Chemistry · General chemistry II · Worked example

Count half-lives in radioactive decay

A radioactive tracer has a half-life of $12.4 h$ . A sample contains $80.0 mg$ of it. How much remains after $37.2 h$ , and what is the decay constant $k$ ?

Use first-order decay

The first example used $k$ and the exponential. Here the half-life is given and the time is a whole number of them, so counting halvings is quicker. Radioactive decay is first order, because each nucleus has the same chance of decaying in the next hour however many others are around.

Count the half-lives

Divide the elapsed time by the half-life. Hours cancel, so $n$ is a pure number.

n = \frac{37.2 h}{12.4 h} = 3.00

Halve the amount $n$ times

Three halvings leave ${(1 / 2)}^{3} = 1 / 8$ of the sample, $80.0$ $\to 40.0$ $\to 20.0$ $\to 10.0 mg$ . Mass works as well as concentration here, because each is proportional to the number of nuclei.

m = \frac{80.0 mg}{2^{3}} = 10.0 mg

Find the decay constant

Rearranging $t_{1 / 2} = ln 2 / k$ gives $k$ from the half-life. Its unit, $h^{- 1}$ , is the unit of a first-order rate constant.

k = \frac{0.693}{12.4 h} = 0.0559 h^{- 1}

Use $k$ between half-lives

With $k$ you are not limited to whole half-lives. After $24.0 h$ , $k t = (0.0559 h^{- 1}) (24.0 h)$ $= 1.34$ , so $m = (80.0 mg) e^{- 1.34}$ $= 20.9 mg$ . That is just above the $20.0 mg$ left at two half-lives, $24.8 h$ , as it should be.

m = (80.0 mg) e^{- 1.34} = 20.9 mg

Result

$10.0 mg$ remains after $37.2 h$ , and $k = 0.0559 h^{- 1}$ .

Your turn

1. One like the example

A tracer has a half-life of $8.00 days$ . How much of a $40.0 mg$ sample is left after $24.0 days$ ?

Show the answer and explanation

$5.00 mg$ .

$24.0 days$ is $24.0 \div 8.00 = 3.00$ half-lives. Three halvings leave one eighth, $40.0$ $\to 20.0$ $\to 10.0$ $\to 5.00 mg$ .

\begin{array}{l} m_{0} = 40.0 mg \\ t_{1 / 2} = 8.00 d \\ n = \frac{24.0 d}{8.00 d} = 3.00 \\ m = \frac{40.0 mg}{8} = 5.00 mg \end{array}

2. A variation

How long does the worked example’s $80.0 mg$ of tracer, with $t_{1 / 2} = 12.4 h$ , take to fall to $5.00 mg$ ?

Show the answer and explanation

$49.6 h$ , four half-lives.

This time the amount left is given and the time is unknown. $80.0 mg \div 5.00 mg = 16.0$ , and $16 = 2^{4}$ , so four half-lives have passed. $n = t / t_{1 / 2}$ gives $t = 4 \times 12.4 h = 49.6 h$ .

\begin{array}{l} t_{1 / 2} = 12.4 h \\ \frac{80.0 mg}{5.00 mg} = 16.0 \\ n = \frac{t}{t_{1 / 2}} \\ n = 4 \\ t = (4) (12.4 h) = 49.6 h \end{array}

3. Somewhere new

Carbon-14 has a half-life of $5730 years$ . A wooden tool contains $25.0 %$ of the carbon-14 found in living wood. How old is it?

Show the answer and explanation

About $11, 500 years$ , two half-lives $(11, 460 years)$ .

$25.0 %$ is one quarter, and one quarter is ${(\frac{1}{2})}^{2}$ , so two half-lives have passed. $t = 2 \times 5730 years$ $= 11, 460 years$ , which the data’s three significant figures give as $1.15 \times 10^{4} years$ . The answer assumes the wood began with the share of carbon-14 that living wood has.

\begin{array}{l} t_{1 / 2} = 5730 yr \\ n = \frac{t}{t_{1 / 2}} \\ n = 2 \\ t = (2) (5730 yr) = 11460 yr \end{array}

Keep exploring

In the Chemistry box, change the last row to $m = 80.0 mg \div (2 \cdot 3)$ $= 13.3 mg$ , dividing by $2 n$ . The row turns to ✗: “After n half-lives, m = m₀ × (½)ⁿ, and the values above give another number.”

Return to the concept →

Sources and scope

Authored study material. Tool results depend on the stated inputs and model assumptions.

After reading

Study it, then try it

Practice the key terms as flashcards, then open the example in its tool, change a value and keep a useful result on your board.

Study the 6 key terms Check the amount in a Chemistry box Check k in a Chemistry box Open worked example on a board Chemistry formulas: kinetics

Your existing work stays on this device. Examples open as editable copies.

Use first-order decay

Count the half-lives

Halve the amount nn times

Find the decay constant

Use kk between half-lives

Result

Your turn

1. One like the example

5.00 mg5.00 mg.

2. A variation

49.6 h49.6 h, four half-lives.

3. Somewhere new

About 11,500 years11,500 years, two half-lives (11,460 years)(11,460 years).

Keep exploring

Study it, then try it

Halve the amount $n$ times

Use $k$ between half-lives

$5.00 mg$ .

$49.6 h$ , four half-lives.

About $11, 500 years$ , two half-lives $(11, 460 years)$ .