The Franck-Hertz Experiment

In 1914, a year before anyone had a working theory of the atom, James Franck and Gustav Hertz built a simple vacuum tube that measured something remarkable: electrons passing through mercury vapour could only lose energy in discrete lumps of about 4.9 electron-volts, never less. Below that threshold an electron might bounce off a mercury atom a thousand times and emerge with all its kinetic energy intact; the moment it carried 4.9 eV it could suddenly hand that exact amount over and stop cold. This is the first direct, non-spectroscopic proof that atoms have quantized internal energy levels. The experiment reads the quantization off a single graph. Electrons are accelerated through mercury vapour by a voltage $V$ and collected on the far side; the collector current $I$ rises smoothly with $V$ until $V\approx4.9$ V, drops sharply, rises again, and drops again at $9.8$ V, $14.7$ V, and so on — a sawtooth whose teeth are spaced by exactly one excitation energy. Each dip is the voltage at which electrons have just enough energy to excite a mercury atom inelastically, dumping their energy and failing to reach the collector. The clincher is that the very same $4.9$ eV shows up a second, completely independent way: mercury vapour excited to that level relaxes by emitting an ultraviolet photon at $253.7$ nm, and $hc/\lambda$ for that line is $4.89$ eV. The energy an electron must deliver to climb the level equals the energy a photon carries when the atom falls back down. This article derives the threshold condition from energy–momentum conservation, shows why the dip spacing is a pure property of the atom, closes the loop between the $4.9$ eV threshold and the $254$ nm emission line numerically, and separates the physics from the famous mistake Franck and Hertz made about what their number meant.

Key takeaways

Electrons accelerated through mercury vapour lose energy only in discrete $\approx4.9$ eV steps; the collector current $I(V)$ shows periodic dips spaced by $4.9$ V.
Below threshold, electron–atom collisions are elastic (fractional energy loss $\sim4m_e/M\approx10^{-5}$ , negligible); at $4.9$ eV they become inelastic, exciting the mercury $6\,^1S_0\to6\,^3P_1$ transition.
Successive dips at $4.9,\,9.8,\,14.7\,\mathrm V$ are electrons suffering one, two, three inelastic collisions in a longer tube — the spacing is one excitation energy, offset-independent.
The same $4.9$ eV appears as the $253.7$ nm ( $2537\,\text{Å}$ ) ultraviolet emission line: $hc/\lambda=4.89$ eV, agreeing to $0.3\%$ — an electron-impact measurement and a photon measurement of the same atomic level.
A classical atom with continuously variable internal energy would show no dips; the sawtooth is the signature of quantized levels.
Key idea: an electron can only excite an atom when its kinetic energy reaches a fixed threshold set by the atom's discrete level spacing — so the current-vs-voltage curve reads atomic quantization directly, without ever looking at light.

The core insight

An electron colliding with an atom can transfer energy only in the exact amounts that match a jump between the atom's discrete internal levels. The lowest such jump in mercury is $4.9$ eV, so the collector current in a Franck–Hertz tube drops precisely each time the accelerating voltage lets electrons reach a multiple of $4.9$ V — a mechanical measurement of energy quantization.

Where this lives today

The Franck–Hertz experiment remains a standard undergraduate teaching lab precisely because it makes quantization tangible without spectroscopy — a knob, a meter, and a sawtooth. The neon variant, with dips near $18.7$ V and glowing orange rings of excited atoms visible through the tube, is a favourite because the excitation is literally visible. Conceptually, the experiment is the electron-impact counterpart of the photoelectric effect: the photoelectric effect shows that light delivers energy in quanta, while Franck–Hertz shows that atoms accept energy in quanta. Together they pinned down the two halves of the light–matter energy exchange years before the Schrödinger equation existed. The modern descendants of this idea — electron energy-loss spectroscopy (EELS) in electron microscopes, and inelastic electron tunnelling spectroscopy — measure excitation spectra of solids and molecules by exactly the same principle: scan the electron energy and watch for the thresholds where inelastic channels switch on.

The physical picture

Picture a heated tube filled with low-pressure mercury vapour. A hot cathode boils off electrons; a positive grid at voltage $V$ accelerates them across the tube, and a collector plate just behind the grid — held at a small retarding voltage — catches whatever electrons still have enough forward energy to climb the last little hill. As $V$ increases, more energetic electrons arrive and the current $I$ rises. But mercury atoms are in the way, and what happens in those collisions is the whole story. At low $V$ , an electron carries too little energy to change the mercury atom's internal state. All it can do is scatter elastically, like a ping-pong ball off a bowling ball: it changes direction but keeps essentially all its kinetic energy, because the fractional energy a light electron can give a heavy atom in an elastic hit is only about $4m_e/M_{\mathrm{Hg}}\approx1.1\times10^{-5}$ . So below threshold the vapour is nearly transparent to energy, and $I$ climbs with $V$ .

Figure

The Franck–Hertz current-vs-voltage curve: periodic dips spaced by one excitation energy

The moment an electron's kinetic energy reaches $4.9$ eV, a new channel opens. It can now scatter inelastically: hand over exactly $4.9$ eV to promote a mercury atom from its ground state $6\,^1S_0$ to the excited $6\,^3P_1$ state, and continue as a slow electron. Having dumped its energy right in front of the grid, it can no longer overcome the retarding voltage, so it never reaches the collector — and the current drops. Raise $V$ further and electrons regain energy after the collision, so $I$ recovers; but at $9.8$ V an electron has enough to excite two atoms in succession, and the current dips again. The dips march at a fixed $4.9$ V spacing.

Figure

Elastic below threshold, inelastic at threshold: the two collision channels

The final piece of evidence is optical. A mercury atom lifted to $6\,^3P_1$ falls back to the ground state by emitting an ultraviolet photon. Franck and Hertz saw exactly one line appear the instant $V$ crossed $4.9$ V: the $253.7$ nm line of mercury. Its photon energy $hc/\lambda=4.89$ eV matches the electron-impact threshold. The atom's internal energy is quantized, and the same level spacing governs how it takes energy from an electron and how it gives energy back to a photon.

The core equations

The threshold is a statement about energy conservation in an inelastic collision. An electron of kinetic energy $K$ hits an atom in its ground state; the atom can absorb energy only in the fixed amount $\Delta E$ that separates two of its levels. The electron emerges with the remainder.

K \;\longrightarrow\; K' + \Delta E, \qquad K' \ge 0 \quad\Rightarrow\quad K \ge \Delta E \qquad (1)

where: $K=eV$ is the kinetic energy an electron gains falling through accelerating voltage $V$ ; $\Delta E$ is the excitation energy of the atom (the gap $6\,^1S_0\to6\,^3P_1$ in mercury, $\Delta E=4.9$ eV); and $K'$ is the electron's leftover kinetic energy. Inequality (1) is the whole threshold: no excitation is possible until $eV$ reaches $\Delta E$ , so the first current dip sits at $V_1=\Delta E/e=4.9$ V. If the tube is long enough, an electron re-accelerated after one inelastic hit can reach $\Delta E$ again and excite a second atom. The $n$ -th dip is therefore

V_n = n\,\frac{\Delta E}{e} + V_0, \qquad V_{n+1}-V_n = \frac{\Delta E}{e} \qquad (2)

where: $n=1,2,3,\dots$ counts the inelastic collisions, $V_0$ is a fixed offset from contact potentials and the retarding voltage, and the spacing $\Delta E/e=4.9$ V is what carries the physics. The offset $V_0$ shifts every dip equally and drops out of the difference, so the spacing is a clean atomic property independent of tube construction. The excited atom returns to the ground state by emitting a photon whose energy is fixed by the same level gap — the Bohr frequency condition.

\Delta E = h\nu = \frac{hc}{\lambda} \qquad (3)

where: $\nu$ and $\lambda$ are the frequency and wavelength of the emitted light, $h$ is Planck's constant, and $c$ is the speed of light. Setting $\Delta E=4.9$ eV gives $\lambda = hc/\Delta E = (1239.84\ \text{eV}\,\cdot\,\text{nm})/(4.9\ \text{eV}) = 253\ \text{nm}$ , the observed $253.7$ nm mercury line. Equations (1) and (3) tie the electron-impact threshold and the photon wavelength to the one number $\Delta E$ . Finally, the reason the collision is either "harmless" or "all-or-nothing" is the mass mismatch. In an elastic head-on collision, the maximum fractional kinetic energy a projectile of mass $m$ transfers to a target of mass $M$ is

\left(\frac{\Delta K}{K}\right)_{\max} = \frac{4mM}{(m+M)^2} \;\approx\; \frac{4m_e}{M_{\mathrm{Hg}}} \qquad (4)

where: $m_e$ is the electron mass and $M_{\mathrm{Hg}}=200.6\,u$ is the mercury atom mass; the approximation uses $m_e\ll M_{\mathrm{Hg}}$ . Numerically this is $1.1\times10^{-5}$ : an elastic collision drains at most a hundred-thousandth of the electron's energy, which is why the vapour is effectively transparent to energy until the inelastic channel opens.

Derivation AWhy the threshold is exactly the atomic level gap

The goal here is to show that the first dip voltage equals the atomic excitation energy divided by $e$ — no more, no less — and to make explicit that the small recoil the mercury atom takes does not shift the threshold in any measurable way. We treat the electron–atom encounter as a one-dimensional collision governed by conservation of energy and momentum, keep the atom's recoil kinetic energy honestly, and then estimate its size. The subtlety worth watching is that "the electron must supply $\Delta E$ " is only exactly true in the limit that the atom is infinitely heavy; the finite-mass correction is what we bound.

Step 1. Write conservation laws for a head-on inelastic collision. Before: electron momentum $p$ , kinetic energy $K=p^2/2m_e$ ; atom at rest. After: electron momentum $p'$ , atom momentum $P$ , and the atom now carries internal energy $\Delta E$ . Momentum: $p=p'+P$ . Energy: $p^2/2m_e = p'^2/2m_e + P^2/2M + \Delta E$ .

Step 2. Find the minimum $K$ that still conserves both. Excitation is just possible when the electron and atom move together afterward (no leftover relative motion), so $p'/m_e = P/M$ . Solving the two conservation laws at that boundary gives the exact threshold $K_{\mathrm{th}} = \Delta E\,\bigl(1 + m_e/M\bigr)$ . Why the "move together" condition: any relative velocity after the collision is extra kinetic energy that had to come from $K$ , so the least-energy way to deposit $\Delta E$ leaves zero relative velocity — the analogue of a perfectly inelastic mechanical collision.

Step 3. Evaluate the correction factor. With $m_e/M_{\mathrm{Hg}} = 9.1\times10^{-31}/(200.6\times1.66\times10^{-27}) = 2.7\times10^{-6}$ , the threshold exceeds $\Delta E$ by fewer than three parts per million. So to the $\sim1\%$ precision of the tube, $K_{\mathrm{th}}=\Delta E$ and $V_1=\Delta E/e=4.9$ V.

Step 4. Cross-check with the recoil energy directly. The recoiling atom carries $P^2/2M$ ; using the electron momentum at $4.9$ eV, $P^2/2M = 1.3\times10^{-5}$ eV, again utterly negligible against $4.9$ eV. The threshold is the atomic gap, and the finite atomic mass is invisible at this resolution. Both quantities — the elastic fractional loss $4m_e/M_{\mathrm{Hg}}=1.1\times10^{-5}$ and the recoil energy $1.3\times10^{-5}$ eV — are far below $10^{-3}$ eV, confirming $K_{\mathrm{th}}=\Delta E$ to well under $1\%$ .

Go deeper · Derivation B · 1 minWhy the dip spacing is a pure atomic property

Derivation BWhy the dip spacing is a pure atomic property

The point of this derivation is to prove that although the absolute position of the first dip depends on messy tube-specific details — contact potentials between dissimilar metals, the retarding voltage, the cathode work function — the spacing between consecutive dips is immune to all of them and equals the atomic excitation energy alone. This is what makes the experiment a clean measurement of a number that belongs to the mercury atom rather than to the apparatus. The reasoning is to write the dip condition with an explicit unknown offset and show it cancels in the difference.

Step 1. Let $V_0$ collect every apparatus-dependent offset: contact potential differences, retarding bias, and the geometric point in the tube where an electron has just reached the grid region with usable energy. The physical requirement for the $n$ -th dip is that the electron accumulates enough energy to make its $n$ -th inelastic collision right at the grid, i.e. $e(V_n - V_0) = n\,\Delta E$ .

Step 2. Solve for the dip voltages: $V_n = V_0 + n\,\Delta E/e$ . Every dip carries the same unknown $V_0$ .

Step 3. Take the difference of consecutive dips: $V_{n+1}-V_n = \Delta E/e$ . The offset $V_0$ has vanished. The spacing is set only by $\Delta E$ , so measuring the spacing — not the first dip's position — yields the atomic level gap free of the additive voltage offset. Why this matters experimentally: the raw first-dip voltage in a real mercury tube is often near $6$ V, not $4.9$ V, precisely because $V_0\approx1$ V of contact potential shifts it; the spacing, however, comes out at $4.9$ V and that is the quotable result.

Step 4. Confirm the offset-independence numerically for two different offsets. With $V_0=0.5$ V the dips fall at $5.4,10.3,15.2,20.1$ V; with $V_0=1.5$ V they fall at $6.4,11.3,16.2,21.1$ V. In both cases, the successive differences are $4.9,4.9,4.9$ V. Dip ladders for retarding offsets $0.5$ V and $1.5$ V both yield a spacing of $4.9$ V, confirming that the spacing is independent of the offset.

Limiting cases

Each limit is a sanity check on the threshold picture.

Case	Result	Meaning
Infinitely heavy atom ( $m_e/M\to0$ )	$K_{\mathrm{th}}\to\Delta E$ exactly	The threshold is purely the level gap; atomic recoil is a $10^{-6}$ correction.
Classical continuous levels	No dips; $I(V)$ rises monotonically	An atom that could absorb any tiny amount would drain electrons at every $V$ ; the sawtooth requires discrete levels.
Very short tube (one gap)	Single dip at $V_1$ , no repeats	Electrons never regain $\Delta E$ after their first inelastic collision, so only $n=1$ appears.
Neon instead of mercury	Dips spaced $\sim18.7$ V, visible orange glow	A larger first-accessible excitation energy and a decay path that emits visible light near $640$ nm.
$V$ far below threshold	Purely elastic, energy loss $\sim10^{-5}$	The vapour is transparent to energy; current tracks the accelerating voltage.

Worked examples

Example 1 — Close the loop: does the $4.9$ eV threshold predict the observed emission line? The current dips are spaced by $4.9$ V. Predict the wavelength of the light that the excited mercury atoms should emit, and compare with the line actually seen.

Step 1. The excitation energy inferred from the dip spacing is $\Delta E=4.9$ eV.

Step 2. Apply the Bohr frequency condition $\lambda = hc/\Delta E$ . Using $hc=1239.84$ eV·nm, $\lambda = 1239.84/4.9 = 253.0$ nm.

Step 3. Compare with the observed mercury line at $253.7$ nm ( $2537\,\text{Å}$ ). The photon energy of the observed line is $1239.84/253.7 = 4.887$ eV, differing from $4.9$ eV by $0.26\%$ . The electron-impact measurement and the photon measurement agree to a fraction of a percent, so the same $4.9$ eV atomic gap governs both — the boundary number that certifies quantization from two independent directions.

Example 2 — How many dips fit in a tube run to $20$ V? A mercury Franck–Hertz tube is swept from $0$ to $20$ V with a negligible contact-potential offset. How many current dips appear, and at what voltages?

Step 1. Dips occur at $V_n = n\times4.9$ V. Compute successive values: $4.9,\,9.8,\,14.7,\,19.6$ V.

Step 2. The next dip would be at $24.5$ V, beyond the $20$ V sweep, so it is not reached.

Step 3. Count: four dips, at $4.9,\,9.8,\,14.7,\,19.6$ V. The number of dips is $\lfloor V_{\max}/4.9\rfloor = \lfloor 20/4.9\rfloor = 4$ . So a $20$ V sweep shows four evenly spaced dips, each an electron completing one more inelastic collision than the last.

Origins · 1 minjames franck and how this was discovered

james franck

1882–1964 · German-American physicist who proved atomic energy levels are quantized by electron impact

James Franck, working with Gustav Hertz at the University of Berlin, built the electron-impact tube that bears their names in 1914. The two initially misread their $4.9$ eV threshold as mercury's ionization energy; it was Niels Bohr who identified it as the first atomic excitation energy, matching his 1913 model. Franck and Hertz shared the 1925 Nobel Prize in Physics "for their discovery of the laws governing the impact of an electron upon an atom." Franck later left Nazi Germany, contributed to the Manhattan Project, and authored the 1945 Franck Report, urging a demonstration rather than military use of the atomic bomb. The Franck–Condon principle in molecular spectroscopy is also his.

History

1913

— Bohr proposes discrete atomic energy levels in his model of hydrogen

1914

— Franck and Hertz report the

4.9

eV mercury threshold (Verh. Dtsch. Phys. Ges. 16, 457), reading it as an ionization potential

1915

— Bohr reinterprets the

4.9

eV as the first excitation energy, tying it to the

253.7

nm line

1920

— Refined tubes resolve multiple higher mercury excitation thresholds

1925

— Franck and Hertz receive the Nobel Prize in Physics

Going further

References. Franck & Hertz, Verh. Dtsch. Phys. Ges. 16, 457 (1914) · Bohr, Phil. Mag. 30, 394 (1915) · Eisberg & Resnick, Quantum Physics, 2nd ed., §4.6 · Griffiths & Schroeter, Introduction to Quantum Mechanics, Ch. 1 (historical) · Melissinos & Napolitano, Experiments in Modern Physics, 2nd ed., Ch. 1. Key numerical checks: (1) $\lambda=hc/\Delta E=253$ nm from $\Delta E=4.9$ eV, with the observed $253.7$ nm line giving $4.887$ eV, agreeing to $0.26\%$ ; (2) the dip ladder $4.9,9.8,14.7,19.6$ V with constant $4.9$ V spacing; (3) the spacing remains $4.9$ V for retarding offsets of both $0.5$ V and $1.5$ V; (4) elastic fractional loss $4m_e/M_{\mathrm{Hg}}=1.1\times10^{-5}$ and Hg recoil energy $1.3\times10^{-5}$ eV; (5) the neon threshold $18.7$ eV and its $\approx1.9$ eV visible glow. Mercury ionization energy $10.4$ eV and the $6\,^1S_0\to6\,^3P_1$ assignment are standard atomic data (NIST).

Builds on: Photoelectric Effect and the Photon · Blackbody Radiation and the Planck Quantum · Bohr's model of atomic energy levels. Leads to: Stationary States and Separation of Variables · Time-Independent Schrödinger Equation · atomic fine structure and selection rules.

Is the

4.9

eV the ionization energy of mercury?

No — and this is the classic trap. The $4.9$ eV is the excitation energy from the ground state $6\,^1S_0$ to the first accessible excited state $6\,^3P_1$ . Mercury's actual ionization energy is $10.4$ eV, more than twice as large. Franck and Hertz themselves made this mistake in 1914; Bohr corrected it in 1915 by pointing out that the accompanying $253.7$ nm emission line is a bound-to-bound transition, which is only possible if the electron lands in an excited bound state, not the ionization continuum. If you run the tube to higher voltage you eventually do see ionization set in near $10.4$ V, but that is a separate feature.

Why does the current recover after each dip instead of staying low?

Because after an electron dumps its $4.9$ eV right at the grid, raising the accelerating voltage further gives it new energy on the remaining path. A little past $4.9$ V it again has enough forward energy to overcome the retarding voltage and reach the collector, so the current climbs back up — until the voltage is large enough for a second inelastic collision at $9.8$ V, when it dips once more. The curve is a sawtooth because each cycle is "gain energy, lose $4.9$ eV, gain energy again."

What sets the spacing between successive dips, physically?

One excitation energy, $\Delta E/e = 4.9$ V. The $n$ -th dip is the voltage at which an electron can just manage its $n$ -th inelastic collision before reaching the grid. Since each collision costs exactly $\Delta E$ , consecutive dips are separated by exactly $\Delta E/e$ . Contact potentials and the retarding voltage shift where the whole ladder sits but add the same offset to every rung, so they cancel out of the spacing. That is why the spacing, not the absolute first-dip voltage, is the clean measurement.

How does this relate to the photoelectric effect?

They are complementary halves of quantized light–matter energy exchange. The photoelectric effect shows that light comes in energy quanta $h\nu$ : a photon either has enough energy to eject an electron or it does not, and no accumulation helps. Franck–Hertz shows that atoms accept energy in quanta: an electron either has enough kinetic energy to bridge an atomic level gap or it does not, and elastic nibbling does not accumulate. One quantizes the field, the other quantizes the atom's internal levels; both point at the same discrete-energy structure of nature.

Why use mercury vapour specifically?

Mercury is convenient: it is liquid at room temperature and has a usefully high vapour pressure when the tube is gently warmed, giving a dense enough gas that electrons collide many times across the tube. Its first excitation energy, $4.9$ eV, sits at an accelerating voltage that ordinary lab supplies reach easily, and it emits a strong, easily identified ultraviolet line at $253.7$ nm. Neon is the other common choice; its higher threshold near $18.7$ eV and visible orange emission make the excited regions glow, which is spectacular for teaching even though the numbers are less tidy.

Would a classical atom show any dips at all?

No, and that is the deepest point of the experiment. If an atom could absorb energy continuously — as a classical charged oscillator would — then an electron of any energy could transfer a little to the atom on every collision. The current would just rise smoothly (or fall smoothly from collisional drag) with no special voltages, no sawtooth. The periodic dips exist only because the atom refuses to accept energy except in the fixed lump $\Delta E$ . The structure of the graph is a direct fingerprint of energy quantization.

Does the neon "glow ring" pattern mean anything quantitative?

Yes. In a neon Franck–Hertz tube you can see distinct luminous bands where electrons have just reached the excitation energy and are exciting neon atoms that then radiate visible orange light. The bands are spaced along the tube by the distance an electron needs to re-accelerate up to the neon threshold, so the number of glowing rings equals the number of inelastic collisions — the spatial analogue of the current dips. Counting rings and knowing the field gives another estimate of the neon excitation energy.