Energy levels of the hydrogen atom

Knowing the radius and velocity of each orbit, we can calculate the most important quantity: the energy of the electron at each level. It is the sum of the kinetic energy (due to motion) and the electric potential energy (due to the attraction with the nucleus):

$E_\text{tot} = \frac{1}{2}m v^2 - \frac{k e^2}{r}$

The minus sign of the potential energy is a reminder that the electron is bound: energy must be supplied to detach it from the nucleus. Substituting the quantised formulas for $v_n$ and $r_n$ and simplifying gives the celebrated levels formula:

$\ev{E_n = -\frac{1}{2}\,\frac{k\,e^2}{r_1}\cdot\frac{1}{n^2} = -\frac{13{,}6\ \text{eV}}{n^2}}$

The energies are negative (bound states) and discrete: only certain values are allowed, one for each integer $n$. The lowest level, $n=1$, equals $-13{,}6$ eV and is the ground state; going up in level the energy increases (becomes less negative) as $-1/n^2$, and the levels crowd together as they approach $0$. The value $E=0$ corresponds to $n=\infty$: the free electron, just torn away from the atom. Ionising hydrogen from the ground state therefore costs exactly $13{,}6$ eV, its ionisation energy.

The first levels:

Diagram of the hydrogen energy levels. The levels crowd together towards $E=0$ (ionisation, $n=\infty$). A transition to a lower level (here $n=3\to n=2$) releases a photon whose energy equals the difference between the two levels.

When an electron jumps from one level to another, the energy difference is emitted or absorbed as a single photon, $E_\text{photon} = |E_\text{fin} - E_\text{init}|$. Since the levels are discrete, the photon energies are discrete too: this is why every element has a characteristic “line” spectrum. (The explicit calculation of one of these lines is worked out in the exercises.)

Selective absorption rule

If a photon strikes the electron of an atom, the electron can jump level only if the photon has energy exactly equal to the difference between the final and initial levels. Otherwise the photon is not absorbed and continues on undisturbed, as if the atom were transparent to its light. This is why every gas absorbs and emits only at very precise frequencies.

This is the radical difference from classical physics: in the classical world it would be enough for the energy to be “at least sufficient”; in the quantum world it must be exactly the right one. Absorption is “all or nothing”.

A 3 eV or 12 eV photon: does it pass or not?

A hydrogen atom in the ground state ($n=1$, $E_1 = -13{,}6$ eV) is struck by a $3$ eV photon. Does anything happen? No: the first available transition, $n=1 \to n=2$, requires exactly $\Delta E = 13{,}6 - 3{,}4 = 10{,}2$ eV. With $3$ eV it doesn’t get there, and the photon passes undisturbed. What about a $12$ eV photon? This isn’t absorbed either: $12 \neq 10{,}2$ (the $n=1\to2$ transition), $12 \neq 12{,}09$ (the $n=1\to3$ transition), $12 \neq 12{,}75$ (the $n=1\to4$ transition)… no transition has exactly a $12$ eV jump. The atom remains transparent even to a photon more energetic than the first jump but “off tune”.

It is precisely this peculiarity — the atom absorbs energy only in precise “steps” — that the Franck-Hertz experiment managed to measure directly, as we shall see.

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Topics: Quantum physics Concepts: Bohr model · Photon · Kinetic energy Objects: Hydrogen atom

Related exercises: Hydrogen levels, Balmer and Lyman · Worked exercise — a red photon from the hydrogen atom · True or false on quantum physics