Q. 5.84 - An Introduction to General, Organic, and Biological Chemistry

Question

A wooden object from the site of an ancient temple has a carbon- 14 activity of 10 counts/man compared with a reference piece of wood cut today that has an activity of 40 counts/min. If the half-life for carbon-14 is 5730 yr, what is the age of the ancient wood object? $(5,3,5.4)$

Step-by-Step Solution

Answer

As a result, the Ce-141 sample's initial activity is  $64.0\mu \mathrm{Ci}$

Step 1: Introduction

Half-life: The time it takes for one-half of a radioisotope to decay is called its half-life.

Ce-141  has a half-life of $32.5$ days.

The activity of the sample is$4.0\mu \mathrm{Ci}$130 days later.

Calculate a sample's starting activity in microcurie.

Step 2: Conversion factors for half-life

Calculate half-life conversion factors.

1 half-life $=32.5$ days

$\frac{32.5\text{ days }}{1\text{ half-life }}$ and $\frac{1\text{ half- life }}{32.5\text{ days }}$

Step 3: Number of half-lives

Calculate the number of half-lives in the given amount of time.

The total number of half-lives $=130$ days $\times \frac{1\text{ half-life }}{32.5\text{ days }}$

$=4$ half-lives

Ce-141 has four half-lives after completing 130 days.

Step 4: The sample present initially

Calculate the amount of sample that was initially present.

After 4 half-lives, Ce-141 present is $4.0\mu \mathrm{Ci}$.

Substitute the value in the following scheme according to the definition of half-life period.

$64.0\mu \mathrm{CiCe}-141\stackrel{1\text{ halfelife }}{⟶}32.0\mu \mathrm{CiCe}-141\stackrel{2\text{ harfines }}{⟶}16.0\mu \mathrm{CiCe}-141$

$\stackrel{3\text{ halflives }}{⟶}8.0\mu \mathrm{CiCe}-141\stackrel{4\text{ half-ies }}{⟶}4.0\mu \mathrm{CiCe}-141$

As a result, the Ce-141 sample's initial activity is $64.0\mu \mathrm{Ci}$.