The Franck-Hertz experiment

The levels formula and the selective absorption rule are elegant, but how can one directly measure that atoms only accept energy in “steps”? The answer came in 1914 from James Franck and Gustav Hertz, with a seemingly trivial experiment — a vacuum tube filled with mercury vapour — that produced the first direct, non-optical proof of the quantisation of atomic levels. The two won the Nobel Prize in 1925.

The apparatus. Inside a glass tube, in sequence, there are:

• a heated cathode, which emits electrons via the thermionic effect;
• a grid (accelerating anode) near the cathode, held at a variable voltage $V$ relative to the cathode;
• a collector (plate) beyond the grid, held at a small retarding voltage $-V_0$ (about $0{,}5$ V) relative to the grid;
• mercury vapour at low pressure, filling the whole tube.

The electrons emitted by the cathode are accelerated by the voltage $V$, cross the mercury vapour, pass the grid and reach the collector — but to get there they must retain enough kinetic energy to overcome the retarding barrier $V_0$. The current $i$ measured at the collector is therefore an indicator of the electrons’ kinetic energy at the grid.

Diagram of the Franck-Hertz apparatus: the cathode emits electrons, the voltage $V$ accelerates them towards the grid through the mercury vapour, and only those still energetic enough overcome the retarding barrier $V_0$ and reach the collector, where an ammeter measures the current $i$.

The key observation. Slowly increasing the accelerating voltage $V$, the current $i$ does not grow monotonically. It grows, then collapses sharply at $V \approx 4{,}9$ V, then grows again, then collapses again at $V \approx 9{,}8$ V, then again at $V \approx 14{,}7$ V, and so on. Maxima and minima follow one another with a regular spacing of $4{,}9$ V.

Current $i$ at the collector as a function of the accelerating voltage $V$: a sequence of regular maxima and collapses, spaced by $4{,}9$ V. Each collapse signals that the electrons have just reached the energy needed to excite a mercury atom.

Interpretation. As long as the electron’s kinetic energy is less than $4{,}9$ eV, the Hg atoms cannot be excited (their first excited level sits $4{,}9$ eV above the ground state): electron-atom collisions are elastic, the electrons cross the vapour losing almost no energy, and the current grows normally. However, when $V$ exceeds $4{,}9$ V, every electron that has reached exactly that energy can give it all up in one go to knock an Hg atom up to its first excited level (inelastic collision): it loses $4{,}9$ eV and is left almost at rest. It can then no longer overcome the $V_0$ barrier and the current drops. As $V$ is increased further, the electrons re-accelerate after the collision and reach the collector again; but at $V = 2\cdot 4{,}9 = 9{,}8$ V they can undergo two inelastic collisions, at $14{,}7$ V three, and so on — hence the periodic collapses.

Principle — The meaning of the result

Atoms cannot absorb an arbitrary amount of energy: they accept it only in predetermined “steps”, of $4{,}9$ eV in the case of mercury. This is the quantisation of energy levels measured directly, without looking at the emitted light and without spectroscopy: a simple oscillating ammeter is already proof that nature, at that scale, counts in discrete steps.

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Topics: Quantum physics Concepts: Bohr model · Kinetic energy · Elastic collision · Inelastic collision

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