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

5.2.4 Half‑life

What is a half‑life?

The half‑life, denoted \$t_{1/2}\$, is the time it takes for half of a sample of a radioactive substance to decay.

Think of a chocolate bar that shrinks by half every 5 minutes – after 5 minutes you have ½ of the bar, after 10 minutes you have ¼, and so on. The same idea applies to radioactive atoms.

Mathematical description

If \$N(t)\$ is the number of undecayed nuclei at time \$t\$, the decay law is

\$\$

N(t)=N_0\,e^{-\lambda t},

\$\$

where \$N_0\$ is the initial number and \$\lambda\$ is the decay constant.

The half‑life is related to \$\lambda\$ by

\$\$

t_{1/2}=\frac{\ln 2}{\lambda}\approx\frac{0.693}{\lambda}.

\$\$

From a decay curve (background not subtracted)

In many experiments the detector records a background count \$B\$ that is always present.

The measured count rate \$R(t)\$ is then

\$\$

R(t)=R_0\,e^{-\lambda t}+B,

\$\$

where \$R_0\$ is the initial net rate (without background).

To find \$t_{1/2}\$ we need to estimate \$\lambda\$ from the data.

Step‑by‑step method

1. Record the count rate at several times \$t_i\$ (e.g. every 2 min).
2. Choose two points where the counts are clearly above the background.

   Let the counts be \$R1\$ at \$t1\$ and \$R2\$ at \$t2\$.

3. Assume the background \$B\$ is roughly constant and estimate it from the late‑time data (when the decay signal is weak).

   For example, take the average of the last three measurements.

4. Subtract \$B\$ from the two chosen points: \$R1' = R1 - B\$, \$R2' = R2 - B\$.
5. Calculate the decay constant:

   \$\$

   \lambda = \frac{\ln(R1'/R2')}{t2 - t1}.

   \$\$

6. Finally compute the half‑life:

   \$\$

   t_{1/2} = \frac{\ln 2}{\lambda}.

   \$\$

Example data

Suppose a Geiger counter records the following counts per minute (cpm) over 10 minutes:

🔍 Background estimate: The last three points are 550, 440, and 330 cpm.

Average background \$B \approx 440\$ cpm.

🧮 Choose points: Use \$t1=0\$ min, \$R1=1250\$ cpm and \$t2=6\$ min, \$R2=680\$ cpm.

Subtract background:

• \$R_1' = 1250 - 440 = 810\$ cpm
• \$R_2' = 680 - 440 = 240\$ cpm

Compute \$\lambda\$:

\$\$

\lambda = \frac{\ln(810/240)}{6-0} = \frac{\ln(3.375)}{6} \approx \frac{1.216}{6} \approx 0.203\ \text{min}^{-1}.

\$\$

Half‑life:

\$\$

t_{1/2} = \frac{\ln 2}{0.203} \approx \frac{0.693}{0.203} \approx 3.4\ \text{min}.

\$\$

Quick check with a graph

Plotting \$\ln(R(t)-B)\$ versus \$t\$ should give a straight line with slope \$-\lambda\$.

If the points line up nicely, your background estimate is good.

Common pitfalls

• Using points that are too close to the background can give large errors.
• Assuming the background is zero when it is not will over‑estimate the decay rate.
• Neglecting statistical fluctuations: counts follow a Poisson distribution, so the error on each count is \$\sqrt{R}\$.

Practice problem

Given the following counts (cpm) at 0, 3, 6, 9, and 12 min: 2000, 1500, 1100, 800, 600.

Assume a constant background of 400 cpm.

Calculate the half‑life.

🧪 Tip: Pick the first and last points for a quick estimate, then refine with a linear fit if needed.