Question

The half-life of francium is 21 minutes. Starting with 4 X 1018 atoms of francium, how many atoms would
disintegrate in 1 hour and 45 minutes? What fraction of the original sample remains?
The half life of a radioactha alemantic 20

Answer 1

Using an exponential equation, it is found that:

• $3.9144 \times 10^{18}$ atoms would disintegrate in 1 hour and 45 minutes.
• 0.0313 = 3.13% of the original sample remains.

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An exponential equation for the proportion of a substance after m minutes is given by:

$P(m) = e^{-km}$

• In which k is the decay rate.

The half-life of francium is 21 minutes.

This means that:

$P(21) = 0.5$

This is used to find k.

$e^{-21k} = 0.5$

$\ln{e^{-21k}} = \ln{0.5}$

$-21k = \ln{0.5}$

$k = -\frac{\ln{0.5}}{21}$

$k = 0.033$

Thus

$P(m) = e^{-0.033m}$

After 1 hour and 45 minutes:

• 105 minutes, thus, the fraction for the proportion that remains is P(105).

$P(105) = e^{-0.033(105)} = 0.0313$

0.0313 = 3.13% of the original sample remains.

• Initially, there was $4 \times 10^{18}$ atoms.

• 0.0313 remained, so 1 - 0.0313 = 0.9687 disintegrated.

The number of atoms that disintegrated is:

$0.9687 \times 4 \times 10^18 = 3.9144 \times 10^{18}$

$3.9144 \times 10^{18}$ atoms would disintegrate in 1 hour and 45 minutes.

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