Chapter 13, Problem 1MCQ

You discover that a 24,000-year-old fossil has one fourth the concentration of a radioactive isotope compared to a living organism.

What is the half-life of this isotope?

1. a. 3000 years
2. b. 6000 years
3. c. 8000 years
4. d. 12,000 years

Summary Introduction

Introduction:

The radioactive isotopes are used for the determining the age of fossils. The half-life of the radioactive isotope is used and the age of the fossil is calculated. The technique of using radioactive isotopes in determining the age of fossils is called radioactive dating.

Answer to Problem 1MCQ

Correct answer:

The half-life of the isotope is $12,000$ years. Therefore, option d. is correct.

Explanation of Solution

Reason for the correct statement:

The formula for the calculation of the half-life of an isotope where A is the remaining amount of the radioactive isotope, ${\text{A}}^{\text{o}}$ is the initial amount of radioactive isotope, it is the age of the fossil, and T is the half-life of the isotope is as follows:

${\text{A = A}}^{\circ }\times {\left(\frac{1}{2}\right)}^{\text{t/T}}$

Here, the concentration of the radioactive isotope is one-fourth that of the original matter. So, the A is equal to one fourth of ${\text{A}}^{\text{o}}$. The age of the fossil is $24,000$ years.

$\begin{array}{l}\text{A = }\frac{1}{4}{\text{A}}^{\circ }\\ \text{t}=\text{ 24,000 years}\end{array}$

$\begin{array}{c}{\text{A = A}}^{\circ }\times {\left(\frac{1}{2}\right)}^{\text{t/T}}\\ \frac{1}{4}{\text{A}}^{\circ }{\text{= A}}^{\circ }\times {\left(\frac{1}{2}\right)}^{\text{24,000/T}}\\ \frac{1}{4}={\left(\frac{1}{2}\right)}^{\text{24,000/T}}\\ {\left(\frac{1}{2}\right)}^2={\left(\frac{1}{2}\right)}^{\text{24,000/T}}\end{array}$

$\begin{array}{c}{\left(\frac{1}{2}\right)}^2={\left(\frac{1}{2}\right)}^{\text{24,000/T}}\\ \text{2 = }\frac{\text{24,000}}{\text{T}}\\ \text{T = }\frac{24,000}{2}\\ \text{T = 12,000}\end{array}$

Option d. is given as “ $12,000$ years”.

As, “the half-life for the radioactive isotope is $12,000$ years, which is present in fossil whose age is $24,000$ years and the concentration of the radioactive isotope is $\frac{1}{4}$ that of the living organism”, is the right answer.

Hence, the option d. is correct.

Reasons for the incorrect statements:

Option a. is given as “ $3000$ years”.

The half-life of the radioactive isotope will be $3000$ years when the age of the fossil is $24,000$ years and the concentration of the radioactive isotope is $\frac{1}{256}$ that of the living organism. So, it is a wrong answer.

Option b. is given as “ $6000$ years”.

The half-life of the radioactive isotope will be $6000$ years when the age of the fossil is $24,000$ years and the concentration of the radioactive isotope is $\frac{1}{16}$ that of the living organism. So, it is a wrong answer.

Option c. is given as “ $8000$ years”.

The half-life of the radioactive isotope will be $8000$ years when the age of the fossil is $24,000$ years and the concentration of the radioactive isotope is $\frac{1}{8}$ that of the living organism. So, it is a wrong answer.

Hence, options a., b., and c. are incorrect.

Conclusion

The half-life for the radioactive isotope is $12,000$ years, which is present in fossil whose age is $24,000$ years and the concentration of the radioactive isotope is $\frac{1}{4}$ that of the living organism.