Radioactive decay follows first-order kinetics . The activity ($A$) of a sample is determined by the product of the decay constant ($\lambda$) and the number of nuclei ($N$), such that $A = \lambda N$. The relationship between the decay constant and half-life ($T_{1/2}$) is given by $\lambda = \frac{\ln 2}{T_{1/2}} \approx \frac{0.693}{T_{1/2}}$ .

Given data:

• $T_{1/2} = 1.4 \times 10^{17}\text{ s}$
• $N = 2.0 \times 10^{21}$

Calculation:

1. First, find $\lambda$: $\lambda = \frac{0.693}{1.4 \times 10^{17}}\text{ s}^{-1}$.
2. Then, find $A$: $A = \left(\frac{0.693}{1.4 \times 10^{17}}\right) \times (2.0 \times 10^{21})$.
3. Simplifying the powers of ten and the coefficients: $A = \left(\frac{0.693}{1.4}\right) \times 2.0 \times 10^{(21-17)} = \left(\frac{0.693}{0.7}\right) \times 10^4 \approx 0.99 \times 10^4\text{ Bq}$.

The activity is nearly equal to $10^4\text{ Bq}$.