Chapter 19, Problem 19.4KSP

Interpretation Introduction

Interpretation:

The age of the rock has to be determined from the given options.

Concept introduction:

• Unstable nuclei emit radiation spontaneously to become stable nuclei by losing energy. This process of emission of radiation by unstable nuclei is known as radioactive decay.
• Radioactive decay is in the first order kinetics. Rate of radioactive decay at a time t is,

$\begin{array}{c}\text{ln}\frac{{\text{N}}_{\text{t}}}{{\text{N}}_{\text{0}}}\text{=}-\text{kt}\\ \text{where,}\text{k=}\text{First}\text{order}\text{rate}\text{constant}\\ {\text{N}}_{\text{0}}\text{=}\text{number of radioactive nuclei at time zero}\\ {\text{N}}_{\text{t}}\text{=}\text{number of radioactive nuclei at a time,}\text{t}\end{array}$

Answer to Problem 19.4KSP

The correct option for the age of rocks is given in option (a) $165millionyears$

Explanation of Solution

Reason for correct option

To find: The age of the rock

The age of the rock is $165millionyears$

The age of the rock is calculated

Given,

$\begin{array}{c}Massof{}^{238}Ucontainsintherock=45.7mg\\ Massof{}^{206}Pbcontainsintherock=1.02mg\\ Atomicmassof{}^{238}U=238g=238\times {10}^{3}mg\\ Atomicmassof{}^{206}Pb=206g=206g\times {10}^{3}mg\\ \text{Rate}\text{constant}of{}^{238}U,k=1.54\times {10}^{-10}y{r}^{-1}\end{array}$

Some of ${}^{238}U$ was converted to ${}^{206}Pb$. Thus the mass of ${}^{238}U$ that converted to ${}^{206}Pb$ is,

$\begin{array}{c}massof{}^{238}U=massof{}^{206}Pb\times \frac{atomicmassof{}^{238}U}{atomicmassof{}^{206}Pb}\\ =1.02mg\times \frac{238\times {10}^{3}mg}{206g\times {10}^{3}mg}\\ =1.178\end{array}$

Hence,

The initial mass of ${}^{238}U$ is $45.7mg+1.178mg=46.878mg$

The age of the rock can be calculated by the equation.

$\text{ln}\frac{{\text{N}}_{\text{t}}}{{\text{N}}_{\text{0}}}\text{=}-\text{kt}$

Therefore,

The age of the rock is,

$\begin{array}{c}t=-\left(\frac{\ln \left(\frac{N_t}{N_0}\right)}{k}\right)\\ =-\left(\frac{\ln \left(\frac{45.7mg}{46.878mg}\right)}{1.54\times {10}^{-10}y{r}^{-1}}\right)\\ =-\left(\frac{\ln 0.9749}{1.54\times {10}^{-10}y{r}^{-1}}\right)\\ =-\left(\frac{-\left(0.02542\right)}{1.54\times {10}^{-10}y{r}^{-1}}\right)\\ =165\times {10}^{6}yr\\ =165millionyears\end{array}$

The age of the rock is $165millionyears$

Hence the correct option for the age of the rock is option a.

Reasons for incorrect options:

Write all the age of the rock given in the options except option a.

(b) $2.50billionyears$

(c) $143millionyears$

(d) $6.49billionyears$

(e) $63.7millionyears$

The age of the rock is found as $165millionyears$.

So, this value of the age of the rock is not matching with any of the above given the age of the rock in remaining options.

Hence the incorrect options are b, c, d and e

Conclusion

The correct option for the age of the rock is identified from the given options. The age of the rock is determined by using the data’s given for the radioactive decay.