Worked example 1
iodine-131 has a half-life of 8.0 days. How much of a 40.0 g sample remains after 32 days?
Try it first: How many 8-day intervals fit into 32 days?
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Nuclear ChemistryHalf-LifeContent level: Core 18 min
Half-Life and Radioactive Decay Calculations
What you'll be able to do: Use half-life to find how much of a sample remains after a given time, and work backwards from the fraction remaining to the elapsed time or the half-life.
Best after: Types of Nuclear Decay and Balancing Nuclear Equations
Introduction
Radioactive decay is the classic first-order process: a fixed fraction decays in every equal interval, never a fixed amount. Once you see that, most half-life questions collapse into repeated halving.
These are recommended, not required. You can start this lesson at any time.
Learning objectives
- State what half-life means and why it is independent of sample size
- Calculate the amount remaining after a whole number of half-lives
- Find the elapsed time or the half-life from the fraction remaining
- Apply the first-order equations k = 0.693 / t₁/₂ and ln(N/N₀) = -kt
- Explain the basis of carbon-14 dating
Lesson
What half-life means
The half-life, t₁/₂, is the time it takes for half of the radioactive nuclei present to decay. It is a constant for a given nuclide. Start with 80 g or 8 g and the half-life is the same; only the amount left differs. Because decay is a nuclear process, heating the sample or bonding the atom into a compound does not change it.
Counting half-lives
If the elapsed time is a whole number of half-lives, do not use logarithms at all. Divide the total time by the half-life to get n, then multiply the starting amount by (1/2)ⁿ. This handles most exam questions in one line.
N = N₀ × (1/2)ⁿ, where n = t / t₁/₂
When the time is not a whole number of half-lives
Radioactive decay is first order, so the integrated rate law applies. First convert the half-life into a rate constant with k = 0.693 / t₁/₂, then use ln(N/N₀) = −kt. The same two equations solve for N, for t or for t₁/₂ depending on which quantity is missing.
k = 0.693 / t₁/₂; ln(N/N₀) = −kt
Keep the units of k and t consistent. If the half-life is in years, k is in yr⁻¹ and t must be in years.
¹⁴C dating
Living things exchange carbon with the atmosphere and hold a steady ¹⁴C level. Once the organism dies, uptake stops and the ¹⁴C decays with a half-life of 5730 years. Measuring the fraction of ¹⁴C remaining and solving for t gives the age. Because roughly ten half-lives leaves under 0.1% of the original, the method runs out at about 50,000 years.
Why a fixed fraction, not a fixed amount
Each nucleus has the same probability of decaying in the next second, so the number of decays per second is proportional to the number of nuclei present. As the sample shrinks the decay rate shrinks with it, which is exactly what makes the decay exponential rather than linear.
Key ideas
Definition
Half-life
The time required for half of the radioactive nuclei in a sample to decay.
Rule
Independence
Half-life does not depend on sample size, temperature, pressure or chemical form.
Rule
Fraction remaining
After n half-lives, the fraction remaining is (1/2)ⁿ.
Equation
First-order link
k = 0.693 / t₁/₂ connects the half-life to the rate constant.
Key concept
Activity tracks amount
The measured activity in decays per second is proportional to the number of undecayed nuclei, so activity halves on the same schedule.
Equation
Fraction remaining
N = N₀ × (1/2)ⁿ
- N₀ = initial amount
- n = number of half-lives elapsed
Equation
Rate constant from half-life
k = 0.693 / t₁/₂
- k = first-order rate constant
- t₁/₂ = half-life
Equation
Integrated first-order law
ln(N/N₀) = −kt
- N = amount remaining at time t
- t = elapsed time
Worked examples
Worked example 2
A wooden artifact retains 25% of the ¹⁴₆C found in living wood. carbon-14 has a half-life of 5730 years. How old is it?
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Worked example 3
strontium-90 has a half-life of 28.8 years. What percentage of a sample remains after 50.0 years?
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Common mistakes
Assuming a fixed mass decays in each half-life, so two half-lives means the sample is gone.
Why it's wrong: Each half-life removes half of what is left, not half of the original, so the amount approaches zero without ever reaching it.
Check instead: Write out the halving sequence explicitly: 100%, 50%, 25%, 12.5%.
Using the half-life directly in place of k in ln(N/N₀) = -kt.
Why it's wrong: The equation requires a rate constant with units of inverse time; the half-life is a time.
Check instead: Always compute k = 0.693 / t₁/₂ as a separate first step.
Expecting heating a sample or forming a compound to speed up decay.
Why it's wrong: Decay happens in the nucleus, which is untouched by chemical bonding or ordinary temperatures.
Check instead: Treat half-life as a fixed property of the nuclide.
Practice this skill
No practice questions are available for this topic yet. You can still practice the whole unit.
What you should now know
The half-life of a nuclide is the time for half of any sample to decay, and it does not depend on how much you start with or on temperature, pressure or chemical form. After n half-lives the fraction remaining is (1/2)ⁿ, so a sample drops to 50%, 25%, 12.5% and so on at equal time intervals. When the elapsed time is not a whole number of half-lives, the first-order equations are used instead: the rate constant k = 0.693 / t₁/₂, and ln(N/N₀) = -kt. carbon-14 dating and medical tracer dosing are both direct applications.
- Half-life is the time for half a sample to decay and is a constant for each nuclide
- After n half-lives the fraction remaining is (1/2)ⁿ
- Whole numbers of half-lives need no logarithms, just repeated halving
- For other times use k = 0.693 / t₁/₂ then ln(N/N₀) = -kt
- Half-life is unaffected by sample size, temperature or chemical form
Sources and further reading
This lesson is original Chem Help content. No external sources were adapted.