2.4: Half-Life

Learning Objectives

• To define half-life.
• To solve word problems involving half-life, time, and mass of a radioactive substance.

Some isotopes are stable indefinitely, while others are radioactive and decay over time. As time passes, less and less of the radioactive isotope will be present, and the level of radioactivity decreases. An interesting and useful aspect of radioactive decay is the half-life. The half-life of a radioactive isotope is the amount of time it takes for one-half of the radioactive isotope to decay. The half-life of a specific radioactive isotope is constant; it is not affected by conditions and is independent of the initial amount of that isotope. This can be represented by 

\[\mathrm{total\: time=number\: of\: half-lives \times length\; of\; half-life} \label{Eq1}\]

where the number of half-lives is an integer, the length of half-life is the duration of each half-life, and the total time how long the decay of the isotope was observed. If the number of half-lives is not an integer a different equation is needed. This book will only use examples in which the number of half-lives is a whole number. 

Consider the following example. Suppose we have 100.0 g of ³H (tritium, a radioactive isotope of hydrogen). It has a half-life of 12.3 years. After 12.3 y, half of the sample will have decayed to ³He by emitting a beta particle, so that only 50.0 g of the original ³H remains. After another 12.3 y—making a total of 24.6 y—another half of the remaining ³H will have decayed, leaving 25.0 g of ³H. After another 12.3 y—now a total of 36.9 y—another half of the remaining ³H will have decayed, leaving 12.5 g of ³H. This sequence of events is illustrated in Figure \(\PageIndex{1}\).

Figure \(\PageIndex{1}\): Radioactive Decay. During each successive half-life, half of the initial amount will radioactively decay.

Please note that "amount remaining" is referring specifically to the amount of ³H that is remaining. As it undergoes beta decay it turns into helium according to this equation: \(^3_1H \rightarrow ^0_{-1}\beta + ^3_2He\). Atoms still exist, but they are helium instead of hydrogen. Throughout this section we will focus on the parent and generally omit information about the daughter. 

We can keep track of the mass of parent that remains by using a table. After each half-life the mass of the parent isotope has decreased by a factor of two. Notice that the initial mass of the parent isotope corresponds to zero half-lives. The initial mass is the mass before the first half-life has occurred.   

Table \(\PageIndex{1}\):Mass of hydrogen-3 remaining after each half-life in its beta decay reaction

Number of Half-Lives	Mass of ^3H remaining
0	100.0 g
1	50.0 g
2	25.0 g
3	12.5 g
4	6.25 g
5	3.13 g
6	1.56 g

Alternatively, we can determine the amount of a radioactive isotope remaining after a given number half-lives by using the following expression:

\[\mathrm{amount\: remaining=initial\: amount\times \left ( \dfrac{1}{2} \right )^n} \label{Eq2}\]

where n is the number of half-lives. This expression works even if the number of half-lives is not a whole number whereas the table (dividing by 2 repeatedly) is only accurate when the number of half-lives is an integer. 

To solve half-life problems involving the initial mass, final mass, half-life, and total time where you are given three of the values and asked to solve for the fourth, the general procedure is: 

1. Calculate the number of half-lives. How you do this depends on whether you were provided with the initial mass and final mass or whether you were provided with the half-life and total time. 
2. Use the number of half-lives that you calculated from step 1 along with the additional information that was provided in the problem (and not yet used) to solve for the requested value. 

Example \(\PageIndex{1}\): Fluorine-20

The half-life of ²⁰F is 11.0 seconds. If a sample initially contains 5.00 g of ²⁰F, how much ²⁰F remains after 44.0 s?

Solution

Step 1: Since two times were provided, but only one mass, use the times to calculate the number of half-lives that pass during this problem. The wording "the half-life is" tells us the length of each half-life; in this case it is 11.0 s. The 44.0 s is the total time. Rearrange \ref{Eq1} to solve for the number of half-lives which will be a small whole number. 

total time = number of half-lives x length of half-life

44.0 s = number of half-lives x 11.0 s

number of half-lives = 4 half-lives

Step 2: Now we know that the fluorine-20 decays to half its mass four times (the number of half-lives). We know the original mass and are solving for the final mass. 

• Option 1: When the initial mass and number of half-lives are known you can start with the initial mass and divide it by 2 once for each half-life as seen in Table \(\PageIndex{1}\). In this case the initial mass is 5.00 g and there are four half-lives, so divide by 2 four times: 5.0 g \(\div\) 2 \(\div\) 2 \(\div\) 2 \(\div\) 2 = 0.313 g. If organizing your work into a table, be sure that the first row matches 5.00 g (the initial mass) with 0 half-lives. 
• Option 2: We can use Equation \ref{Eq2} with \(n = 4\). Substituting and solving results in the following:

\[ \begin{align*} \mathrm{amount\: remaining} &=5.00\: g \times \left(\dfrac{1}{2}\right)^4 \\[4pt] &=5.00\: g \times \dfrac{1}{16} \\[4pt] &=0.313\: g \end{align*}\]

Using either option we find that 0.313 g ²⁰F remains.

Exercise \(\PageIndex{1}\): Titanium-44

The half-life of ⁴⁴Ti is 60.0 y. A sample initially contains 0.600 g of ⁴⁴Ti. How much ⁴⁴Ti remains after 180.0 y?

Answer

0.075 g.

Half-lives of isotopes range from fractions of a microsecond to billions of years. Table \(\PageIndex{1}\) lists the half-lives of some isotopes.

Table \(\PageIndex{1}\): Half-Lives of Various Isotopes

Isotope	Half-Life
^3H	12.3 y
^14C	5,730 y
^40K	1.26 × 10^9 y
^51Cr	27.70 d
^90Sr	29.1 y
^131I	8.04 d
^222Rn	3.823 d
^235U	7.04 × 10^8 y
^238U	4.47 × 10^9 y
^241Am	432.7 y
^248Bk	23.7 h
^260Sg	4 ms

Example \(\PageIndex{2}\): Iodine-131

The isotope \(\ce{I}\)-131 is used in treatment for thyroid cancer and has a half-life of 8.05 days. If a clinic initially has 400 mg, how long will it take for the sample to decay to only 25 mg of \(\ce{I}\)-131?

Solution

We begin by determining how many half-lives it would take for the sample to decay from 400 mg to 25 mg of \(\ce{I}\)-131. There are several ways to do this! We know that for each half-life the mass of \(\ce{I}\)-131 is reduced to one half of the previous value so we could solve this conceptually: 

\[ \begin{align*} \dfrac{400 \: \text{mg}}{2} = 200 \: \text{mg} \\ \dfrac{200 \: \text{mg}}{2} = 100 \: \text{mg} \\ \dfrac{100 \: \text{mg}}{2} = 50 \: \text{mg} \\ \dfrac{50 \: \text{mg}}{2} = 25 \: \text{mg} \end{align*}\]

We had to divide the original mass by 2 four times to get the final mass. Therefore, it took four half-lives to get from 400 mg to 25 mg of \(\ce{I}\)-131. 

Once the number of half-lives has been determined, we can calculate the total time. 

\[ \begin{align*} \text{total time} = \text{half-life}\times \text{number of half-lives}  \\ = 8.05 \: \text{days} \times 4 \: \text{half-lives} = 32.2 \: \text{days} \end{align*} \]

 

A more sophisticated way to solve the problem is to make use of percentages or fractions. The graph below shows the percentage of \(\ce{I}\)-131 remaining after each half-life. These percentages are always the same, regardless of the initial mass of \(\ce{I}\)-131. 

Typical radioactive decay curve.

The graph above illustrates a typical decay curve. Since the half-life for \(\ce{I}-131\) is 8.05 days, the x-axis has units of days. The y-axis indicates the percentage of sample remaining. Initially, (at elapsed time = 0), 100 % of the sample remains. After one half-life (8.05 days for \(\ce{I}-131\)), 50 % remains. After two half-lives, 25 % of the initial sample remains. We can use this pattern as an alternate way of calculating the number of half-lives when the initial and final masses are known. 

Alternate method of solving: 

\[\dfrac{\text{final mass}}{\text{initial mass}}\times 100 = \text{percentage remaining} \]

\[\dfrac{25 \: \text{mg}}{400 \: \text{mg}}\times 100 = 6.25 \text{%} \nonumber \]

This result, 6.25 % remaining, corresponds to four half-lives. Careful! Don't count 100 % as the first half-life. At time = 0, nothing has happened yet. 50 % remaining = 1 half-life, 25 % remaining = 2 half-lives, 12.5 % remaining = 3 half-lives, and 6.25 % remaining = 4 half-lives. These percentages are the same for all isotopes so you could calculate them once and then use them for multiple problems. 

To finish this problem, multiply the number of half-lives by the length of each half-life to get the total time. 

Exercise \(\PageIndex{2}\)

A sample of \(\ce{Ac}\)-225 originally contained 80 grams and after 50 days only 2.5 grams of the original \(\ce{Ac}\)-225 remain. What is the half life of \(\ce{Ac}\)-225?

Answer

10 days. Notice that the phrase "what is the half-life?" means "how long is the half-life?".

Looking Closer: Half-Lives of Radioactive Elements

Many people think that the half-life of a radioactive element represents the amount of time an element is radioactive. In fact, it is the time required for half—not all—of the element to decay radioactively. Occasionally, however, the daughter element is also radioactive, so its radioactivity must also be considered.

The expected working life of an ionization-type smoke detector is about 10 years. In that time, americium-241, which has a half-life of about 432 y, loses less than 4% of its radioactivity. A half-life of 432 y may seem long to us, but it is not very long as half-lives go. Uranium-238, the most common isotope of uranium, has a half-life of about 4.5 × 10⁹ y, while thorium-232 has a half-life of 14 × 10⁹ y.

On the other hand, some nuclei have extremely short half-lives, presenting challenges to the scientists who study them. The longest-lived isotope of lawrencium, ²⁶²Lr, has a half-life of 3.6 h, while the shortest-lived isotope of lawrencium, ²⁵²Lr, has a half-life of 0.36 s. As of this writing, the largest atom ever detected has atomic number 118, mass number 293, and a half-life of 120 ns (nanoseconds). Can you imagine how quickly an experiment must be done to determine the properties of elements that exist for so short a time?

Key Takeaways

• Natural radioactive processes are characterized by a half-life, the time it takes for half of the material to decay radioactively.
• The amount of material left over after a certain number of half-lives can be easily calculated.

Concept Review Exercises

1. Define half-life.
2. Describe a way to determine the amount of radioactive isotope remaining after a given number of half-lives.

Answers

1. Half-life is the amount of time needed for half of a radioactive material to decay.
2. take half of the initial amount for each half-life of time elapsed