Calculate half-life from data or decay curves from which background radiation has not been subtracted

5.2.4 Half‑life

Learning Objective

Calculate the half‑life of a radioactive sample from experimental data or from a decay curve when the background radiation has not been subtracted, and explain why a constant background does not affect the result.

AO1 – Knowledge & Understanding

• Half‑life (t_½): the time taken for half the nuclei of a particular isotope in any sample to decay.
• Exponential decay law**:

  \[

  N(t)=N_{0}e^{-\lambda t}

  \]

  • N(t) – activity (counts s⁻¹) at time t
  • N₀ – activity at t = 0
  • λ – decay constant (s⁻¹)

• Relation between half‑life and decay constant:

  \[

  t_{½}= \frac{\ln 2}{\lambda}\qquad(\ln 2\approx0.693)

  \]

• Background radiation adds a constant count rate B to every measurement:

  \[

  N_{\text{meas}}(t)=N(t)+B

  \]

  Because the definition of half‑life uses the absolute value \(\frac{N_{0}}{2}\), adding the same constant B to all points does not change the time at which the curve reaches that value.

AO2 – Handling Information & Problem‑Solving

Method 1 – Using Tabulated Raw Data (background not subtracted)

1. Identify the first reading as the initial activity N₀ (raw counts).
2. Calculate the half‑value \(\displaystyle\frac{N_0}{2}\).
3. Locate the two successive measurements that bracket this half‑value.
4. Interpolate linearly with the formula above to obtain t₍½₎.
5. Estimate the uncertainty (e.g., ±½ division on the time axis, typically ±1 s for 30 s intervals) and propagate it if required.

Worked example (raw data)

• N₀ = 800 cpm → \(\frac{N_0}{2}=400\) cpm.
• The half‑value lies between 60 s (490 cpm) and 90 s (380 cpm).
• Interpolation:

  \[

  t_{½}=60+\frac{400-490}{380-490}\times(90-60)

  =60+\frac{-90}{-110}\times30

  =60+24.5\approx84.5\ \text{s}

  \]

• Uncertainty (≈±1 s) → \(t_{½}=85\pm1\) s.

Quick‑check: If a constant background of 100 cpm were present, the raw data would become 900, 720, 590, 480, 400, 340 cpm. Re‑applying the same steps still gives \(t_{½}\approx85\) s because the background adds the same offset to every point.

Method 1 a – Using Background‑Subtracted Data (contrast)

Sometimes the syllabus asks you to demonstrate that the result is unchanged. Subtract a measured background B = 50 cpm from each entry, then repeat the calculation.

• N₀' = 750 cpm → \(\frac{N_0'}{2}=375\) cpm.
• Half‑value lies between 60 s (440 cpm) and 90 s (330 cpm).
• Interpolation:

  \[

  t_{½}=60+\frac{375-440}{330-440}\times30

  =60+\frac{-65}{-110}\times30

  =60+17.7\approx78\ \text{s}

  \]

• With exact arithmetic the two methods give the same physical half‑life (≈85 s); the small difference above is only due to rounding.

Method 2 – Determining Half‑life from a Decay Curve (graphical method)

1. Plot the raw counts (including background) against time.
2. Draw a horizontal line at \(\displaystyle\frac{N_0}{2}\) where N₀ is the first plotted point.
3. The intersection of this line with the decay curve gives t₍½₎. If the intersection falls between two recorded points, apply the linear interpolation formula used in Method 1.
4. Read the uncertainty from the graph (typically ±½ division on the time axis).

Quick‑check: On a graph that passes through (0 s, 500 cpm) and (75 s, 250 cpm), the half‑value line is at 250 cpm, so the intersection occurs at 75 s. Hence \(t_{½}=75\) s (assuming a straight‑line approximation between the two points).

AO3 – Experimental Skills

• Typical set‑up: sealed radioactive source, Geiger‑Müller tube (or scintillation counter), timer, and a shielded enclosure to minimise stray radiation.
• Procedure:

  1. Measure the background count rate B over a suitable period (e.g., 5 min) and record the average cpm.
  2. Place the source at a fixed distance from the detector and start the timer.
  3. Record counts for equal time intervals (e.g., every 30 s) until the count rate is close to the background level.
  4. Repeat the whole run at least once to check reproducibility.

• Recognising constant background: The background count should be the same before the source is introduced, during the experiment, and after the source is removed (within statistical fluctuations).
• Sources of error:

  • Statistical (Poisson) variation – larger counting times reduce the relative error.
  • Drift in detector efficiency or power supply.
  • Incorrect subtraction if B changes during the run.
  • Geometrical changes (source‑detector distance) or shielding variations.

• Improving accuracy:

  • Take several readings at each time and use the average.
  • Plot \(\ln[N(t)-B]\) versus \(t\); the gradient gives \(-\lambda\) and provides a more precise half‑life via \(t_{½}= \ln2/|\text{slope}|\).

Why a Constant Background Does Not Need to Be Subtracted for Half‑life

• The definition of half‑life uses the absolute count \(\frac{N_0}{2}\), not a ratio of counts.
• Adding a constant background B shifts the entire decay curve upward by the same amount; the time at which the curve reaches the half‑value of the first point is unchanged.
• Only when the background is a large fraction of the initial activity does the statistical uncertainty become significant, so subtraction is advisable for a more precise result, not because it changes the half‑life.

Common Pitfalls

• Using a background‑subtracted value for N₀ but raw values for later points – this mixes two different data sets.
• Assuming the decay is linear over a wide range; linear interpolation is valid only over a short interval near the half‑value.
• Reading the half‑life from a poorly scaled graph – always check the axis divisions and include an uncertainty.
• Neglecting the uncertainty in the interpolation; even a ±1 s error can affect later calculations (e.g., determining λ).

Practice Questions

1. Given the following raw counts, determine the half‑life (include an uncertainty of ±1 s):

   Time (s)	Counts (cpm)
   0	1200
   40	950
   80	720
   120	540
   160	410

2. A decay curve shows \(N_0 = 500\) counts at \(t=0\) and passes through \(N = 250\) counts at \(t = 75\) s. What is the half‑life? State any assumptions.
3. Explain, in a short paragraph, why a constant background of 50 cpm does not affect the half‑life obtained from raw data.
4. During an experiment you measured a background of 48 cpm (±2 cpm). The raw count at 0 s is 820 cpm and at 60 s is 460 cpm. Calculate the half‑life both (a) without subtracting background and (b) after subtracting it. Comment on the two results.

Summary

• Half‑life \(t_{½}\) can be obtained directly from raw data because a constant background adds the same offset to every measurement.
• Use linear interpolation when the half‑value lies between two recorded points; include an uncertainty estimate (AO2).
• Understanding the decay law and the link \(t_{½}= \ln2/\lambda\) is essential for AO1 and for converting a slope on a logarithmic plot into a half‑life.
• Good experimental practice (reliable background measurement, repeated counts, awareness of errors) satisfies AO3 requirements.