E3-3. The Law of Radioactive Decay & Half-Life HL

1. Activity and the Decay Constant

Because radioactive decay is a random quantum process, we cannot predict when a specific atom will decay. However, for a large sample containing $N$ undecayed nuclei, the rate of decay is strictly proportional to the number of nuclei currently present.

  • Activity ($A$): The rate at which nuclei decay, or the number of disintegrations per second. It is measured in Becquerels (Bq), where $1\text{ Bq} = 1\text{ decay per second}$. $$A = -\frac{dN}{dt}$$
  • Decay Constant ($\lambda$): The probability of decay of a single nucleus per unit time. Its unit is $\text{s}^{-1}$ (or $\text{hr}^{-1}$, $\text{yr}^{-1}$, etc.).
  • The fundamental differential equation governing radioactive decay connects these two concepts:
    $$A = \lambda N$$

2. The Exponential Law of Radioactive Decay

By applying calculus and integrating the differential equation $\frac{dN}{dt} = -\lambda N$, we derive the Exponential Law of Radioactive Decay. It shows that the number of undecayed nuclei decreases exponentially over time.

$$N = N_0 e^{-\lambda t}$$
  • $N_0$: Initial number of undecayed nuclei at time $t = 0$.
  • $N$: Number of undecayed nuclei remaining at time $t$.

Because Activity ($A$) and detected Count Rate ($C$) are directly proportional to $N$, they follow the exact same exponential drop-off:

$$A = A_0 e^{-\lambda t} \quad \text{and} \quad C = C_0 e^{-\lambda t}$$

3. Half-Life ($T_{1/2}$) Relationship

The Half-life ($T_{1/2}$) is defined as the time taken for exactly half of the radioactive nuclei in a sample to decay (meaning $N = \frac{1}{2} N_0$).

Exponential Radioactive Decay Number of Undecayed Nuclei (N) N₀ N₀ / 2 N₀ / 4 N₀ / 8 Time (t) 1 Half-life 2 Half-lives 3 Half-lives

We can mathematically link the half-life to the decay constant using the exponential law. If we substitute $N = \frac{N_0}{2}$ at $t = T_{1/2}$:

$$\frac{N_0}{2} = N_0 e^{-\lambda T_{1/2}} \implies \frac{1}{2} = e^{-\lambda T_{1/2}}$$ Take the natural logarithm ($\ln$) of both sides: $$\ln(0.5) = -\lambda T_{1/2} \implies -0.693 = -\lambda T_{1/2}$$
$$T_{1/2} = \frac{\ln(2)}{\lambda}$$

4. Advanced HL Examples

Example 1: Graphical Linearisation of Decay Data (Paper 3 / Data Analysis)

Problem: A student measures the activity $A$ of a radioactive isotope over 10 hours. To determine the half-life accurately, they plot a graph of $\ln(A)$ against time $t$ in hours. The resulting graph is a straight line with a gradient of $-0.138\text{ hr}^{-1}$ and a y-intercept of $5.40$.
(a) Show mathematically why this plot produces a straight line.
(b) Calculate the half-life of the isotope in hours.
(c) Determine the initial activity $A_0$ in Becquerels.

Solution:

Part (a): Linearisation:
Start with the decay equation: $A = A_0 e^{-\lambda t}$. Take the natural logarithm of both sides: $$\ln(A) = \ln(A_0 e^{-\lambda t}) = \ln(A_0) + \ln(e^{-\lambda t})$$ $$\ln(A) = -\lambda t + \ln(A_0)$$ This matches the equation of a straight line $y = mx + c$, where $y = \ln(A)$, $x = t$, the gradient $m = -\lambda$, and the y-intercept $c = \ln(A_0)$.

Part (b): Half-Life:
Since the gradient $m = -\lambda$, the decay constant is $\lambda = 0.138\text{ hr}^{-1}$. $$T_{1/2} = \frac{\ln(2)}{\lambda} = \frac{0.693}{0.138} = \mathbf{5.02\text{ hours}}$$

Part (c): Initial Activity:
The y-intercept is $\ln(A_0) = 5.40$. Apply the exponential function to both sides: $$A_0 = e^{5.40} \approx \mathbf{221\text{ Bq}}$$

Example 2: Radiocarbon Dating Mechanics

Problem: A wooden artifact found in an ancient tomb contains $2.0 \times 10^{10}$ atoms of Carbon-14. A fresh sample of living wood of the exact same mass contains $8.5 \times 10^{10}$ atoms of Carbon-14. The half-life of Carbon-14 is 5730 years. Calculate the age of the wooden artifact.

Solution:

Step 1: Calculate the decay constant $\lambda$.
$$\lambda = \frac{\ln(2)}{T_{1/2}} = \frac{0.693}{5730\text{ yr}} \approx 1.21 \times 10^{-4}\text{ yr}^{-1}$$

Step 2: Apply the exponential decay law and solve for $t$.
$$N = N_0 e^{-\lambda t} \implies 2.0 \times 10^{10} = (8.5 \times 10^{10}) e^{-(1.21 \times 10^{-4})t}$$ $$\frac{2.0}{8.5} = e^{-(1.21 \times 10^{-4})t} \implies 0.2353 = e^{-(1.21 \times 10^{-4})t}$$ $$\ln(0.2353) = -(1.21 \times 10^{-4})t \implies -1.447 = -1.21 \times 10^{-4} t$$ $$t = \frac{1.447}{1.21 \times 10^{-4}} \approx \mathbf{11,960\text{ years}}$$

Example 3: Medical Tracers (Mass/Activity Conversion)

Problem: A hospital sources an iodine-131 (${}^{131}\text{I}$) tracer. The sample must have a radioactive activity of exactly $7.4 \times 10^7\text{ Bq}$ at the time of injection. The half-life of Iodine-131 is 8.02 days. Determine the exact mass of pure Iodine-131, in micrograms ($\mu\text{g}$), required to achieve this activity.

Solution:

Step 1: Find the decay constant in standard SI units ($\text{s}^{-1}$).
$$T_{1/2} = 8.02\text{ days} \times 24 \times 3600 = 692,928\text{ s}$$ $$\lambda = \frac{\ln(2)}{692,928} \approx 1.00 \times 10^{-6}\text{ s}^{-1}$$

Step 2: Use $A = \lambda N$ to find the required number of atoms ($N$).
$$7.4 \times 10^7 = (1.00 \times 10^{-6}) \times N \implies N = 7.4 \times 10^{13}\text{ atoms}$$

Step 3: Convert atoms to mass.
$$n = \frac{N}{N_A} = \frac{7.4 \times 10^{13}}{6.02 \times 10^{23}} = 1.23 \times 10^{-10}\text{ moles}$$ $$m = n \times M = (1.23 \times 10^{-10}\text{ mol}) \times 131\text{ g/mol} = 1.61 \times 10^{-8}\text{ g}$$ Convert to micrograms ($1\text{ g} = 10^6\ \mu\text{g}$): $m = \mathbf{0.0161\ \mu\text{g}}$.