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**Problem**: Rubidium-87 decays by β-particle production to strontium-87 with a half-life of 4.7 x 1010 years. What is the age of a rock sample that contains 109.7 μg of 87Rb and 3.1 μg of 87Sr? Assume that no 87Sr was present when the rock was formed. The atomic masses for 87Rb and 87Sr are 86.90919 u and 86.90888 u, respectively.

###### FREE Expert Solution

###### FREE Expert Solution

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**

^{-1}**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**

**k =**

**1.4748**

**x10**^{-11}**yr**

^{–1}[N]_{t} = ^{87}Rb final = **7.6012x10 ^{17} atoms**

**t = ???**

Solving for **t**:

###### Problem Details

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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