Chapter 19, Problem 76CWP

Interpretation Introduction

Interpretation: The half life of Rubidium- $87$ is given. A rock containing $109.7\text{μg}$ of ${}^{\text{87}}\text{Rb}$ and $3.1\text{μg}$ of ${}^{\text{87}}\text{Sr}$ is given. Assuming no ${}^{\text{87}}\text{Sr}$ was originally present; the age of the rock is to be calculated.

Concept introduction: A process through which an unstable nuclide loses its energy due to excess of protons or neutrons is known as radioactive decay.

To determine: The age of the rock.

Answer to Problem 76CWP

Solution

The age of rock is $\underset{\_}{1.9\times 10^{9}years}$.

Explanation of Solution

Explanation

Given

The value of half life of ${}^{\text{87}}\text{Rb}$ is $4.5\times {10}^9\text{years}$.

The decay constant is calculated by the formula,

$\lambda =\frac{0.693}{t_{1/2}}$

Where

• $t_{1/2}$ is the half life of ${}^{\text{87}}\text{Rb}$.
• $\lambda$ is decay constant.

Substitute the value of half life in the above formula.

$\begin{array}{c}\lambda =\frac{0.693}{4.7\times {10}^{10}}{\text{years}}^{-1}\\ =\underset{\_}{1.4\times 10^{-11}year{s}^{-1}}\end{array}$.

The decay constant is $\underset{\_}{1.4\times 10^{-11}year{s}^{-1}}$.

Explanation

The number of moles of ${}^{\text{87}}\text{Rb}$ is $\underset{\_}{1.262\times 10^{-6}mol}$.

The number of moles of ${}^{\text{87}}\text{Sr}$ is $\underset{\_}{0.035\times 10^{-6}mol}$.

Given

The mass of ${}^{\text{87}}\text{Rb}$ in the rock sample is $109.7\text{μg}$

The mass of ${}^{\text{87}}\text{Sr}$ in the rock sample is $3.1\text{μg}$

The conversion of $\text{μg}$ into $\text{g}$ is done as,

$\text{1}\text{μg}={10}^{-6}\text{g}$

Therefore, the conversion of $109.7\text{μg}$ into $\text{g}$ is,

$109.7\text{μg}=109.7\times {10}^{-6}\text{g}$

Similarly, the conversion of $3.1\text{μg}$ is done as,

$3.1\text{μg}=3.1\times {10}^{-6}\text{g}$

Molar mass of ${}^{\text{87}}\text{Rb}$ is $\text{86}\text{.90919 g/mol}$.

Number of moles of ${}^{\text{87}}\text{Rb}$ is calculated by the formula,

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Substitute the value of given mass and molar mass of ${}^{\text{87}}\text{Rb}$ in the above equation.

$\begin{array}{c}\text{Number of moles}=\frac{109.7\times {10}^{-6}\text{g}}{\text{86}\text{.90919 g/mol}}\\ =\underset{\_}{1.262\times 10^{-6}mol}\end{array}$

Molar mass of ${}^{\text{87}}\text{Sr}$ is $\text{86}\text{.90888 g/mol}$.

Number of moles of ${}^{\text{87}}\text{Sr}$ is calculated by the formula,

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Substitute the value of given mass and molar mass of ${}^{\text{87}}\text{Sr}$ in the above equation.

$\begin{array}{c}\text{Number of moles}=\frac{3.1\times {10}^{-6}\text{g}}{\text{86}\text{.90888 g/mol}}\\ =\underset{\_}{0.035\times 10^{-6}mol}\end{array}$

Explanation

The age of rock is $\underset{\_}{1.9\times 10^{9}years}$.

One mole of ${}^{\text{87}}\text{Rb}$ decays to one mole of ${}^{\text{87}}\text{Sr}$ in the beta decay, therefore, the total amount of ${}^{\text{87}}\text{Rb}$ in the rock sample initially present $\left(N_0\right)$ is calculated by the formula,

${\text{Amount of}}^{\text{87}}{\text{Rb and}}^{\text{87}}\text{Sr present initially}={\text{Amount of }}^{\text{87}}\text{Rb}+{\text{Amount of }}^{\text{87}}\text{Sr }$

Substitute value of amount of ${}^{\text{87}}\text{Rb}$ and ${}^{\text{87}}\text{Sr}$ in the above equation.

$\begin{array}{c}{\text{Amount of}}^{\text{87}}{\text{Rb and}}^{\text{87}}\text{Sr present initially}=1.262\times {10}^{-6}\text{ mol}+0.035\times {10}^{-6}\text{ mol}\\ =1.297\times {10}^{-6}\text{ mol}\end{array}$

The amount of ${}^{\text{87}}\text{Rb}$ present now $\left(N\right)$ is $1.262\times {10}^{-6}\text{ mol}$.

The age of the rock is calculated by the formula,

$\ln \left(\frac{N}{N_0}\right)=-\lambda t$

Where,

• $N$ is the amount of ${}^{\text{87}}\text{Rb}$ and ${}^{\text{87}}\text{Sr}$ present now.
• $N_0$ is the amount of ${}^{\text{87}}\text{Rb}$ present originally.
• $t$ is the decay time.
• $\lambda$ is the decay constant.

Substitute the values of $N$, $N_0$, $\lambda$ in the above equation.

$\begin{array}{c}\ln \left(\frac{N}{N_0}\right)=-\lambda t\\ \ln \frac{1.262\times {10}^{-6}}{1.297\times {10}^{-6}}=-1.4\times {10}^{-11}\times t\\ -0.027=-1.4\times {10}^{-11}\times t\\ t=\underset{\_}{1.9\times 10^{9}years}\end{array}$

Therefore, the age of rock is $\underset{\_}{1.9\times 10^{9}years}$.

Conclusion

Conclusion

The age of rock is $\underset{\_}{1.9\times 10^{9}years}$.