The Geiger–Müller counter: plateau, counting statistics, inverse square law, absorption, half-life and dead time

Theory

How a Geiger–Müller tube counts

The tube is a gas-filled cylinder (cathode) with a thin central wire (anode) at a few hundred volts. The field near the wire, \(E(r) = V / (r\ln(b/a))\), is so strong that a single ion pair starts a Townsend avalanche. In the Geiger region, UV photons from the avalanche spread it along the whole wire, so every ionising particle gives the same large pulse, whatever its energy. A halogen (or organic) quench gas stops the discharge, and the slow positive-ion sheath then drifts to the cathode.

Plateau

Below the starting voltage the pulses are too small to cross the discriminator. Above it the count rate rises steeply, then levels off on the plateau, where it changes only slowly because of spurious after-pulses. Past the plateau, the tube goes into continuous discharge, which damages it. The plateau slope is quoted as

$$ S = \frac{R_2 - R_1}{\bar R}\,\frac{100}{V_2 - V_1}\times 100\ \%\ \text{per }100\ \text{V},$$

and the operating voltage is chosen about one third of the way along the plateau.

Counting statistics

Decays are independent random events, so the number of counts N in a fixed time follows the Poisson distribution, \(P(N) = \bar N^{N} e^{-\bar N}/N!\), whose variance equals its mean: \(\sigma_N = \sqrt{N}\). The rate \(R = N/t\) therefore has \(\sigma_R = \sqrt N/t\), and a background-corrected rate has \(\sigma = \sqrt{\sigma_R^2 + \sigma_b^2}\).

Inverse square law and solid angle

A point source emits isotropically, so the fraction entering a window of radius a at distance D is \(\Omega/4\pi = \tfrac12\left(1 - D/\sqrt{D^2+a^2}\right) \approx a^2/4D^2\). The particles are counted somewhere inside the tube, not at the window, so \(D = d + d_0\):

$$ R - R_b = \frac{k}{(d+d_0)^2} \quad\Rightarrow\quad \frac{1}{\sqrt{R-R_b}} = \frac{d + d_0}{\sqrt k}. $$

A plot of \(1/\sqrt{R-R_b}\) against d is a straight line whose intercept on the d axis is \(-d_0\). At short distances the exact solid angle falls below \(1/D^2\), which you can see in the data.

Absorption

β particles have a continuous energy spectrum. Their transmission is close to exponential in the mass thickness \(x_m = \rho x\) (g cm⁻²), \(R = R_0 e^{-\mu_m x_m}\), up to a maximum range \(R_\beta\). Empirically \(\mu_m \approx 17\,E_{max}^{-1.14}\) cm² g⁻¹ (E in MeV) and the Katz–Penfold range is \(R_\beta = 0.412\,E^{\,1.265 - 0.0954\ln E}\) g cm⁻². Both give the end-point energy. γ rays are attenuated exactly exponentially, with a half-value thickness \(x_{1/2} = \ln 2/\mu\). Pb at 662 keV has \(\mu_m = 0.110\) cm² g⁻¹. α particles have a short, sharp range: about 4 cm of air, or one sheet of paper.

Half-life

The ¹³⁷Cs/^137mBa isotope generator gives a fresh, short-lived sample of ^137mBa (T½ = 2.552 min) when it is eluted. Then \(R - R_b = R_0 e^{-\lambda t}\), so \(\ln(R-R_b)\) against t is a straight line of slope \(-\lambda\), and \(T_{1/2} = \ln 2/\lambda\).

Dead time

After each discharge the tube is dead for a time \(\tau_d\) while the ion sheath drifts away, and then recovers gradually. Pulses arriving during recovery are smaller and are counted only if they cross the discriminator. The resulting resolving time τ causes counting losses, \(n = R/(1 - R\tau)\). With two halves of a source, measured separately, together, and with no source:

$$ \tau \approx \frac{R_A + R_B - R_{AB} - R_b}{R_{AB}^2 - R_A^2 - R_B^2}. $$

Inside this simulation

Nothing is drawn from a formula for the count rate. The model works event by event: decays are generated as a Poisson process from the source activity, emission lines, solid angle, air, window, absorber, detector efficiency and background. Each event then goes through a tube model with a hard dead time, exponential recovery of the pulse height, a discriminator and voltage-dependent after-pulsing. The plateau slope, the counting losses, the scatter and the oscilloscope picture all emerge from that model.

Procedure

1 · Plateau

1. Place Cs-137 or Sr-90 about 5 cm below the window. Turn the HV to zero.
2. Raise the HV in 20–25 V steps (halogen tube) and take a 30 s count at each step. The starting voltage is where counts first appear.
3. Stop as soon as the rate climbs steeply again past the plateau, to avoid continuous discharge.
4. Plot R against V, find the plateau, its slope, and the operating voltage.

2 · Statistics

1. At the operating voltage, choose a distance that gives about 10–50 counts per preset time.
2. Take 100 repeated counts (Auto-run, 10 s preset, 60× clock).
3. Compare the histogram with the Poisson and Gaussian curves and check that s² ≈ N̄.

3 · Inverse square law

1. Measure the background. Use the γ-rich Co-60 source to avoid β absorption in air.
2. Count at d = 2, 3, …, 14 cm. Plot \(1/\sqrt{R-R_b}\) against d and read \(d_0\).

4 · Absorption

1. β: Sr-90 or Tl-204 with aluminium foils from 0 to 5 mm. Fit ln(R − R_b) against mass thickness and convert μ_m to E_max.
2. γ: Cs-137 or Co-60 with lead from 0 to 30 mm. The first millimetre of lead removes the β particles, then the slope gives μ for the γ rays.
3. α: Am-241 at 1–2 cm. Try a single sheet of paper.

5 · Half-life

1. Choose the Ba-137m eluate, measure the background, then press Milk the generator.
2. Take consecutive 10 s or 20 s counts for about 15 minutes (Auto-run at 10×).
3. Fit ln(R − R_b) against time.

6 · Dead time

1. Choose the split Sr-90 pair. Without moving anything, count A, then A + B, then B, then neither, using 100 s counts.
2. The table computes τ. Compare with the oscilloscope: the gap after each pulse is the dead time.

Precautions

• Never run the tube in the continuous-discharge region. Reduce the HV before switching off.
• Keep the geometry fixed during a series. Handle sources with tongs, follow ALARA (time, distance, shielding), and keep them in the lead castle.
• Always subtract background, and make counts long enough that \(\sqrt N/N\) is small.

Viva questions

Why are GM pulses all the same size, whatever the particle energy?

In the Geiger region the discharge spreads along the whole anode and is limited by the space charge of the positive ions, not by the initial ionisation.

Why does the plateau have a slope at all?

At higher voltages, more spurious after-pulses get through. Positive ions reaching the cathode, or incomplete quenching, can release new electrons. The sensitive volume also grows slightly.

Why is the GM tube a poor γ detector?

γ rays interact mainly in the tube wall, through Compton scattering and the photoelectric effect. Only about 1 % of them give an electron that reaches the gas.

Why can't you use 1/d² at very short distances?

The window and source are not points: the solid angle saturates. The effective depth \(d_0\) also matters more when d is small.

What is the difference between dead time, recovery time and resolving time?

Dead time: no discharge is possible. Recovery time: until pulses regain full height. Resolving time: the minimum separation at which the counting electronics registers two pulses. It depends on the discriminator level.

Why is the half-life experiment done with Ba-137m?

Its half-life of 2.55 min is short enough to watch in one lab session. The generator can be milked repeatedly because the 30-year Cs-137 parent keeps producing it.