Half-life Knowledge-Based Study

Study notes

1. What is Half-Life?

   Radioactive decay is a spontaneous nuclear process in which an unstable atomic nucleus loses energy by emitting ionizing radiation such as alpha particles, beta particles, or gamma photons. Because radioactive decay is governed by quantum mechanics, it is fundamentally random on an individual nucleus level: it is impossible to predict precisely when a specific nucleus will disintegrate.

   However, when dealing with a statistically significant sample containing a vast number of identical unstable nuclei, the decay follows a predictable, statistical rate. The primary parameter used to quantify this rate is the half-life ($T_{1/2}$).

   Definition: The half-life of a radioisotope is the time required for half of the radioactive parent nuclei in a given sample to decay into daughter products.

   • After 1 half-life ($t = T_{1/2}$), $\frac{1}{2}$ ($50\%$) of the original parent nuclei remain.
   • After 2 half-lives ($t = 2T_{1/2}$), $\frac{1}{4}$ ($25\%$) of the original parent nuclei remain.
   • After 3 half-lives ($t = 3T_{1/2}$), $\frac{1}{8}$ ($12.5\%$) of the original parent nuclei remain.
   • After $n$ half-lives, the fraction remaining is $\left(\frac{1}{2}\right)^n$.

   The Decay Law and Decay Constant

   The rate of decay, or activity ($A$), is directly proportional to the number of radioactive nuclei ($N$) present at time $t$:

   $$\frac{dN}{dt} = -\lambda N$$

   where $\lambda$ is the decay constant, representing the probability of decay per nucleus per unit time (expressed in $\text{s}^{-1}$).

   Integrating this differential equation yields the standard exponential decay law:

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

   where $N_0$ is the initial number of parent nuclei at $t = 0$.

   Connecting Half-Life to the Decay Constant

   By definition, when $t = T_{1/2}$, the remaining population is $N = \frac{N_0}{2}$. Substituting this into the decay equation:

   $$\frac{N_0}{2} = N_0 e^{-\lambda T_{1/2}}$$

   Taking the natural logarithm of both sides gives:

   $$\ln\left(\frac{1}{2}\right) = -\lambda T_{1/2} \implies -\ln 2 = -\lambda T_{1/2}$$

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

   This relationship shows that the half-life depends solely on the inherent nuclear structure of the isotope and is independent of initial mass, temperature, chemical state, or pressure.

   Activity and Practical Calculations

   The activity ($A$) of a radioactive source is the number of decays occurring per second:

   $$A(t) = \lambda N(t)$$

   The SI unit of activity is the becquerel ($\text{Bq}$), where $1\text{ Bq} = 1\text{ decay per second}$. Activity decays following the exact same half-life exponential pattern as the number of nuclei:

   $$A(t) = A_0 \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}}$$

   Worked Example

   Problem: A sample containing $1.20 \times 10^{20}$ atoms of iodine-131 has a half-life of $8.0\text{ days}$. How many undecayed iodine-131 nuclei remain after $24.0\text{ days}$?

   Solution:

   1. Determine the number of elapsed half-lives ($n$):
      $n = \frac{t}{T_{1/2}} = \frac{24.0\text{ days}}{8.0\text{ days}} = 3$
   2. Apply the fractional decay relation:
      $N = N_0 \left(\frac{1}{2}\right)^n = (1.20 \times 10^{20}) \times \left(\frac{1}{2}\right)^3 = (1.20 \times 10^{20}) \times \frac{1}{8}$
   3. Calculate the final value:
      $N = 1.50 \times 10^{19}\text{ nuclei}$

   Thus, $1.50 \times 10^{19}$ radioactive iodine-131 nuclei remain, while the remaining $1.05 \times 10^{20}$ nuclei have transformed into stable xenon-131.