Radioactive Decay Law

First find the decay constant from the half-life: $\lambda{} =\frac{ \ln{}(2)}{T}\frac{1}{2}$. Then insert λ and t into the exponential; both must carry the same unit of time. Example iodine-131 (T½ = 8.02 d): $\lambda{} =\frac{ 0.693}{8.02} \approx{} 0.0864$ per day. After 24.1 days $N = N_{0}\cdot{}e^{(-0.0864 \times{} 24.1)} \approx{} 0.125\cdot{}N_{0}$ remains, one eighth. You can check the result: 24.1 d is exactly three half-lives, and $(\frac{1}{2})^{3} =\frac{ 1}{8}$. For neat multiples of the half-life the power form $N = N_{0}\cdot{}(\frac{1}{2})^{(\frac{t}{T}\frac{1}{2})}$ is usually quicker than the exponential.

The half-life T½ is the time after which, on average, half of the originally present nuclei have decayed. It is linked to the decay constant via $T_{1/2} =\frac{ \ln{}(2)}{\lambda{}}$ and is characteristic for each nuclide: C-14 has 5,730 years, iodine-131 about 8 days, polonium-214 only fractions of a millisecond. The multiplicative character is important: after each further half-life the stock halves again; after two, 25% remains, after four, 6.25%; the substance is never exactly "gone". For a single nucleus T½ is only a probability statement: it decays with 50% probability within one half-life, but it does not age; the probability always stays the same.

Living organisms constantly exchange carbon with the atmosphere and thus maintain a known C-14 fraction. With death the exchange stops, and the incorporated C-14 decays with $T_{1/2} = 5\,730$ years while the stable C-12 remains. Measuring the ratio today, the remaining fraction reveals the age: $t = -\frac{\ln{}(\frac{N}{N_{0}})}{\lambda{}}$. Example: if a sample still shows 25% of the original C-14, two half-lives have passed, about 11,460 years. In practice the method works up to about 50,000 years; after that too little C-14 is left. For older finds one uses longer-lived clocks such as potassium-argon (T½ = 1.25 billion years) with the same calculation principle.

The decay constant λ is the decay probability per nucleus and unit time, a fixed material property with unit $\frac{1}{s}$. The activity A, by contrast, is the actual number of decays per second in a concrete sample: $A = \lambda{}\cdot{}N$, measured in becquerels (1 Bq = 1 decay/s). It therefore depends on the sample size and decreases exponentially with time just like N: $A(t) = A_{0}\cdot{}e^{(-\lambda{}t)}$. Example: $10^{15}$ nuclei with $\lambda{} = 0.0864 d^{-1} = 10^{-6} s^{-1}$ have $A = 10^{9} Bq$. In measurement problems you almost always work with the activity, because counters register decays per time; the decay law holds for A and N in identical form.

No. Radioactive decay is a fundamentally random quantum process. For a single nucleus only a probability can be stated: per unit time it decays with probability λ, regardless of how long it has already existed. Nuclei do not age; an "old" nucleus is as ready to decay as a freshly created one. Only for very many nuclei does chance turn into a precise law: with $10^{20}$ particles the relative fluctuations are vanishingly small, and $N(t) = N_{0}\cdot{}e^{(-\lambda{}t)}$ describes the stock practically exactly. It is the same principle as rolling dice: a single roll is unpredictable, the average of millions of rolls very accurate. No external influence, temperature, pressure or chemical bonding, changes λ measurably.