The Franck–Hertz Experiment

The Franck–Hertz experiment survives as one of the great teaching experiments — nearly every university physics course still runs the mercury or neon tube, because there is no more visceral way to see energy quantisation than a current that collapses on cue every 4.9 volts. But its logic reaches far beyond the classroom. Electron energy-loss spectroscopy (EELS), built into modern transmission electron microscopes, is the Franck–Hertz idea industrialised: fire electrons of known energy at a thin sample, measure precisely how much energy they lose, and read off the material's excitation energies — mapping chemical bonds and electronic structure atom by atom. The same principle of quantised inelastic scattering underlies how we probe plasmas, how gas-discharge lamps and fluorescent tubes work (electrons exciting mercury or noble-gas atoms, which then emit light), and how physicists measure excitation and ionisation cross-sections for everything from fusion plasmas to the upper atmosphere. Every one of these rests on the fact Franck and Hertz nailed down: an atom is a ladder, and a passing electron can climb it only one whole rung at a time.

Picture an evacuated glass tube with a little mercury sealed inside; warm it gently and a thin mercury vapour fills the space. At one end a hot filament (the cathode) boils off electrons by thermionic emission. A wire mesh grid G sits near the far end, held at a positive accelerating voltage $V$ relative to the cathode, so the electrons are pulled toward it, gaining kinetic energy as they fall through the voltage. Just beyond the grid is a collecting plate P, and here is the clever twist: the plate is held very slightly negative with respect to the grid, by a small retarding voltage of about half a volt. An electron only reaches the plate — and registers as current on the sensitive ammeter — if it arrives at the grid with at least that much energy to spare.

Now think about the journey of one electron. Pulled from the cathode, it accelerates toward the grid, and along the way it keeps bumping into mercury atoms. What happens in each bump is the entire story. If the electron is carrying less than the mercury atom's first excitation energy, it cannot excite the atom — there is no lower rung to lift the atom onto, and quantum mechanics forbids the atom from taking a fraction of a rung. So the electron simply bounces off in an elastic collision, like a ping-pong ball off a cannonball, and — as the derivation below makes numerical — it barely loses any energy at all. It keeps almost all its kinetic energy and races on. An electron can suffer thousands of these gentle elastic bounces and still arrive at the grid with essentially the full energy $eV$ that the accelerating voltage gave it. Everything changes the instant the electron has accumulated the first excitation energy of mercury, which is 4.9 eV. Now a completely new option is available: an inelastic collision in which the electron hands over exactly 4.9 eV, promoting the mercury atom from its ground state to its first excited state, and continues on with almost all of its energy gone. Where the electron happens to reach this 4.9 eV threshold decides everything about the current, because an electron that has just paid 4.9 eV to a mercury atom right in front of the grid arrives at the grid nearly at rest — and cannot climb the small retarding voltage to the plate. It drops out of the count. That single fact — that inelastic collisions are impossible until the electron carries a specific quantum of energy, and then suddenly drain it — is what turns a smooth expectation into the jagged, repeating current curve that made the experiment famous.

Slowly raise the accelerating voltage $V$ from zero and watch the plate current. The first thing an electron picks up between cathode and grid is kinetic energy equal to the charge times the voltage it has fallen through:

where: $e = 1.60\times10^{-19}\ \mathrm{C}$ is the electron's charge and $V$ is the accelerating voltage in volts. The physical reading is the tidy fact that an electron accelerated through $V$ volts arrives carrying exactly $V$ electronvolts of energy — the electronvolt is defined as the energy one electron gains crossing one volt. So "the electron has 4.9 eV" and "the electron has fallen through 4.9 V" are the same statement, and that identity is why the dips will land at whole-number multiples of 4.9 volts. At low voltage every collision is elastic, the electrons sail through to the grid with energy to spare, and the current climbs steadily as more of them are pulled across per second. But once $V$ reaches about 4.9 V, electrons arriving near the grid have just enough energy to excite a mercury atom. The condition for that inelastic channel to open is simply that the electron's kinetic energy has reached the atom's first excitation energy $E_1$:

where: $E_1$ is the energy gap between mercury's ground state and its first excited state. When this is met, electrons dump their 4.9 eV just before the grid, arrive nearly stopped, fail to overcome the retarding voltage, and the collector current drops sharply. Push $V$ higher and each electron has spare energy again after its one inelastic collision, so the current recovers — until, near 9.8 V, an electron can afford two inelastic collisions (one around the mid-point of the tube, a second just before the grid), and the current dips again. The pattern repeats: a dip every time the electrons can squeeze in one more 4.9 eV collision before the grid.

The spacing of the dips is the whole result. Reading the voltages of successive minima and subtracting, Franck and Hertz found them evenly spaced, and the spacing measured the level gap directly:

where: $n$ counts how many inelastic collisions an electron can afford before the grid. Because the electronvolt ties volts to energy one-for-one, the numerical spacing in volts, $E_1/e = 4.9$, equals the excitation energy in electronvolts, 4.9. There is nothing smooth or gradual about this curve — a wave-like or continuous transfer of energy would give a current that simply rose and levelled off. The sawtooth exists only because the atom accepts energy in one fixed lump and refuses everything smaller.

Everything so far used no light whatsoever — just electrons, a voltage, and an ammeter. That is the point: the discreteness has been measured mechanically, by collision. But the argument gets its final seal when we look at what the excited mercury atoms do next. An atom cannot stay excited; within nanoseconds it falls back to the ground state and releases the 4.9 eV it absorbed as a single photon. By the Bohr relation the photon's energy equals the level gap, and a photon of energy $E$ has frequency $f$ and wavelength $\lambda$ given by

where: $h = 6.63\times10^{-34}\ \mathrm{J\,s}$ is Planck's constant and $c = 3.00\times10^{8}\ \mathrm{m\,s^{-1}}$. Solving for the wavelength with the board shortcut $hc = 1240\ \mathrm{eV\,nm}$ gives $\lambda = 1240/4.9 = 253\ \mathrm{nm}$, an ultraviolet line, and a frequency $f = E_1/h = (4.9\times1.60\times10^{-19})/(6.63\times10^{-34}) = 1.18\times10^{15}\ \mathrm{Hz}$. When Franck and Hertz raised the voltage past 4.9 V, their tube began to glow with exactly this 253 nm ultraviolet line — and mercury's spectrum, measured years earlier by entirely optical means, has a line at 254 nm (253.7 nm). The number the electrons measured by collision and the number the light carries are the same number. The energy channel and the photon channel are two windows onto one identical rung of the ladder.

Board level — the mercury line from the dip spacing (JEE / A-Level AQA pattern)

In a Franck–Hertz tube filled with mercury vapour the plate current dips at 4.9 V, 9.8 V and 14.7 V. Take $h = 6.63\times10^{-34}\ \mathrm{J\,s}$, $c = 3.00\times10^{8}\ \mathrm{m\,s^{-1}}$, $e = 1.60\times10^{-19}\ \mathrm{C}$ and $hc = 1240\ \mathrm{eV\,nm}$. Find (a) the first excitation energy of mercury, (b) the wavelength of the light the excited atoms emit, and (c) its frequency.

Step 1. The dips are evenly spaced by 4.9 V, and an electron falling through 4.9 V gains 4.9 eV (equation 1). So the first excitation energy is $E_1 = \boxed{4.9\ \mathrm{eV}}$. The equal spacing confirms one and the same quantum is being paid off repeatedly.

Step 2. The excited atom emits a photon of this energy, so $\lambda = hc/E_1 = 1240/4.9 = \boxed{253\ \mathrm{nm}}$ — ultraviolet, matching mercury's known 254 nm resonance line.

Step 3. The frequency: first convert the energy, $E_1 = 4.9\times1.60\times10^{-19} = 7.84\times10^{-19}\ \mathrm{J}$, then $f = E_1/h = (7.84\times10^{-19})/(6.63\times10^{-34}) = \boxed{1.18\times10^{15}\ \mathrm{Hz}}$. Cross-check with $c/\lambda = (3.00\times10^{8})/(253\times10^{-9}) = 1.19\times10^{15}\ \mathrm{Hz}$, agreeing to rounding.

Competitive level — counting the collisions (JEE Main and Advanced pattern)

Electrons are accelerated from rest through a potential difference of 15.0 V in mercury vapour whose first excitation energy is 4.9 eV. Assuming an electron makes as many inelastic collisions as its energy allows, find (a) the maximum number of mercury atoms one electron can excite, (b) the kinetic energy it retains on reaching the grid, and (c) the wavelength of the photons the excited atoms emit.

Step 1. An electron arrives at the grid carrying $eV = 15.0\ \mathrm{eV}$. Each inelastic collision costs 4.9 eV, so the number of collisions it can afford is the whole-number part of $15.0/4.9 = 3.06$, giving $\boxed{3}$ excitations. It cannot manage a fourth because $4\times4.9 = 19.6\ \mathrm{eV} > 15.0\ \mathrm{eV}$.

Step 2. After three inelastic collisions the electron has spent $3\times4.9 = 14.7\ \mathrm{eV}$, leaving $15.0 - 14.7 = \boxed{0.3\ \mathrm{eV}}$ of kinetic energy — barely moving, which is why the current is deep in a dip near 14.7 V.

Step 3. Each excited mercury atom decays by emitting a photon of $\lambda = 1240/4.9 = \boxed{253\ \mathrm{nm}}$, the same ultraviolet line regardless of how many collisions occurred — every excitation is to the same first level, so every emitted photon has the identical energy. Timing note: about 60 seconds. The only real step is the division $15.0/4.9 = 3.06$ and taking its whole part; parts (b) and (c) are then a subtraction and the single stored fact $1240/4.9 = 253\ \mathrm{nm}$. In an exam, resist the urge to round $3.06$ up — the electron cannot pay for a collision it cannot afford.

Competitive level — the method reads any atom (JEE Advanced pattern)

A Franck–Hertz experiment is run with a hypothetical vapour whose plate-current dips are spaced 2.0 V apart. Find (a) the first excitation energy of this atom and (b) the wavelength of the light it would emit on de-excitation. Compare with mercury and comment on where the emission lands in the spectrum.

Step 1. The dip spacing in volts equals the first excitation energy in electronvolts (equation 3), so $E_1' = \boxed{2.0\ \mathrm{eV}}$ — less than half of mercury's 4.9 eV, so this atom's first rung sits much lower.

Step 2. The emitted wavelength is $\lambda' = hc/E_1' = 1240/2.0 = \boxed{620\ \mathrm{nm}}$. Because a smaller energy gap means a longer wavelength, this atom glows in the visible — an orange-red line — rather than mercury's invisible 253 nm ultraviolet. This is exactly why the classroom neon version of the experiment shows a visible orange glow: neon's relevant level gaps are small enough to emit visible light, while mercury's larger 4.9 eV gap pushes its line into the ultraviolet. Timing note: about 40 seconds. Both parts are one line each once you see that "dip spacing in volts = excitation energy in eV" and "$\lambda = 1240/E$". The comment — smaller gap gives longer wavelength — is worth stating explicitly because graders reward the reasoning, not just the number.

Calling 4.9 eV an ionisation energy. The first dip measures the first excitation energy, not ionisation (mercury ionises at about 10.4 eV). Franck and Hertz made this very slip in 1914; do not repeat it. If asked "what does the 4.9 V spacing represent?", the answer is the energy gap to the first excited state.

Rounding the collision count up. With 15 V and a 4.9 eV cost, $15/4.9 = 3.06$, so the electron excites 3 atoms, not 4 — you take the floor, because it cannot pay for a fourth collision it cannot afford. Every year students round to the nearest integer and lose the mark.

Confusing volts and electronvolts. The dip spacing is 4.9 volts; the energy gap is 4.9 electronvolts. They are numerically equal only because the electronvolt is defined that way. Keep the units straight: an electron through $V$ volts has $V$ eV, which is $V\times1.6\times10^{-19}$ joules.

Wavelength–energy inequality flipped. A larger energy gap gives a shorter wavelength ($\lambda = hc/E$). Mercury's big 4.9 eV gap gives short-wave ultraviolet (253 nm); a small gap gives long-wave visible light. Students routinely pair "high energy" with "long wavelength" and lose easy marks.

Forgetting the elastic collisions matter. The reason the curve rises at all between dips — and the reason an electron can accumulate energy over many collisions — is that below-threshold collisions are elastic and drain almost nothing (about $10^{-5}$ per hit). A common conceptual error is to think every collision costs energy; only inelastic ones do, and only above threshold.

Board scope. The Franck–Hertz experiment is examinable on CBSE (qualitatively, as confirmation of Bohr's discrete levels), JEE Main and JEE Advanced, IB HL (Theme E — atomic energy levels and spectra, with the quantised-level maths HL-only and the Franck–Hertz apparatus illustrative rather than a named guide item; Standard Level does not carry the Bohr maths), and some A-Level specifications — notably AQA, which examines electron-collision excitation and the electron-volt, with the Franck–Hertz apparatus itself in AQA's optional Turning Points unit (it is not on CIE 9702). It is not part of the core NEET syllabus and is not on AP Physics 2 (which carries atomic energy levels and line spectra but not this apparatus), nor on ICSE or IGCSE at this level. Do not cite it as required for a board that does not list it.

Texts. NCERT Physics Class 12, Chapter 12 ("Atoms"), for Bohr's stationary states and energy levels · H. C. Verma, Concepts of Physics Vol. 2, the "Bohr's Model and Physics of the Atom" chapter · Eisberg and Resnick, Quantum Physics of Atoms, Molecules, Solids, Nuclei and Particles, for the excitation and ionisation cross-section treatment · Krane, Modern Physics, for the electron-collision picture.

Provenance. Locked anchor values for mercury. Recompute: first excitation $E_1 = 4.9\ \mathrm{eV} = 7.84\times10^{-19}\ \mathrm{J}$; emitted wavelength $\lambda = 1240/4.9 = 253\ \mathrm{nm}$ (SI cross-check $hc/E_1 = 254\ \mathrm{nm}$); frequency $f = E_1/h = 1.18\times10^{15}\ \mathrm{Hz}$, with $f\lambda = 2.99\times10^{8}\ \mathrm{m\,s^{-1}}$; dip voltages $V_n = n\times4.9 = 4.9, 9.8, 14.7\ \mathrm{V}$; collision count at 15 V $= \lfloor 15.0/4.9\rfloor = 3$, energy left $= 15.0 - 14.7 = 0.3\ \mathrm{eV}$; hypothetical 2.0 eV atom $\lambda = 1240/2.0 = 620\ \mathrm{nm}$; elastic energy-loss fraction $4m/M = 1.09\times10^{-5}$ with $M/m = 3.66\times10^{5}$; electron speed at 4.9 eV $v = \sqrt{2E_1/m} = 1.31\times10^{6}\ \mathrm{m\,s^{-1}}$.

Prerequisites. Energy Levels & the Hydrogen Spectrum — where the ladder of quantised levels and the relation $\Delta E = hf$ are established; the Franck–Hertz experiment is the collision-based proof that those levels are physically real. Leads to. For enrichment, the university seam The Franck–Hertz Experiment (go-deeper only, never an exam dependency here), which treats the excitation and ionisation cross-sections and the full transport picture inside the tube.