energy levels observed in the hydrogen spectrum (Lymann series):
Screen shot 2013-03-22 at 5.51.39 PM

A wavelength of 121.566 nm corresponds to an energy of: E_{photon} = h\nu = \frac{hc}{\lambda} = 10.1968 eV. Not bad agreement! Bohr’s energy levels inversely depend upon the square of the principle quantum number, n. The model is derived from the consideration of a lone electron (of a hydrogen atom or a helium anion) and therefore didn’t work well to describe the spectral lines of the many-electron atoms. This was something that vexed him. As Max Tegmark and John Archibald Wheeler describe in their wonderful “100 Years of Quantum Mysteries”: “Back in Copenhagen, Bohr got a letter from Rutherford telling him he had to publish his results. Bohr wrote back that nobody would believe him unless he explained the spectra of all the elements. Rutherford replied: Bohr, you explain hydrogen and you explain helium, and everyone will believe all the rest.”

In a later post, at some point, it would be interesting to see if we could derive the general square dependence of the Bohr radius on the principle quantum number, n. This has been done for the specific case above, but for a general derivation, it would likely be necessary to determine how the principle quantum number arises from the wave equation. I have never taken a formal quantum course so need to do some reading before I can post about this. Also, it would be fun to try to derive the hydrogen wavefunctions too!

Overall, I think this wraps up my experiments with the overlap between quantum and multivariable calculus and physics for some time. I had a blast and I hope you got to share in the fun too!

*Note: I wonder if we could calculate another version of the kinetic energy using a deBroglie relationship: E_{k,avg}=\int_V{\hbar\omega}\psi^2dV, treating the electron as a wave.

I am a math and science teacher at a boarding school in Delaware.

Posted in Advanced Chemistry, Multivariable Calculus
One comment on “Energy of electrons described by hydrogen wavefunctions
  1. […] have used to write many entries on this blog and am still learning how to use it. It can be fun AND empowering to express your ideas the way you […]

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