Chapter 24.2, Problem 18SSC

Interpretation Introduction

Interpretation:

The mass of the sample of polonium-214 is 1.0 g. The percentage of mass remaining after one, two and three half-lives needs to be determined. The percentage remaining versus the number of half-lives should be plotted. The time required for 20% of the original sample to remain needs to be determined.

Concept introduction:

The half lifetime is defined as the time required reducing the amount of a reactant to half of its initial value.

Explanation of Solution

The mass of the sample remaining after 1 half-life will be half of the initial mass that is 0.5 g. now, after another half-life, the mass becomes half of 0.5 g that is 0.25 g. After the third half-life, the mass becomes half of 0.25 g that is 0.125 g.

Therefore, the mass of the sample remaining after one, two and three half-lives is 0.5 g, 0.25 g and 0.125g respectively.

Now, the initial mass is 1 g thus, the percentage of masses will be 100%, 50%, 25% and 12.5%.

The graph can be plotted from the percentage of mass remaining versus the number of half-lives as follows:

Percentage of mass remaining	Number of half-lives
100 %	0
50%	1
25%	2
12.5%	3

When 20% of the original mass remains, the value of $A_t$ can be calculated as follows:

$A_t=20\%A_o=\frac{20}{100}\times 1\text{ g}=\text{0}\text{.20}$

Now, from the graph, the elapsed time for 20% of the remaining mass will be approximately 2.25 half-lives.