Chapter 30, Problem 75GP

(a)

To determine

To find: The percentage of $\text{U}$ after 100 million years ago

Answer to Problem 75GP

  $1.63\%$

Explanation of Solution

Given data:

Half-life of ${}_{92}^{235}\text{U}$ is $7.04\times {10}^8\text{yr}$

Half-life of ${}_{92}^{238}\text{U}$ is $4.468\times {10}^9\text{yr}$

Formula used:

Total number N of radioactive (parent) nucleipresent:

  $\begin{array}{l}N=N_0{e}^{-\lambda t}\\ \Rightarrow N_0=N{e}^{\lambda t}=N{e}^{\frac{t}{T_{1/2}}\ln 2}\end{array}$

Calculation:

Today per 100 gram of total uranium, mass of ${}_{92}^{235}\text{U}$ is $0.720\text{g}$ ,

and ${}_{92}^{238}\text{U}$ is $100.000-0.720=99.280\text{g}$

Relate the amounts today to the amounts $1.0\times {10}^9$ years ago.

  ${\left(N_0\right)}_{235}=\left(N_{235}\right)e^{\frac{t}{T_{1/2}}\ln 2}=(0720\text{g})e^{\frac{\left(1.0\times {10}^9\text{yr}\right)}{\left(7.04\times {10}^8\text{yr}\right)}\ln 2}=1.927\text{g}$

  ${\left(N_0\right)}_{238}=\left(N_{238}\right)e^{\frac{t}{T_{1/2}}\ln 2}=(99.280\text{g})e^{\frac{\left(1.0\times {10}^9\right)}{\left(4.468\times {10}^9\right)}\ln 2}=115.94\text{g}$

  ${\text{ The percentage of }}_{92}^{235}\text{U was }\frac{1.927}{1.927+115.94}\times 100\%=1.63\%$

Conclusion:

The percentage of $\text{U}$ after 100 million years is $1.63\%$

(b)

To determine

To find:The percentage of $\text{U}$ after 100 million years from now

Answer to Problem 75GP

  $0.663\%$

Explanation of Solution

Given data:

Half-life of ${}_{92}^{235}\text{U}$ is $7.04\times {10}^8\text{yr}$

Half-life of ${}_{92}^{238}\text{U}$ is $4.468\times {10}^9\text{yr}$

Formula used:

Total number N of radioactive (parent) nucleipresent:

  $\begin{array}{l}N=N_0{e}^{-\lambda t}\\ \Rightarrow N_0=N{e}^{\lambda t}=N{e}^{\frac{t}{T_{1/2}}\ln 2}\end{array}$

Calculation:

Today per 100 gram of total uranium, mass of ${}_{92}^{235}\text{U}$ is $0.720\text{g}$ ,

and ${}_{92}^{238}\text{U}$ is $100.000-0.720=99.280\text{g}$

Relate the amounts today to the amounts $100\times {10}^6$ years from now.

  $N=N_0{e}^{-\lambda t}\to \left(N_{235}\right)={\left(N_0\right)}_{235}{e}^{-\frac{t}{T_{1/2}}\ln 2}=(0.720\text{g})e^{-\frac{\left(100\times {10}^6\text{yr}\right)}{\left(7.04\times {10}^8\text{yr}\right)}\ln 2}=0.6525\text{g}$

  $\left(N_{238}\right)={\left(N_0\right)}_{238}{e}^{-\frac{t}{T_{1/2}}\ln 2}=(99.280\text{g})e^{-\frac{\left(100\times {10}^6\text{yr}\right)}{\left(4.468\times {10}^9\text{yr}\right)}\ln 2}=97.752\text{g}$

The percentage of ${}_{92}^{235}\text{U}$ will be $\frac{0.6525}{0.6525+97.752}\times 100\%=0.663\%$

Conclusion:

The percentage of $\text{U}$ after 100 million years from now $0.663\%$