How Do You Use the Radioactive Decay Law? (HL)

At HL, use the radioactive decay law to calculate the number of undecayed nuclei remaining after time $t$: $N=N_0e^{-\lambda t}$. Substitute consistent values for $N_0$, $\lambda$, and $t$, then evaluate the exponential.

This is part of IB Physics E.3 Radioactive decay, specifically the Additional higher level content.

The negative exponent means that $N$ decreases exponentially. A larger decay constant means a greater probability of decay per unit time and therefore faster decay.

The law follows from the fact that the rate of decrease of $N$ is proportional to $N$: $\frac{dN}{dt}=-\lambda N$. Integrating this relationship produces the exponential model. Consequently, equal time intervals remove equal fractions, not equal numbers, of nuclei because is constant.

For example, suppose $N_0=8.0\times10^{20}$, $\lambda=0.025\ \text{day}^{-1}$, and :

$$N=(8.0\times10^{20})e^{-(0.025)(40)}=2.9\times10^{20}$$

Therefore, $2.9\times10^{20}$ nuclei remain. The fraction remaining is $N/N_0=e^{-1}=0.368$, or . The number decayed is , or . Always identify whether the question asks for remaining or decayed nuclei.

To find elapsed time, rearrange using natural logarithms:

$$t=\frac{1}{\lambda}\ln\left(\frac{N_0}{N}\right)$$

The half-life is $T_{1/2}=\frac{\ln 2}{\lambda}$. Here, $$, so 40 days is about 1.44 half-lives. Since , activity also follows .

When activity data are given, first subtract any background count rate before applying the model. Because activity is proportional to $N$, the same remaining fraction applies: $A/A_0=N/N_0$. For example, after one half-life, both and corrected activity fall to one-half of their initial values, although individual decays remain unpredictable.

A common misconception is that the law predicts when one nucleus will decay. It does not: decay is random and spontaneous, while the law predicts a large sample statistically.

Exam technique: For a Calculate question, show the equation, substitution, and answer. Check that $\lambda t$ is dimensionless, and use $\ln$, not base-10 logarithms.