A certain substance has a half-life of $5.0$ hours. How many nuclei of the substance are required to give an initial activity of $6.0 \mu \mathrm { Ci }$ ? $\left( 1 \mathrm { Ci } = 3.7 \times 10 ^ { 10 } \mathrm {~Bq} \right)$

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The activity $A$ of a radioactive substance is given by $A = \lambda N$, where $\lambda$ is the decay constant and $N$ is the number of nuclei. The decay constant $\lambda$ can be found using the half-life formula $\lambda = \frac{\ln(2)}{T_{1/2}}$, where $T_{1/2}$ is the half-life. Given $T_{1/2} = 5.0$ hours and $1 \mathrm{Ci} = 3.7 \times 10^{10} \mathrm{Bq}$, and the initial activity $A = 6.0 \mu \mathrm{Ci} = 6.0 \times 10^{-6} \mathrm{Ci} = 6.0 \times 10^{-6} \times 3.7 \times 10^{10} \mathrm{Bq}$, we can solve for $N$. Rearranging the activity formula gives $N = \frac{A}{\lambda}$. Substituting the values and solving for $N$ yields approximately $5.8 \times 10^{9}$ nuclei.

A Certain Substance Has a Half-Life Of 5.05.05.0 Hours How Many Nuclei of the Substance Are Required to Many

A Certain Substance Has a Half-Life Of $5.0$ Hours How Many Nuclei of the Substance Are Required to Many