In my Barron's SAT Physics book, it states that ionization energy is equal to the absolute value of the ground state energy. This doesn't make sense to me because ionization energy is the minimum amount of energy required to eject an electron from an atom, which I would think would require an ionization energy greater than the absolute value of the ground state energy. If the ionization energy is equal to the absolute value of the ground state energy, then wouldn't that mean the electron is now situated on the outer most energy level with energy value 0 eV (electron still in atom)? I speculate that the reason for why I am wrong is that the 0 eV energy level is actually imaginary, which I have concluded based the 0 eV ring being dashed in my book's diagram of the energy levels as shown: enter image description here

However, in this Khan Academy video https://www.khanacademy.org/science/physics/quantum-physics/atoms-and-electrons/v/atomic-energy-levels at 9:22 the narrator states that if an electron has more than zero energy then it is ejected, which is what I would think would be true.

  • $\begingroup$ A slightly tangential point: I've heard that that book is useless: people say it isn't a good representation of what you're expected to know for that test, and apparently there are tonnes of potentially misleading, loosely-framed points. (Don't ask me which alternative books are good: I have no idea) $\endgroup$
    – user191954
    Commented Jul 13, 2018 at 7:32
  • 1
    $\begingroup$ Look at those numbers carefuly: They have a minus sign in front of them. Negative energy is asociated with bound states, positive total energy with free states. Your book doesn't contradict what the great Khan says. $\endgroup$ Commented May 19, 2019 at 1:41

3 Answers 3


The "absolute" energy of the ground state is simply the energy of the ground state with respect to the vacuum. So the 0eV state you describe is a free election that is at rest in the vacuum, it is not a bound state, since its radius is essentially infinity (in comparison to the ground state radius).

In other words, the ionization energy is the energy required to remove an electron from the ground state to the vacuum state where it has zero kinetic energy.

  • $\begingroup$ Wait so then for the photoelectric effect, photons must have an energy greater than the ionization energy? I am thinking this because the photoelectric effect converts photon energy into electric energy, so if the electron has 0 kinetic energy, then it is not producing a current (thus no photoelectric effect) $\endgroup$
    – 54284User
    Commented Jul 13, 2018 at 6:44
  • $\begingroup$ Yes. There is an onset of the photoelectric current that starts at the ionization energy (or Work function as it is called for bulk solids). This actually matches the plot of current vs photon energy, that you can see in experiment. $\endgroup$
    – KF Gauss
    Commented Jul 13, 2018 at 10:39

In the hydrogen atom of your illustration, the electron is assumed in the ground state, i.e. it has fallen into the potential releasing a photon of 13.6eV . To separate the electron and proton from the potential, energy has to be supplied equal to 13.6eV by a photon ( since this is the electromagnetic force that binds atoms). Once supplied you would have an ion ( the proton) and the electron at rest with each other, and unless they interacted with some other particle, they would fall back to the ground state, releasing a 13.6eV photon.

If an incoming photon with higher energy than 13.6 eV hits a neutral hydrogen atom there is a quantum mechanical probability that the hydrogen atom will split to an ion and an electron carrying away positive energy and possibly a degraded photon. This is a quantum mechanical problem that has been studied as you wills see if you google "hydrogen photon scattering" and is not simple, even for such a simple atom.

The ground state energy is a level in the potential, so energy has to be supplied either to go to higher allowed states our break away from each other.

I would disagree with the statement : "if an electron has more than zero energy" because in principle it is a system of two particles that will be gaining or losing energy, the proton and the electron. One in quantum mechanics speaks of energy levels of the bound electron, but it is the system that has the energy. If more than the ground state energy is supplied to the system the two particles will separate. The electron and the proton are equally ejected from the bound state, except that the proton mass gives it small motion with the equal and opposite momentum of the electron , necessary for momentum conservation, since its mass is so much larger.


Your book's claim is correct, and both of the definitions you quote are equivalent.

We normally fix the zero level of energy by specifying that, at zero energy, the electron is no longer bound to the nucleus, and it is therefore free to escape - but it has zero remaining kinetic energy when it (asymptotically) reaches infinity; in the Kepler problem, this corresponds to a parabolic orbit. Electrons in such a state are definitely unbound, and if you add a static extraction field (as in photoelectric-effect experiments) then that field will take the electron away.

The core role of the ionization energy is as a threshold. It is defined as the difference between the ground-state energy and the zero-energy level, and thus

  • if you supply more energy than that, then the electron will be released and it will reach infinity with kinetic energy to spare, while
  • if you supply less energy than that, then the electron might reach some high-lying Rydberg state, but it will remain bound to the nucleus.

To that, your natural next question is almost certainly "and what if you supply the electron with exactly the ionization energy?", to which the answer is: you can't.

  • For one thing, there's no such thing as "exactly" in physics, and all measurements and quantities are always a bit "thick". In this case, it means that you'll be depositing energy to either side of the ionization threshold, i.e. you'll be putting some population in high-lying bound Rydberg states, and some population in low-but-positive-energy unbound continuum states.
  • In the more specific case of transitions in quantum mechanics, attempting to deliver exactly the ionization energy means that you're attempting a process with $\Delta E=0$, and by the Heisenberg Uncertainty Principle this means that the process must take $\Delta t=\infty$, i.e. it is impossible to achieve an infinitely narrow bandwidth in a finite amount of time.

That said, mathematically speaking, there is a state at zero energy, described in more detail in this MathOverflow thread - but really, I would leave the question of what happens there until you've had a proper go at the full mathematical toolkit of quantum mechanics.

As to why your book has drawn it as a dashed line - it's because the ionization threshold acts as an accumulation point for the Rydberg-state energies, which 'bunch up' as $n$ increases. Instead of drawing all of them (which is just not possible) we often draw their limit point as the dashed line in your book. How you went from a dashed line to "that level is imaginary" is a bit beyond me, though.


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