use λ = 0.693 / t

Radioactive Decay

Learning Objective

By the end of this lesson you should be able to calculate the decay constant (λ) of a radionuclide using the relationship

$$\lambda = \frac{0.693}{t_{1/2}}$$

where $t_{1/2}$ is the half‑life of the nuclide.

1. Introduction to Radioactivity

Radioactivity is the spontaneous transformation of an unstable nucleus into a more stable configuration. The process is random for individual nuclei but follows a predictable statistical law for a large collection.

2. The Decay Law

The number of undecayed nuclei, $N$, at any time $t$ is given by the exponential decay law:

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

• $N_0$ – initial number of nuclei at $t=0$
• $\lambda$ – decay constant (s\(^{-1}\))
• $t$ – elapsed time (s)

3. Half‑Life and the Decay Constant

The half‑life, $t_{1/2}$, is the time required for half of the original nuclei to decay. Setting $N(t_{1/2}) = \tfrac{1}{2}N_0$ in the decay law gives:

$$\frac{1}{2}N_0 = N_0 e^{-\lambda t_{1/2}} \;\;\Rightarrow\;\; e^{-\lambda t_{1/2}} = \frac{1}{2}$$

Taking natural logarithms leads to the useful relationship:

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

4. Types of Radioactive Decay

Decay Mode	Particle Emitted	Change in Nucleus	Typical Energy (MeV)
Alpha (α) decay	Helium nucleus ($^4_2\text{He}$)	$A \rightarrow A-4,\; Z \rightarrow Z-2$	4–9
Beta‑minus (β⁻) decay	Electron ($e^-$) + antineutrino	$A \rightarrow A,\; Z \rightarrow Z+1$	0.1–3
Beta‑plus (β⁺) decay / Positron emission	Positron ($e^+$) + neutrino	$A \rightarrow A,\; Z \rightarrow Z-1$	0.5–2
Electron capture	Capture of an inner‑shell electron	$A \rightarrow A,\; Z \rightarrow Z-1$	\overline{0} (no kinetic particle)
Gamma (γ) decay	High‑energy photon	No change in $A$ or $Z$ (de‑excitation)	0.1–10

5. Practical Calculations

1. Determine the half‑life $t_{1/2}$ of the radionuclide (from tables or experiment).
2. Calculate the decay constant using $\lambda = 0.693/t_{1/2}$.
3. Use $N(t)=N_0 e^{-\lambda t}$ to find the remaining nuclei after a given time.
4. Alternatively, calculate the activity $A = \lambda N$ (in becquerels, Bq).

6. Example Problem

Problem: A sample contains $2.0 \times 10^{20}$ atoms of $^{60}$Co, whose half‑life is $5.27$ years. Find the activity after $10$ years.

Solution:

• Convert half‑life to seconds: $t_{1/2}=5.27\;\text{yr}\times 3.156\times10^{7}\;\text{s yr}^{-1}=1.66\times10^{8}\;\text{s}$.
• Decay constant: $\displaystyle \lambda = \frac{0.693}{1.66\times10^{8}\;\text{s}} = 4.18\times10^{-9}\;\text{s}^{-1}$.
• Number of atoms after $10$ yr ($t = 3.156\times10^{8}\;\text{s}$): $$N = N_0 e^{-\lambda t}=2.0\times10^{20} e^{-4.18\times10^{-9}\times3.156\times10^{8}} \approx 1.0\times10^{20}.$$
• Activity: $A = \lambda N = 4.18\times10^{-9}\times1.0\times10^{20}=4.2\times10^{11}\;\text{Bq}$.

7. Summary

• The decay constant $\lambda$ quantifies the probability per unit time that a nucleus will decay.
• It is directly related to the half‑life by $\lambda = 0.693/t_{1/2}$.
• Radioactive decay follows an exponential law, allowing prediction of remaining nuclei and activity.
• Understanding decay modes helps interpret the changes in atomic number and mass number.