Step 1: Recall the relation between activity and number of atoms.
The activity $A$ of a radioactive substance is the number of disintegrations per second. It is related to the number of atoms $N$ present and the decay constant $\lambda$ by \[ A = \lambda N \] Rearranging to find $N$: $N = \frac{A}{\lambda}$.
Step 2: Write the relation between decay constant and half-life.
The decay constant and half-life $T_{1/2}$ are related by \[ \lambda = \frac{0.693}{T_{1/2}} \] This tells us how quickly the substance decays: a larger $\lambda$ means faster decay and shorter half-life.
Step 3: Convert the half-life to seconds.
Given $T_{1/2} = 10^8$ years. We need to convert this to seconds since the activity is in Bq (disintegrations per second). \[ 1\ \text{year} = 365 \times 24 \times 3600 = 3.15 \times 10^7\ \text{s} \] \[ T_{1/2} = 10^8 \times 3.15 \times 10^7 = 3.15 \times 10^{15}\ \text{s} \]
Step 4: Calculate the decay constant.
\[ \lambda = \frac{0.693}{3.15 \times 10^{15}} \approx 2.2 \times 10^{-16}\ \text{s}^{-1} \]
Step 5: Calculate the number of atoms.
Given activity $A = 10^4$ Bq, \[ N = \frac{A}{\lambda} = \frac{10^4}{2.2 \times 10^{-16}} \approx 4.5 \times 10^{19} \]
Step 6: State the final answer.
The number of atoms of the radioactive substance present is \[ \boxed{N \approx 4.5 \times 10^{19}} \]