Franck–Hertz Experiment

Fire electrons through mercury or neon vapour, read the collector current as you raise the voltage, and discover for yourself that atoms accept energy only in fixed packets — the first direct proof of Bohr's energy levels (1914).

Prepared by Mr. Manoj Devkota
Kamala Science Campus, Dhungrebas, Sindhuli

Do the experiment

JAMES FRANCK & GUSTAV HERTZ · 1914 · NOBEL PRIZE 1925

Slowly raise the accelerating voltage U₁ (or press “Auto sweep”) and watch the ammeter. Every reading is plotted on the graph. Where does the current suddenly fall, and why?

Collector current I_A0.00 nA

Live readings

In simple words
Basic principle

An atom cannot take any amount of energy. A mercury atom ignores an electron that brings less than 4.9 eV — the electron just bounces off. If the electron brings 4.9 eV or more, the atom takes exactly 4.9 eV and no other amount.

How it works

Electrons from a hot cathode are speeded up by U₁ through mercury vapour. A small reverse voltage in front of the collector stops slow electrons. Whenever U₁ is just enough for an electron to give away one more 4.9 eV packet before the grid, it arrives too slow to pass the reverse voltage, and the current dips. The dips repeat every 4.9 V.

Conclusion

Atomic energy levels are discrete, as Bohr proposed. The excited atoms give back the same 4.9 eV as ultraviolet light of 253.7 nm (E = hc/λ). The experiment measures an atomic energy level with a voltmeter and an ammeter.

How the experiment works

  1. A heated cathode K emits electrons; the voltage U₁ between K and the grid G accelerates them through the vapour.
  2. Elastic collisions (energy below 4.9 eV): the electron changes direction but loses almost nothing, only about \(2m_e/M\approx5\times10^{-6}\) of its energy.
  3. Inelastic collisions (energy ≥ 4.9 eV): the electron gives exactly \(E_2-E_1=4.9\) eV to the atom and starts again almost from rest.
  4. Between G and the collector A a small retarding voltage U_r turns back electrons with less than eU_r of energy, so only fast electrons are counted.
  5. Dips in I_A appear whenever the electrons have just lost one more packet near the grid:
\[ e\,\Delta U = E_2-E_1 = h\nu = \frac{hc}{\lambda}\quad\Rightarrow\quad \lambda=\frac{hc}{e\,\Delta U}=\frac{1239.8\ \text{nm·V}}{\Delta U} \]

Hg: ΔU ≈ 4.9 V → λ ≈ 253.7 nm (UV), seen in the light emitted by the tube. Ne: ΔU ≈ 18.7 V; the light (585–640 nm) is visible orange.

Try this

  • Measure: keep the theoretical curve hidden. Sweep U₁ from 0 to 30 V and watch the graph build up. Count the maxima.
  • Analyse: read the voltages of successive maxima (or minima), find the mean spacing ΔU and compute \(\lambda=hc/e\Delta U\). The read-out does this for you once three maxima have been recorded: compare with 253.7 nm.
  • Contact potential: untick it. The first maximum moves, but the spacing does not change. Why?
  • Retarding voltage: set U_r = 0. The dips become shallow, because even electrons that just lost energy still reach the collector.
  • Temperature: cool the oven to 120 °C. Fewer atoms → electrons gain too much energy before colliding → peaks smear out and, above 10.4 eV, mercury is ionised and the current shoots up.
  • Neon: switch gas and raise U₁ slowly. One orange glow layer appears near the grid at about 19 V, a second at about 38 V, a third near 57 V — each layer is a place where electrons lose 18.7 eV.

What it proves, and finer points

  • Energy is transferred to atoms in discrete amounts: the first electrical proof of quantised levels, independent of spectroscopy. Franck and Hertz received the 1925 Nobel Prize.
  • The first maximum is not at 4.9 V: the contact potential between cathode and grid (≈ 2 V) shifts every peak equally. Only the spacing matters.
  • In real tubes the spacing grows slightly with order (e.g. 4.9, 5.0, 5.1 V): electrons travel a little beyond the threshold before colliding (finite mean free path) and overshoot more at higher voltage. Lower the oven temperature to about 150 °C to see the peaks shift and broaden.
  • Mercury's lowest excited level (6³P₀, 4.67 eV) is hardly seen; the dominant excitation is 6³P₁ at 4.89 eV, which decays by emitting 253.7 nm.
  • Franck and Hertz first interpreted 4.9 V as the ionisation energy; the correct reading, as an excitation, came soon after (ionisation of Hg is 10.4 eV).

Your measurements

Every reading you take is stored here (U₁ in volts, I_A in nA). Download them as CSV to plot in a spreadsheet for your practical record.

#U₁ (V)I_A (nA)

Analysis of your data

Record a sweep from 0 V to the top of the range to analyse it.

Maxima are found automatically after light smoothing. Use at least three of them for a good average.