Geiger Counter Decay Curve with a Tail
A laboratory class activates a foil and follows its beta count rate with a Geiger–Müller tube. The background, measured for 20 min with the source removed, is 18.0 counts per minute. Each source measurement is a 5.00 min count; the recorded gross rates are 860 cpm at t = 0, 610 cpm at 30.0 min, 450 cpm at 60.0 min, 330 cpm at 90.0 min, and 205 cpm at 150.0 min. The counter has a specified dead time of 250 µs. The instructor suspects either a single isotope with a half-life near one hour plus a long-lived contaminant, or a systematic problem with background. The class must decide whether the first four points are consistent with one exponential, whether dead-time correction is needed, and whether the 150 min point changes the conclusion.
1. Subtract background and test whether the first four net rates follow a single exponential. Estimate the half-life.
Read the worked answer
The net rates are 860 − 18 = 842 cpm, 610 − 18 = 592 cpm, 450 − 18 = 432 cpm, and 330 − 18 = 312 cpm. The successive ratios over equal 30 min intervals are 592/842 = 0.703, 432/592 = 0.730, and 312/432 = 0.722. These are close to a constant ratio, so the first four points are consistent with a single exponential within roughly the counting noise.
Using the first and fourth points over 90 min gives λ = −ln(312/842)/90 = −ln(0.3706)/90 = 0.992/90 = 0.0110 min⁻¹. The half-life is T = ln2/λ = 0.693/0.0110 = 63 min. Fitting the intermediate points gives essentially the same value, about 63 min, not the 12 h that might be guessed from a quick glance.
2. Assess dead-time correction. Is it needed for these rates? Show the size.
Read the worked answer
The highest gross rate is 860 cpm = 14.3 counts/s. For a nonparalyzable counter, the fractional loss is approximately nτ = 14.3 × 250 × 10⁻⁶ = 3.6 × 10⁻³, or 0.36%. The corrected rate would be 860/(1 − 0.0036) = 863 cpm, only about 3 cpm higher.
At the 150 min point the rate is 205 cpm = 3.4 counts/s, so the loss is about 0.09%, even smaller. Correcting for dead time changes the half-life by well under 0.1 min. Dead time is not the explanation for any discrepancy in this data set; it can be mentioned as a small correction but not used to rescue a bad fit.
3. Include the 150 min point. Does it support a single isotope plus background? What alternative and follow-up test do you recommend?
Read the worked answer
Using λ = 0.0110 min⁻¹ from the first four points, the predicted net rate at 150 min is 842 exp(−0.0110 × 150) = 842 exp(−1.65) = 161 cpm. Adding background gives a predicted gross rate of about 179 cpm. The observed gross rate is 205 cpm, so the net excess is about 187 − 161 = 26 cpm.
The 5 min count at 150 min contains about 1025 gross counts, so the counting standard deviation is sqrt(1025) = 32 counts, or 6.4 cpm. The background subtraction adds about 1 cpm. The 26 cpm excess is therefore roughly four standard deviations, which is unlikely to be a statistical fluctuation.
The first four points fit a single isotope, but the 150 min point is too high for that fit. The likely explanation is a small long-lived contaminant, or a background that is higher during the later measurement than the 18 cpm assumed. Repeat the background with a longer count immediately after removing the source, and follow the decay for several more half-lives. A two-component fit should show a constant tail from the contaminant; if no tail appears, investigate background stability and counter contamination.