Recall that ** radioactive/nuclear decay of isotopes** follows first-order kinetics, and the integrated rate law for first-order reactions is:

$\overline{){\mathbf{ln}}{\mathbf{}}{\mathbf{\left[}\mathbf{N}\mathbf{\right]}}_{{\mathbf{t}}}{\mathbf{=}}{\mathbf{-}}{\mathbf{kt}}{\mathbf{+}}{\mathbf{ln}}{\mathbf{}}{\mathbf{\left[}\mathbf{N}\mathbf{\right]}}_{{\mathbf{0}}}}$

where:

**[N] _{t}** = concentration at time t

**k** = decay constant

**t** = time

**[N] _{0}** = initial concentration.

Also, recall that ** half-life** is the time needed for the amount of a reactant to decrease by 50% or one-half.

The half-life of a first-order reaction is given by:

$\overline{){{\mathbf{t}}}_{\raisebox{1ex}{$\mathbf{1}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{2}$}\right.}{\mathbf{=}}\frac{\mathbf{ln}\mathbf{}\mathbf{2}}{\mathbf{k}}}$

We first need to calculate for the decay constant using the given half-life of ^{87}Rb, ** 4.7 x 10^{10} years**:

$\mathbf{k}\mathbf{=}\frac{\mathbf{ln}\mathbf{}\mathbf{2}}{{\mathbf{t}}_{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}\mathbf{=}\frac{\mathbf{ln}\mathbf{}\mathbf{2}}{\mathbf{4}\mathbf{.}\mathbf{7}\mathbf{\times}{\mathbf{10}}^{\mathbf{10}}\mathbf{}\mathbf{yr}}$

**k = 1.4748x10 ^{-11} yr**

**Calculate the initial amount of ^{87}Rb atoms present:**

**initial amount of ^{87}Rb = final amount of ^{87}Rb + amount of ^{87}Sr**

*amu is approximately equal to molar mass (g/mol)*

${}^{\mathbf{87}}\mathbf{Rb}\mathbf{}\mathbf{atoms}\mathbf{=}\mathbf{109}\mathbf{.}\mathbf{7}\mathbf{}\overline{)\mathbf{\mu g}}\mathbf{\times}\frac{{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{}\overline{)\mathbf{g}}}{\mathbf{1}\mathbf{}\overline{)\mathbf{\mu g}}}\mathbf{\times}\frac{\mathbf{1}\mathbf{}\overline{)\mathbf{mol}}}{\mathbf{86}\mathbf{.}\mathbf{90919}\mathbf{}\overline{)\mathbf{g}}}\mathbf{\times}\frac{\mathbf{6}\mathbf{.}\mathbf{022}\mathbf{\times}{\mathbf{10}}^{\mathbf{23}}\mathbf{}\mathbf{atoms}}{\mathbf{1}\mathbf{}\overline{)\mathbf{mol}}}$

^{87}Rb **= 7.6012x10 ^{17} atoms**

${}^{\mathbf{87}}\mathbf{Sr}\mathbf{}\mathbf{atoms}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{1}\mathbf{}\overline{)\mathbf{\mu g}}\mathbf{\times}\frac{{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{}\overline{)\mathbf{g}}}{\mathbf{1}\mathbf{}\overline{)\mathbf{\mu g}}}\mathbf{\times}\frac{\mathbf{1}\mathbf{}\overline{)\mathbf{mol}}}{\mathbf{86}\mathbf{.}\mathbf{90888}\mathbf{}\overline{)\mathbf{g}}}\mathbf{\times}\frac{\mathbf{6}\mathbf{.}\mathbf{022}\mathbf{\times}{\mathbf{10}}^{\mathbf{23}}\mathbf{}\mathbf{atoms}}{\mathbf{1}\mathbf{}\overline{)\mathbf{mol}}}$

^{87}Sr **= 2.148x10 ^{16} atoms**

**initial amount of ^{87}Rb = 7.6012x10^{17} atoms + 7.6012x10^{17} atoms**

**initial amount of ^{87}Rb = 7.816x10^{17} atoms**

**Calculate the age of a rock sample:**

[N]_{0} = ^{87}Rb initial = **7.816x10 ^{17} atoms**

[N]_{t} = ^{87}Rb final = **7.6012x10 ^{17} atoms**

Solving for **t**:

Rubidium-87 decays by *β*-particle production to strontium-87 with a half-life of 4.7 x 10^{10} years. What is the age of a rock sample that contains 109.7 μg of ^{87}Rb and 3.1 μg of ^{87}Sr? Assume that no ^{87}Sr was present when the rock was formed. The atomic masses for ^{87}Rb and ^{87}Sr are 86.90919 u and 86.90888 u, respectively.

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Our data indicates that this problem or a close variation was asked in Chemistry: An Atoms First Approach - Zumdahl Atoms 1st 2nd Edition. You can also practice Chemistry: An Atoms First Approach - Zumdahl Atoms 1st 2nd Edition practice problems.