Chapter 19, Problem 79CP

Interpretation Introduction

Interpretation: Half life of ${}^{\text{238}}\text{U}$ and ${}^{\text{235}}\text{U}$ and their relative abundances is given. Their relative abundances when earth was formed $4.5$ billion years ago are to be calculated.

Concept introduction: A process through which an unstable nuclide looses its energy due to excess of protons or neutrons is known as radioactive decay. Decay constant is the quantity that expresses the rate of decrease of number of atoms of a radioactive element per second. Half life of radioactive sample is defined as the time required for the number of nuclides to reach half of the original value.

To determine: The relative abundances of ${}^{\text{238}}\text{U}$ and ${}^{\text{235}}\text{U}$ when earth was formed.

Answer to Problem 79CP

Answer

The relative abundance of ${}^{\text{238}}\text{U}$ when earth was formed is $\underset{\_}{76\%}$.

The relative abundance of ${}^{\text{235}}\text{U}$ when earth was formed is $\underset{\_}{24\%}$.

Explanation of Solution

Explanation

The decay constant is calculated by the formula given below.

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

Where

• $t_{1/2}$ is the half life of nuclide.
• $\lambda$ is decay constant.

The value of $t_{1/2}$ for ${}^{\text{238}}\text{U}$ is $4.5\times {10}^9\text{years}$.

Substitute the value of half life in the above formula.

$\lambda =\frac{0.693}{4.5\times {10}^9}{\text{years}}^{\text{-1}}$

The decay constant is $\frac{\text{0}\text{.693}}{\text{4}{\text{.5×10}}^{\text{9}}}{\text{years}}^{\text{-1}}$.

The decay constant is calculated by the formula given below.

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

Where

• $t_{1/2}$ is the half life of nuclide.
• $\lambda$ is decay constant.

The value of $t_{1/2}$ for ${}^{\text{235}}\text{U}$ is $7.1\times {10}^8\text{years}$.

Substitute the value of half life in the above formula.

$\lambda =\frac{0.693}{7.1\times {10}^8}{\text{years}}^{\text{-1}}$.

The decay constant is $\frac{\text{0}\text{.693}}{\text{7}{\text{.1×10}}^{\text{8}}}{\text{years}}^{\text{-1}}$.

The relative abundance of ${}^{\text{238}}\text{U}$ is calculated by the formula,

$\ln \left(\frac{n}{n_0}\right)=-\lambda \cdot t$

Where

• $n_0$ is the amount of ${}^{\text{238}}\text{U}$ when earth was formed.
• $n$ is the amount of ${}^{\text{238}}\text{U}$ after $4.5\times {10}^9$ years.
• $t$ is the decay time.

The value of $t$ is $4.5\times {10}^9$ years.

Substitute the values of $t$ and decay constant in the above equation.

$\begin{array}{c}\ln \left(\frac{n}{n_0}\right)=-\lambda \cdot t\\ \ln \left(\frac{n}{n_0}\right)=-\frac{0.693\times 4.5\times {10}^9}{4.5\times {10}^9}\\ \ln \left(\frac{n}{n_0}\right)=-0.693\\ \left(\frac{n}{n_0}\right)=0.50\end{array}$

The value of $\left(\frac{n}{n_0}\right)$ for ${}^{\text{238}}\text{U}$ is $0.50$.

The relative abundance of ${}^{\text{235}}\text{U}$ is calculated by the formula,

$\ln \left(\frac{n}{n_0}\right)=-\lambda \cdot t$

Where

• $n_0$ is the amount of ${}^{\text{235}}\text{U}$ when earth was formed.
• $n$ is the amount of ${}^{\text{235}}\text{U}$ after $4.5\times {10}^9$ years.
• $t$ is the decay time.

The value of $t$ is $7.1\times {10}^8$ years.

Substitute the values of $t$ and decay constant in the above equation.

$\begin{array}{c}\ln \left(\frac{n}{n_0}\right)=-\lambda \cdot t\\ \ln \left(\frac{n}{n_0}\right)=-\frac{0.693\times 4.5\times {10}^9}{7.1\times {10}^8}\\ \left(\frac{n}{n_0}\right)={\text{e}}^{-4.39}\\ \left(\frac{n}{n_0}\right)=0.012\end{array}$

The value of $\left(\frac{n}{n_0}\right)$ for ${}^{\text{235}}\text{U}$ is $0.012$.

The natural composition of ${}^{\text{238}}\text{U}$ is $99.28\%=\frac{9928}{1000}$

Therefore, $9928$ nuclei of ${}^{\text{238}}\text{U}$ are present in $10000$ uranium nuclei.

Therefore, the value of $n$ for ${}^{\text{238}}\text{U}$ is $9928$.

Substitute the value of $n$ in the equation,

$\begin{array}{c}\left(\frac{n}{n_0}\right)=0.50\\ n_0=\frac{n}{0.50}\\ n_0=\frac{9928}{0.50}=\text{1}{\text{.9×10}}^{\text{4}}{}^{\text{238}}\text{U}\text{nuclei}\end{array}$

The amount of ${}^{\text{238}}\text{U}$ when earth was formed is $\text{1}{\text{.9×10}}^{\text{4}}{}^{\text{238}}\text{U}\text{nuclei}$

The natural composition of ${}^{\text{235}}\text{U}$ is $0.72\%=\frac{72}{1000}$

Therefore, $72$ nuclei of ${}^{\text{235}}\text{U}$ are present in $10000$ uranium nuclei.

Therefore, the value of $n$ for ${}^{\text{235}}\text{U}$ is $72$.

Substitute the value of $n$ in the equation,

$\begin{array}{c}\left(\frac{n}{n_0}\right)=0.012\\ n_0=\frac{n}{0.012}\\ n_0=\frac{72}{0.012}={\text{6×10}}^{\text{3}}{}^{\text{235}}\text{U}\text{nuclei}\end{array}$

The amount of ${}^{\text{235}}\text{U}$ when earth was formed is ${\text{6×10}}^{\text{3}}{}^{\text{235}}\text{U}\text{nuclei}$.

The composition of ${}^{\text{238}}\text{U}$ from $4.5$ billion years ago is calculated by the formula,

$\begin{array}{c}\text{Composition of}{}^{\text{238}}\text{U}=\frac{{\text{Amount of }}^{\text{238}}\text{U when earth was formed}}{\text{Amount}\text{of}\left({}^{\text{238}}{\text{U+}}^{\text{235}}\text{U}\right)\text{ when}\text{earth}\text{was}\text{formed}}\times 100 \\ =\frac{1.9\times {10}^4}{1.9\times {10}^4+6\times {10}^3}\times 100=76\%\end{array}$

The composition of ${}^{\text{238}}\text{U}$ from $4.5$ billion years ago is $\underset{\_}{76\%}$ .

The composition of ${}^{\text{235}}\text{U}$ from $4.5$ billion years ago is calculated by the formula,

$\begin{array}{c}\text{Composition of}{}^{\text{235}}\text{U}=\frac{{\text{Amount of}}^{\text{235}}\text{U when earth was formed}}{\text{Amount}\text{of}\left({}^{\text{238}}{\text{U+}}^{\text{235}}\text{U}\right)\text{when}\text{earth}\text{was}\text{formed}}\times 100 \\ =\frac{6\times {10}^3}{1.9\times {10}^4+6\times {10}^3}\times 100=24\%\end{array}$

The composition of ${}^{\text{238}}\text{U}$ from $4.5$ billion years ago is $\underset{\_}{24\%}$ .

Conclusion

Conclusion

The relative abundance of ${}^{\text{238}}\text{U}$ when earth was formed is $\underset{\_}{76\%}$ .

The relative abundance of ${}^{\text{235}}\text{U}$ when earth was formed is $\underset{\_}{24\%}$ .