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Chapter 21: Problem 33

What percentage of a sample's original radioactivity remains after three half- lives?

Short Answer

Expert verified
Answer: 12.5%

Step by step solution

01

Understand the half-life concept

A half-life is the time it takes for half of the radioactivity in a substance to decay. In other words, after one half-life, only half of the original radioactive material remains.
02

Apply the exponential decay formula

After each half-life, the remaining radioactivity is multiplied by a factor of one-half or 0.5. This can be expressed as \(Remaining_{n} = (0.5)^n \times Remaining_{0}\), where \(Remaining_{n}\) is the remaining radioactivity after \(n\) half-lives, and \(Remaining_{0}\) is the original radioactivity.
03

Calculate the remaining radioactivity after three half-lives

We can simplify the formula for this specific case, by setting \(n = 3\). Thus the formula will be \(Remaining_{3} = (0.5)^3 \times Remaining_{0}\).
04

Calculate the remaining percentage

To obtain the percentage of remaining radioactivity, we will divide the remaining radioactivity after three half-lives by the initial radioactivity and multiply by 100%. This can be written as \(Percentage = \frac{Remaining_{3}}{Remaining_{0}} \times 100\).
05

Calculate the final percentage

First, calculate \((0.5)^3\): \((0.5)^3 = 0.5 \times 0.5 \times 0.5 = 0.125\) Now, plug this into the percentage formula: \(Percentage = \frac{0.125 \times Remaining_{0}}{Remaining_{0}} \times 100 = 0.125 \times 100 = 12.5\%\) Thus, 12.5% of the sample's original radioactivity remains after three half-lives.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Half-life
The term "half-life" refers to the time required for half of the radioactive atoms in a sample to decay. Understanding half-life is crucial in predicting how much of a radioactive substance will remain over time. If a substance has a half-life of 5 years, for instance, it means that every 5 years, the amount of radioactive material is halved. To break it down:
  • After one half-life, 50% of the original material remains.
  • After two half-lives, 25% remains (half of 50%).
  • With three half-lives completed, only 12.5% (half of 25%) of the original substance remains.
This consistent halving feature is what makes the concept of half-life very useful for understanding exponential decay processes.
Exponential Decay Formula
Exponential decay describes the process where a quantity decreases at a constant rate over time. For radioactive decay, this is often expressed with a formula that makes it easier to compute the remaining quantity of a substance.For any given number of half-lives, the exponential decay formula is:\[Remaining_n = (0.5)^n \times Remaining_{0}\]where:
  • \(Remaining_n\) is the amount left after \(n\) half-lives,
  • \(0.5\) represents the halving effect of each half-life,
  • \(Remaining_{0}\) is the starting quantity of the substance.
For example, after three half-lives, using this formula, you would compute the remaining amount as \((0.5)^3 \times Remaining_0\), which simplifies to just 12.5% of the original substance.
Percentage Calculation
After computing the remaining amount of a substance through exponential decay, the next step involves determining what percentage of the original quantity remains. This is where percentage calculation is important.To find this percentage, divide the remaining radioactivity by the initial radioactivity and multiply by 100%. Using the formula:\[Percentage = \frac{Remaining_n}{Remaining_0} \times 100\]For three half-lives, you calculate \((0.5)^3\), which equals 0.125. Therefore:
  • \(Percentage = \frac{0.125 \times Remaining_0}{Remaining_0} \times 100 = 12.5\%\).
This means that only 12.5% of the original sample remains, illustrating how percentages can easily convey the scale of decay over multiple half-lives.

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Most popular questions from this chapter

Spent fuel removed from nuclear power stations contains plutonium-\(239\left(t_{1 / 2}=2.41 \times 10^{4} years) \right.\) How long will it take a sample of this radionuclide to reach a level of radioactivity that is \(2.5 \%\) of the level it had when it was removed from a reactor?

Thorium-232 slowly decays to bismuth-212 \(\left(t_{1 / 2}=1.4 \times 10^{10} \mathrm{yr}\right) .\) Bismuth-212 decays to lead-208 by two pathways: first \(\beta\) and then \(\alpha\) decay, or \(\alpha\) and then \beta decay. The intermediate nuclide in the second pathway is thallium-208. The thallium-208 can be separated from a sample of thorium nitrate by passing a solution of the sample through a filter pad containing ammonium phosphomolybdate. The radioactivity of \(^{208}\) T1 trapped on the filter is measured as a function of time. In one such experiment, the following data were collected: $$\begin{array}{cc}\text { Time (s) } & \text { Counts/min } \\\\\hline 60 & 62 \\\\\hline 120 & 40 \\\\\hline 180 & 35 \\\\\hline 240 & 22 \\\\\hline 300 & 16 \\\\\hline 360 & 10 \\\\\hline\end{array}$$ Use the data in the table to determine the half-life of \(^{208} \mathrm{T} 1 .\)

Strontium-90 in Milk In the years immediately following the explosion at the Chernobyl nuclear power plant, the concentration of \({ }^{90} \mathrm{Sr}\) in cow's milk in southern Europe was slightly elevated. Some samples contained as much as \(1.25 \mathrm{~Bq} / \mathrm{L}\) of \({ }^{90} \mathrm{Sr}\) radioactivity. The half-life of strontium-90 is 28.8 years. a. Write a balanced nuclear equation describing the decay of \({ }^{90} \mathrm{Sr}\). b. How many atoms of \({ }^{90} \mathrm{Sr}\) are in a \(200 \mathrm{~mL}\) glass of milk with \(1.25 \mathrm{~Bq} / \mathrm{L}\) of \({ }^{90} \mathrm{Sr}\) radioactivity? c. Why would strontium-90 be more concentrated in milk than other foods, such as grains, fruits, or vegetables?

Bombardment of a \({ }^{239} \mathrm{Pu}\) target with \(\alpha\) particles produces \({ }^{242} \mathrm{Cm}\) and another particle. a. Use a balanced nuclear equation to determine the identity of the missing particle. b. The synthesis of which other nuclide described in this chapter involves the same subatomic particles?

In \(1999,\) the U.S. Environmental Protection Agency set a maximum radon level for drinking water at \(4.0 \mathrm{pCi}\) per milliliter. a. How many decay events occur per second in a milliliter of water for this level of radon radioactivity? b. If the above radioactivity were due to the decay of \(^{222} \operatorname{Rn}\left(t_{1 / 2}=3.8 \text { days }\right),\) how many \(^{222} \mathrm{Rn}\) atoms would there be in \(1.0 \mathrm{mL}\) of water?
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