The decay constant for a radioactive nuclide is 1.5 $\times$ 10$^{-5}$ s$^{-1}$. Atomic weight of the substance is 60 g mole$^{-1}$, ($N_A=6\times10^{23}$). The activity of 1.0 $\mu$g of the substance is ___________ $\times$ 10$^{10}$ Bq.

<p>The activity of a radioactive substance is defined as the rate of decay or disintegration of the substance. It is given by the following formula:</p> <p>$A = \lambda N$</p> <p>where $A$ is the activity, $\lambda$ is the decay constant, and $N$ is the number of radioactive atoms present.</p> <p>We can use this formula to find the activity of 1.0 $\mu$g (or $10^{-6}$ g) of the substance. First, we need to find the number of radioactive atoms present in 1.0 $\mu$g of the substance. We can use the following formula to do this:</p> <p>$N = \frac{m}{M} N_A$</p> <p>where $m$ is the mass of the substance, $M$ is its molar mass, and $N_A$ is Avogadro&#39;s number.</p> <p>Substituting the given values, we get:</p> <p>$$N = \frac{1.0 \times 10^{-6} \, \text{g}}{60 \, \text{g/mol}} \times 6 \times 10^{23} \approx 10^{16} \text{ atoms}$$</p> <p>Now, we can use the formula for activity:</p> <p>$$A = \lambda N = (1.5 \times 10^{-5} \, \text{s}^{-1}) (10^{16}) = 1.5 \times 10^{11} \, \text{Bq}$$</p> <p>Therefore, the activity of 1.0 $\mu$g of the substance is 15 $\times$ 10$^{10}$ Bq.</p>