All detectors are connected to a counting unit (digital counter or computer interface) that records the number of ionising events, N, in a chosen counting time, t.
| Symbol | Meaning | Unit |
| N | Number of counts (raw, background or corrected) | counts |
| t | Counting time | seconds (s) or minutes (min) |
| R | Count rate | cps (counts s⁻¹) or cpm (counts min⁻¹) |
| ε | Detector efficiency (fraction of emitted particles recorded) | dimensionless (0 – 1) |
| A | Activity of the sample | Bq (decays s⁻¹) |
| σ | Standard uncertainty | same as the quantity it refers to |
1. Raw (sample + background) count rate
\$ R{\text{raw}} = \frac{N{\text{raw}}}{t_{\text{raw}}}\quad\text{(cps or cpm)} \$
2. Background count rate (measured separately)
\$ R{\text{bg}} = \frac{N{\text{bg}}}{t_{\text{bg}}} \$
Best practice: Use the same counting time for both measurements (normally 30 s or 1 min). If the times differ, calculate the two rates first and then subtract.
Subtract the background contribution:
\$ R{\text{corr}} = R{\text{raw}} - R_{\text{bg}} \$
When the counting times are identical, you can work directly with the raw counts:
\$ N{\text{corr}} = N{\text{raw}} - N_{\text{bg}} \$
| Measurement | Counts N | Time t (s) | Rate R (cps) |
|---|---|---|---|
| Raw sample | 1 200 | 30 | 40.0 |
| Background | 200 | 30 | 6.7 |
| Corrected | 1 000 | 30 | 33.3 |
Suppose the background is measured for 20 s while the sample is measured for 30 s.
\$ R_{\text{bg}} = \frac{200}{20}=10\;\text{cps} \$
\$ R_{\text{raw}} = \frac{1\,200}{30}=40\;\text{cps} \$
\$ R_{\text{corr}} = 40 - 10 = 30\;\text{cps} \$
\$ \sigma_N = \sqrt{N} \$
\$ \sigma_R = \frac{\sqrt{N}}{t} \$
\$ \sigma{R{\text{corr}}}= \sqrt{\sigma{R{\text{raw}}}^{2}+\sigma{R{\text{bg}}}^{2}} \$
\$ \sigma{R{\text{raw}}}= \frac{\sqrt{1\,200}}{30}= \frac{34.64}{30}=1.15\;\text{cps} \$
\$ \sigma{R{\text{bg}}}= \frac{\sqrt{200}}{30}= \frac{14.14}{30}=0.47\;\text{cps} \$
\$ \sigma{R{\text{corr}}}= \sqrt{1.15^{2}+0.47^{2}}=1.24\;\text{cps} \$
Result: Rcorr = 33.3 ± 1.2 cps.
The activity, A, of the sample is related to the corrected count rate by the detector efficiency, ε:
\$ A = \frac{R_{\text{corr}}}{\varepsilon}\qquad\text{(Bq)} \$
Important note: ε is specific to the detector‑radiation combination (e.g. a GM tube may have ε ≈ 0.30 for β‑particles from a thin source but ε ≈ 0.05 for γ‑rays). Always use the efficiency value supplied by the teacher or obtained from a calibrated source.
\$ A = \frac{33.3}{0.25}=133.2\;\text{Bq} \$
Uncertainty (propagating only the rate uncertainty):
\$ \sigmaA = \frac{\sigma{R_{\text{corr}}}}{\varepsilon}= \frac{1.24}{0.25}=4.96\;\text{Bq} \$
Hence, A = 133 ± 5 Bq (rounded to 2 s.f.).
| Sample | Background | Corrected |
|---|---|---|
| cpm | cpm | cpm |
Insert your measured values in the blanks; the “Corrected” column is simply “Sample – Background”. This format is commonly expected in Cambridge exam answers.
The syllabus often uses the command words calculate and explain for this topic. In your answer:

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