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

What is Radioactive Decay?

In radioactive decay, an unstable nucleus transforms into a more stable one by emitting particles or energy. Think of it like a ticking clock that keeps on ticking until it stops – the ticking is the decay.

The Exponential Law

All radioactive decay follows the same mathematical rule:

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

Where:

  • \$x\$ = quantity remaining after time \$t\$ (e.g., number of undecayed nuclei, activity)
  • \$x_0\$ = initial quantity at \$t = 0\$
  • \$\lambda\$ = decay constant (probability per unit time that a nucleus decays)
  • \$t\$ = elapsed time

Because the rate of decay is proportional to the amount still present, the graph of \$x\$ vs. \$t\$ is a smooth, downward‑sloping curve that never quite reaches zero.

Key Parameters

ParameterDefinitionFormula
Decay ConstantProbability per unit time that a nucleus decays.\$\lambda\$
Half‑LifeTime taken for half the nuclei to decay.\$T_{1/2} = \dfrac{\ln 2}{\lambda}\$
ActivityNumber of decays per unit time.\$A = \lambda N\$

Sketching the Decay Curve

Imagine a steep hill that gradually flattens out. The hill’s slope represents the rate of decay. As time goes on, the hill becomes less steep because fewer nuclei remain to decay.

Key points to label on your sketch:

  1. Initial point \$(0, x_0)\$
  2. Half‑life point \$(T{1/2}, x0/2)\$
  3. Long‑term asymptote approaching \$0\$ but never touching it.

Use a smooth curve – no sharp corners – to show the continuous nature of the process.

Practical Example

Suppose we have \$N0 = 1.0 \times 10^{12}\$ atoms of a radioactive isotope with a half‑life of \$T{1/2} = 5\$ years.

  1. Find \$\lambda\$:

\$\lambda = \dfrac{\ln 2}{T_{1/2}} = \dfrac{0.693}{5\,\text{yr}} = 0.1386\,\text{yr}^{-1}\$

  1. After \$t = 10\$ years, how many atoms remain?

\$N = N_0 e^{-\lambda t} = 1.0 \times 10^{12} e^{-0.1386 \times 10} \approx 2.5 \times 10^{11}\$

  1. What is the activity at \$t = 10\$ years?

\$A = \lambda N \approx 0.1386 \times 2.5 \times 10^{11} \approx 3.5 \times 10^{10}\,\text{decays yr}^{-1}\$

Remember: Activity is often expressed in becquerels (decays per second). Convert if needed.

Exam Tips 📚

  • Always write the decay law in the form \$x = x_0 e^{-\lambda t}\$ before manipulating.
  • When converting between half‑life and decay constant, use \$T_{1/2} = \dfrac{\ln 2}{\lambda}\$.
  • Check units carefully – \$\lambda\$ is in s⁻¹ or yr⁻¹, \$t\$ must match.
  • For activity questions, remember \$A = \lambda N\$ and that \$A\$ is proportional to the number of undecayed nuclei.
  • Sketches: label axes, include key points (initial, half‑life, asymptote) and show the curve as a smooth exponential decay.
  • Use the analogy of a “decaying candle” or “fading light” to explain why the curve never reaches zero.