## Half-Life ($T_{1/2}$) The time required for half of the original amount of a radioactive substance to decay into a more stable form. It is a characteristic property of each radioactive element and describes the rate at which a radioactive substance decays. A longer half-life implies a slower rate of decay and less radiation emitted over time. ## Decay Constant ($\lambda$) A constant that represents the probability per unit time for a nucleus to decay. It is directly related to the rate of decay. ## Relation between Decay Constant and Half-Life The number of un-decayed nuclei ($N$) at time ($t$) from an initial number ($N_o$) is given by the exponential decay law: $$N = N_o e^{-\lambda t}$$ At one half-life ($t = T_{1/2}$), the number of nuclei remaining is $N = N_o/2$. Substituting these values into the decay law: $$ \begin{aligned} \frac{N_o}{2} &= N_o e^{-\lambda T_{1/2}} \\ \frac{1}{2} &= e^{-\lambda T_{1/2}} \\ 2 &= e^{\lambda T_{1/2}} \end{aligned} $$ Taking the natural logarithm of both sides: $$\ln 2 = \lambda T_{1/2}$$ Thus, the relationship is: $$T_{1/2} = \frac{\ln 2}{\lambda} \quad \text{or} \quad T_{1/2} = \frac{0.693}{\lambda}$$ ## Graphical Representation of Nuclear Decay Figure 22.9 illustrates the exponential decrease in the number of un-decayed nuclei over time. The half-life can be visually determined from such a graph by finding the time it takes for the number of nuclei to drop to half of its initial value. ## Activity ($A$) The rate of decay of a radioactive sample. It is given by the formula $A = \lambda N$, where $\lambda$ is the decay constant and $N$ is the number of radioactive nuclei present at that instant. Units of activity include Becquerel (Bq) and Curie (Ci). ## Possible Questions and Answers **Q:** Cesium-137, a beta emitter, has a half-life of $9.5 \times 10^8$ s. What is its decay constant? **A:** Given $T_{1/2} = 9.5 \times 10^8$ s. Using the formula $\lambda = \frac{0.693}{T_{1/2}}$: $$\lambda = \frac{0.693}{9.5 \times 10^8 \text{ s}} = 7.3 \times 10^{-10} \text{ s}^{-1}$$ **Q:** For the Cesium-137 in the previous question, if we have one mole ($N_o = 6.02 \times 10^{23}$) at the start, what will be its initial activity? **A:** We found $\lambda = 7.3 \times 10^{-10}$ s$^{-1}$. Using the formula $A = \lambda N$: $$A = 7.3 \times 10^{-10} \text{ s}^{-1} \times 6.02 \times 10^{23} = 4.395 \times 10^{14} \text{ Bq}$$ (approximately $1.188 \times 10^4$ Ci) **Q:** The half-life of strontium (Sr-91) is 9.70 hours. Find its decay constant. **A:** Given $T_{1/2} = 9.70$ hours. Using the formula $\lambda = \frac{0.693}{T_{1/2}}$: $$\lambda = \frac{0.693}{9.70 \text{ hours}} \approx 0.0714 \text{ hours}^{-1}$$ Alternatively, converting to seconds: $T_{1/2} = 9.70 \times 3600 = 34920$ s: $$\lambda = \frac{0.693}{34920 \text{ s}} \approx 1.98 \times 10^{-5} \text{ s}^{-1}$$ ## Main Ideas - Half-life ($T_{1/2}$) is the time required for half of a radioactive substance to decay, acting as a characteristic property of the element. - The decay constant ($\lambda$) quantifies the rate or probability of nuclear decay. - Half-life and decay constant are inversely proportional, related by the formula $T_{1/2} = \frac{\ln 2}{\lambda}$. - Radioactive decay follows an exponential decrease in the number of un-decayed nuclei over time. - Activity ($A$) is the rate of radioactive decay, calculated as the product of the decay constant and the number of nuclei present ($A = \lambda N$). | Formula | Description | | :------ | :----------- | | $N = N_o e^{-\lambda t}$ | Exponential decay law (number of nuclei) | | $T_{1/2} = \frac{\ln 2}{\lambda}$ | Relation between half-life and decay constant | | $T_{1/2} = \frac{0.693}{\lambda}$ | Practical form of half-life relation | | $A = \lambda N$ | Activity of a radioactive sample | **Significance:** Understanding half-life is crucial for predicting the longevity and decay behavior of radioactive materials. It has wide applications, from medical treatments and industrial measurements to geological dating and managing nuclear waste.