Question

If$\mathbf{2}\mathbf{.8}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{10}}\mathit{\%}$ of the atoms of a radioactive isotope disintegrate in$\mathbf{1}\mathbf{.0}\mathbf{\text{yr}}$ , what is the decay constant of the process?

Short answer

The decay constant value is$2.8\times {10}^{-12}{\text{year}}^{-1}.$

Step-by-step solution

Definition of decay constant

The size of a population of radioactive atoms and the pace at which the population diminishes due to radioactive decay are proportionate.

Calculation of the decay constant

A: Activity, the amount of particles that have degraded during time.

N: Number of particles

k: decay constant

Every day, one trillionth of the atoms dissolve.

$\begin{array}{c}\text{A}=2.8\times {10}^{-10}\%\text{decays}/\text{year}\cdot \text{N}\\ =\frac{2.8\times {10}^{-10}\text{decays}/\text{year}}{100}\cdot \text{N}\\ =2.8\times {10}^{-12}\text{decays}/\text{year}\cdot \text{N.}\end{array}$

Now, using this equation, compute the decays constant.

$\begin{array}{c}\text{A}=\text{k}\cdot \text{N}\\ \text{k}=\frac{\text{A}}{\text{N}}\\ =\frac{2.8\times {10}^{-12}\text{decays/year}\cdot \text{N}}{\text{N}}\\ =2.8\times {10}^{-12}{\text{year}}^{-1}.\end{array}$

Therefore, the computed value is $2.8\times {10}^{-12}{\text{year}}^{-1}.$