Radioactive Decay and Half-Life
Radioactive Decay and Half-Life
Radioactive decay is the spontaneous transformation of an unstable atomic nucleus into a more stable configuration by emitting particles or electromagnetic radiation. This process is governed by one of the most important mathematical laws in science — the exponential decay law — and it is the foundation of nuclear engineering, medical imaging, carbon dating, and radiation safety.
Every radioactive isotope has a characteristic half-life, denoted t one-half. This is the time required for exactly half of the atoms in a sample to decay. Carbon-14 has a half-life of 5,730 years, making it ideal for archaeological dating. Iodine-131, used in thyroid treatment, has a half-life of about 8 days. Uranium-238, the most abundant uranium isotope, has a half-life of 4.5 billion years — comparable to the age of the Earth.
The mathematical description is elegantly simple. If N-zero atoms are present at time zero, then the number remaining at time t is N of t equals N-zero times e raised to the power of negative lambda t, where lambda is the decay constant. The decay constant is related to the half-life by lambda equals the natural log of 2 divided by the half-life. This means that regardless of how many atoms you start with, the fraction remaining depends only on how many half-lives have elapsed.
The activity of a sample — the number of decays per second — is A equals lambda times N, measured in becquerels (one decay per second) or curies (3.7 times 10 to the 10 decays per second). Activity also decays exponentially with the same time constant. Understanding these relationships is essential for handling radioactive materials safely, designing shielding, planning medical treatments, and managing nuclear waste.
The decay constant $\lambda$ is the probability per unit time that a given nucleus will decay:
$$\lambda = \frac{\ln 2}{t_{1/2}} = \frac{0.6931}{t_{1/2}}$$
$$N(t) = N_0 \, e^{-\lambda t}$$
where $N_0$ is the initial number of atoms, $\lambda$ is the decay constant, and $t$ is elapsed time.
Equivalently in terms of half-lives: $N(t) = N_0 \left(\frac{1}{2}\right)^{t/t_{1/2}}$
Activity $A$ is the rate of decay (disintegrations per second):
$$A(t) = \lambda N(t) = A_0 \, e^{-\lambda t}$$
Units: 1 Bq = 1 decay/s. 1 Ci = $3.7 \times 10^{10}$ Bq.
Cobalt-60 has a half-life of 5.27 years. A medical source starts with $10^{15}$ atoms. How many remain after 15 years?
Step 1: Find the decay constant.
$$\lambda = \frac{\ln 2}{5.27} = \frac{0.6931}{5.27} = 0.1315 \text{ yr}^{-1}$$
Step 2: Apply the decay law.
$$N(15) = 10^{15} \cdot e^{-0.1315 \times 15} = 10^{15} \cdot e^{-1.9725}$$
Step 3: Evaluate.
$$e^{-1.9725} \approx 0.1391$$
$$N(15) \approx 1.39 \times 10^{14} \text{ atoms}$$
About 13.9% of the original atoms remain — just under 3 half-lives have elapsed.
A sample has an initial activity of 800 Bq. After 24 hours, the activity is 200 Bq. Find the half-life.
Step 1: Use $A = A_0 e^{-\lambda t}$ to find $\lambda$.
$$200 = 800 \, e^{-\lambda \cdot 24} \implies e^{-24\lambda} = 0.25$$
Step 2: Take the natural log.
$$-24\lambda = \ln(0.25) = -1.3863 \implies \lambda = 0.05776 \text{ hr}^{-1}$$
Step 3: Find the half-life.
$$t_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.6931}{0.05776} = 12.0 \text{ hours}$$
How long until only 1% of a radioactive sample remains?
Step 1: Set $N/N_0 = 0.01$.
$$0.01 = e^{-\lambda t} \implies \lambda t = -\ln(0.01) = 4.605$$
Step 2: Express in half-lives.
$$t = \frac{4.605}{\lambda} = \frac{4.605}{\ln 2} \cdot t_{1/2} \approx 6.64 \, t_{1/2}$$
After about 6.64 half-lives, only 1% of the original material remains.
Practice Problems
Show Answer Key
1. $\lambda = 0.6931/8.02 = 0.0864$ day$^{-1}$.
2. $\lambda = 0.6931/28.8 = 0.02407$ yr$^{-1}$. $N/N_0 = e^{-0.02407 \times 100} = e^{-2.407} \approx 0.0900$. Mass $= 500 \times 0.090 = 45.0$ g.
3. $A = A_0 / 2^3 = 1200/8 = 150$ Bq.
4. $0.001 = (1/2)^n \Rightarrow n = \log_2(1000) \approx 9.97 \approx 10$ half-lives.
5. $0.25 = (1/2)^{t/5730} \Rightarrow t/5730 = 2 \Rightarrow t = 11{,}460$ years.
6. $1/32 = (1/2)^5$. Five half-lives.
7. $\lambda = 0.6931/(24110 \times 365.25 \times 24 \times 3600) = 9.11 \times 10^{-13}$ s$^{-1}$. $A = \lambda N = 9.11 \times 10^{-13} \times 10^{24} = 9.11 \times 10^{11}$ Bq $ \approx 24.6$ Ci.
8. $(1/2)^{30/12.3} = (1/2)^{2.439} \approx 0.184$. About 18.4% remains.
9. $A_A = A_0 e^{-\ln 2 \cdot 10/2} = A_0/32$. $A_B = A_0 e^{-\ln 2 \cdot 10/10} = A_0/2$. Ratio $= (1/32)/(1/2) = 1/16$.
10. Set $N(t_{1/2}) = N_0/2$: $N_0/2 = N_0 e^{-\lambda t_{1/2}} \Rightarrow 1/2 = e^{-\lambda t_{1/2}} \Rightarrow \ln(1/2) = -\lambda t_{1/2} \Rightarrow t_{1/2} = \ln 2/\lambda$. ✓