Q24.38 P - Chemistry: Molecular Nature Of Matter And Change

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?

Step-by-Step Solution

Answer

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

Step 1: 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.

Step 2: 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 $2.8\times {10}^{-12}{\text{ year }}^{-1}.$