understand the exponential nature of radioactive decay, and sketch and use the relationship x = x0e–λt, where x could represent activity, number of undecayed nuclei or received count rate

Radioactive Decay

Radioactive decay is a random, spontaneous process in which an unstable nucleus transforms into a more stable configuration. The number of undecayed nuclei, the activity, or the count rate measured by a detector all follow the same exponential law.

Exponential Decay Law

The fundamental relationship is

\$x = x_0 e^{-\lambda t}\$

where

• \$x\$ – quantity at time \$t\$ (number of nuclei \$N\$, activity \$A\$, or count rate \$C\$)
• \$x_0\$ – initial quantity at \$t = 0\$
• \$\lambda\$ – decay constant (s\(^{-1}\))
• \$t\$ – elapsed time (s)

Derivation from First Principles

1. Assume the probability that a given nucleus decays in a short interval \$dt\$ is proportional to \$dt\$: \$dP = \lambda\,dt\$.
2. The expected change in the number of undecayed nuclei \$N\$ is \$dN = -\lambda N\,dt\$.
3. Rearranging gives the differential equation \$\displaystyle\frac{dN}{dt} = -\lambda N\$.
4. Integrating from \$N0\$ at \$t=0\$ to \$N\$ at time \$t\$ yields \$\displaystyle\ln\!\left(\frac{N}{N0}\right) = -\lambda t\$.
5. Exponentiating both sides gives \$N = N0 e^{-\lambda t}\$, which is the same form as \$x = x0 e^{-\lambda t}\$.

Key Parameters

Graphical Representation

The plot of \$x\$ versus \$t\$ is a decreasing exponential curve. A semi‑log plot (log \$x\$ against \$t\$) yields a straight line with gradient \$-\lambda\$.

Using the Decay Equation

Typical problems require rearranging the equation to solve for a particular variable.

• Finding the remaining quantity after a given time:

  \$x = x_0 e^{-\lambda t}\$

• Determining the time required for a quantity to reach a specified value:

  \$t = \frac{1}{\lambda}\ln\!\left(\frac{x_0}{x}\right)\$

• Calculating the decay constant from a known half‑life:

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

Worked Example

Problem: A sample of \$^{60}\$Co has an initial activity of \$A_0 = 3.0 \times 10^6\ \text{Bq}\$. Its half‑life is \$5.27\$ years. Calculate the activity after \$12\$ years.

1. Convert the half‑life to seconds (optional, but not required if \$\lambda\$ is expressed per year):

   \$t_{1/2}=5.27\ \text{yr}\$

2. Calculate the decay constant:

   \$\lambda = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{5.27\ \text{yr}} = 0.1315\ \text{yr}^{-1}\$

3. Insert into the decay law:

   \$A = A_0 e^{-\lambda t}=3.0\times10^{6}\,e^{-0.1315\times12}\$

4. Evaluate:

   \$A = 3.0\times10^{6}\,e^{-1.578}=3.0\times10^{6}\times0.206 \approx 6.2\times10^{5}\ \text{Bq}\$

Thus the activity after 12 years is approximately \$6.2\times10^{5}\ \text{Bq}\$.

Common Mistakes to Avoid

• Confusing the decay constant \$\lambda\$ with the half‑life \$t_{1/2}\$; they are inversely related, not equal.
• Using linear scales for exponential decay without recognising the curvature; a semi‑log plot is often clearer.
• Neglecting unit consistency, especially when \$t\$ is given in minutes, hours, or years.
• Assuming the count rate measured by a detector is the same as activity; the detector efficiency \$\epsilon\$ must be considered: \$C = \epsilon A\$.

Summary Checklist

1. Write the exponential law \$x = x_0 e^{-\lambda t}\$.
2. Identify whether \$x\$ represents \$N\$, \$A\$, or \$C\$.
3. Calculate \$\lambda\$ from \$t_{1/2}\$ or use a given \$\lambda\$.
4. Rearrange the equation to solve for the required variable.
5. Check units and, if needed, convert to a common time unit.
6. Use a semi‑log plot to verify linearity and extract \$\lambda\$ experimentally.