Decay Calculations

Radioactive decay calculations allow us to determine how the amount or activity of a radioactive isotope changes with time.

The most important idea is that after every half-life, the amount of radioactive parent material remaining is divided by:

2

For example:

800 → 400 → 200 → 100 → 50

Each step represents one half-life.

Decay calculations can involve:

• number of radioactive nuclei
• mass of a radioactive isotope
• activity
• corrected count rate
• elapsed time
• original quantity
• percentage or fraction remaining

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The Basic Half-Life Pattern

The fraction remaining follows this pattern:

This table can solve many introductory decay problems without using logarithms.

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Finding the Number of Half-Lives

If the elapsed time and half-life are known:

n = t/t₁/₂

where:

• n = number of half-lives
• t = elapsed time
• t₁/₂ = half-life

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Example: Number of Half-Lives

An isotope has a half-life of:

6 hours

How many half-lives occur during:

24 hours?

Use:

n = t/t₁/₂

n = 24/6

n = 4

Therefore:

4 half-lives have elapsed.

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Example: Days

An isotope has a half-life of:

8 days.

How many half-lives occur in:

40 days?

n = 40/8

n = 5 half-lives

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Keep Units Consistent

Before dividing, make sure the units are the same.

For example:

half-life = 3 hours

elapsed time = 1 day

Convert:

1 day = 24 hours

Then:

n = 24/3

n = 8 half-lives

Do not calculate:

1/3

because days and hours are different units.

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Calculating the Remaining Quantity

After n half-lives:

remaining quantity = initial quantity × (1/2)ⁿ

For the number of nuclei:

N = N₀(1/2)ⁿ

where:

• N = number remaining
• N₀ = initial number
• n = number of half-lives

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Example: Number of Nuclei

A radioactive sample initially contains:

12,000 radioactive nuclei.

Three half-lives pass.

Use:

N = 12,000(1/2)³

N = 12,000 × 1/8

N = 1500 nuclei

You can also halve repeatedly:

12,000 → 6000 → 3000 → 1500

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Example: Mass Remaining

A sample initially contains:

320 mg

of a radioactive isotope.

After four half-lives:

320 → 160 → 80 → 40 → 20

Therefore:

20 mg remains

Using the equation:

m = 320(1/2)⁴

m = 320/16

m = 20 mg

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Example: Activity Remaining

A source initially has an activity of:

9600 Bq.

After three half-lives:

9600 → 4800 → 2400 → 1200

Therefore:

A = 1200 Bq

Remember:

1 Bq = 1 decay/s

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Example: Corrected Count Rate

A source has an initial corrected count rate of:

4800 counts/min.

After five half-lives:

4800(1/2)⁵

= 4800/32

= 150 counts/min

The same half-life mathematics applies because corrected count rate is proportional to activity when the detector setup remains unchanged.

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Combining Time and Remaining Quantity

Many questions give:

• initial quantity
• half-life
• elapsed time

The best strategy is:

Step 1: Find the number of half-lives.

Step 2: Halve the quantity that many times.

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Worked Example

A radioactive isotope has:

initial mass = 640 mg

half-life = 5 years

Find the remaining mass after:

20 years.

Find the number of half-lives

n = 20/5

n = 4

Halve four times

640 → 320 → 160 → 80 → 40

Therefore:

40 mg remains.

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Worked Example: Activity

A source has:

initial activity = 16,000 Bq

half-life = 3 hours

elapsed time = 12 hours

Number of half-lives:

n = 12/3

n = 4

Activity:

16,000 → 8000 → 4000 → 2000 → 1000

Therefore:

A = 1000 Bq

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Finding the Original Amount

Sometimes the amount remaining is known, but the original amount is unknown.

In this case, work backwards.

Instead of halving:

double once for each half-life.

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Example: Finding Original Mass

After three half-lives:

25 mg remains.

Work backwards:

25 → 50 → 100 → 200

Therefore:

original mass = 200 mg

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Example: Finding Original Activity

A source has an activity of:

375 Bq

after four half-lives.

Work backwards:

375 → 750 → 1500 → 3000 → 6000

Therefore:

original activity = 6000 Bq

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Original Quantity Equation

Starting with:

N = N₀(1/2)ⁿ

we can rearrange:

N₀ = N × 2ⁿ

For example:

N = 500

n = 4

N₀ = 500 × 2⁴

N₀ = 500 × 16

N₀ = 8000

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Finding Elapsed Time

Sometimes the question gives:

• initial amount
• final amount
• half-life

but asks for:

elapsed time

First determine how many times the amount has halved.

Then:

t = n × t₁/₂

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Example: Finding Time

A radioactive sample decreases from:

640 mg

to:

40 mg.

Count the halvings:

640 → 320 → 160 → 80 → 40

That is:

4 half-lives

If:

t₁/₂ = 7 days

then:

t = 4 × 7

t = 28 days

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Example: Activity and Time

A source decreases from:

6400 Bq

to:

800 Bq.

Halve:

6400 → 3200 → 1600 → 800

That is:

3 half-lives

If:

t₁/₂ = 12 hours

then:

t = 3 × 12

t = 36 hours

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Finding the Half-Life

Sometimes elapsed time and the change in quantity are known.

First determine how many half-lives occurred.

Then:

t₁/₂ = t/n

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Example: Finding Half-Life

A radioactive sample decreases from:

4800 Bq

to:

600 Bq

during:

18 hours.

Find the number of halvings:

4800 → 2400 → 1200 → 600

Therefore:

n = 3

Now:

t₁/₂ = 18/3

t₁/₂ = 6 hours

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Another Example

A sample decreases from:

160 mg

to:

10 mg

during:

32 days.

Halving:

160 → 80 → 40 → 20 → 10

Therefore:

n = 4

Half-life:

t₁/₂ = 32/4

t₁/₂ = 8 days

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Fraction Remaining

The fraction remaining after n half-lives is:

(1/2)ⁿ

For example:

after 4 half-lives:

(1/2)⁴ = 1/16

So:

1/16 of the original radioactive material remains.

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Percentage Remaining

Convert the fraction to a percentage:

percentage remaining = (1/2)ⁿ × 100%

After 4 half-lives:

percentage remaining:

= 1/16 × 100%

= 6.25%

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Percentage Decayed

Do not confuse:

percentage remaining

with:

percentage decayed

Use:

percentage decayed = 100% − percentage remaining

After four half-lives:

percentage remaining = 6.25%

Therefore:

percentage decayed:

= 100 − 6.25

= 93.75%

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Example: Five Half-Lives

After five half-lives:

fraction remaining:

= (1/2)⁵

= 1/32

Percentage remaining:

= 3.125%

Percentage decayed:

= 100 − 3.125

= 96.875%

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Finding Half-Lives from a Fraction

Sometimes the fraction remaining is given.

For common fractions, simply recognize the pattern.

If:

fraction remaining = 1/2

then:

1 half-life

If:

fraction remaining = 1/4

then:

2 half-lives

If:

fraction remaining = 1/8

then:

3 half-lives

If:

fraction remaining = 1/16

then:

4 half-lives

If:

fraction remaining = 1/32

then:

5 half-lives

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Example: Percentage Remaining

A sample has:

12.5%

of its original radioactive isotope remaining.

Recognize:

12.5% = 1/8

and:

1/8 = (1/2)³

Therefore:

3 half-lives have elapsed.

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Exponential Decay

Radioactive decay follows an exponential relationship.

For a large population:

N = N₀e^(−λt)

where:

• N = number remaining
• N₀ = initial number
• λ = decay constant
• t = elapsed time

The equivalent activity equation is:

A = A₀e^(−λt)

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Why Radioactive Decay Is Exponential

The rate of radioactive decay depends on how many radioactive nuclei remain.

At the beginning:

many radioactive nuclei are present.

Therefore:

many decays occur.

Later:

fewer radioactive nuclei remain.

Therefore:

fewer decays occur during the same length of time.

This produces a curved exponential decay graph.

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Constant Fraction Rather Than Constant Amount

Consider:

1600 → 800 → 400 → 200 → 100

The amount lost during each interval is:

800

400

200

100

The amount lost is not constant.

But the fraction lost is always:

1/2

Therefore radioactive decay is:

exponential

rather than:

linear

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Linear vs Exponential Decay

A linear decrease might look like:

1000 → 900 → 800 → 700 → 600

The same amount is removed each time:

100

Radioactive exponential decay might look like:

1000 → 500 → 250 → 125 → 62.5

The same fraction is removed:

1/2

This distinction is essential when interpreting decay graphs.

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Decay Constant

The decay constant, λ, describes the probability of decay per unit time.

Half-life and decay constant are related:

λ = 0.693/t₁/₂

and:

t₁/₂ = 0.693/λ

A larger λ means:

• faster decay
• shorter half-life

A smaller λ means:

• slower decay
• longer half-life

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Example: Decay Constant

An isotope has:

t₁/₂ = 10 days

Then:

λ = 0.693/10

λ = 0.0693 day⁻¹

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Example: Finding Half-Life from λ

Suppose:

λ = 0.035 h⁻¹

Then:

t₁/₂ = 0.693/0.035

t₁/₂ ≈ 19.8 hours

So the half-life is approximately:

20 hours

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Calculations with Partial Half-Lives

Not every problem involves an exact whole number of half-lives.

Suppose:

t₁/₂ = 8 hours

but:

t = 20 hours

Then:

n = 20/8

n = 2.5 half-lives

We can still use:

remaining = initial × (1/2)ⁿ

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Example: Partial Half-Life

Initial mass:

100 mg

Half-life:

8 hours

Elapsed time:

20 hours

Number of half-lives:

n = 20/8

= 2.5

Then:

m = 100(1/2)^2.5

(1/2)^2.5 ≈ 0.1768

Therefore:

m ≈ 100 × 0.1768

m ≈ 17.7 mg

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Half-Life Form of the Decay Equation

Instead of using λ, decay can be written directly using half-life:

N = N₀(1/2)^(t/t₁/₂)

This equation works for:

• whole numbers of half-lives
• fractional numbers of half-lives

Similarly:

A = A₀(1/2)^(t/t₁/₂)

and:

m = m₀(1/2)^(t/t₁/₂)

when m represents the amount of radioactive parent isotope.

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Worked Example: Partial Half-Life

A radioactive isotope has:

initial activity = 2000 Bq

half-life = 12 hours

Find its expected activity after:

30 hours.

Use:

A = A₀(1/2)^(t/t₁/₂)

A = 2000(1/2)^(30/12)

A = 2000(1/2)^2.5

A ≈ 2000 × 0.1768

A ≈ 354 Bq

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Finding Time Using Logarithms

For more advanced problems:

N = N₀(1/2)^(t/t₁/₂)

Divide by N₀:

N/N₀ = (1/2)^(t/t₁/₂)

Taking logarithms allows t to be calculated:

t = t₁/₂ × ln(N/N₀) / ln(1/2)

Because both logarithms are negative when N < N₀, the calculated time is positive.

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Example: Finding Time with a Non-Simple Fraction

A sample begins with:

1000 mg

and decreases to:

300 mg.

Half-life:

5 years.

Use:

t = 5 × ln(300/1000) / ln(1/2)

t = 5 × ln(0.3) / ln(0.5)

t ≈ 5 × 1.737

t ≈ 8.68 years

So approximately:

8.7 years have elapsed.

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Finding the Original Amount Using the Equation

Suppose:

remaining mass = 18 mg

half-life = 4 years

elapsed time = 12 years

First:

n = 12/4

n = 3

Use:

m₀ = m × 2ⁿ

m₀ = 18 × 2³

m₀ = 18 × 8

m₀ = 144 mg

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Count Rate Calculations

Radiation experiments often provide count rate instead of activity.

Remember:

measured count rate = source count rate + background count rate

Therefore:

corrected count rate = measured count rate − background count rate

Use the corrected rate for decay calculations.

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Worked Example with Background Radiation

A detector initially records:

820 counts/min

Background:

20 counts/min

Therefore:

initial corrected rate:

820 − 20 = 800 counts/min

After some time, the detector records:

120 counts/min

Corrected:

120 − 20 = 100 counts/min

Now compare:

800 → 400 → 200 → 100

That represents:

3 half-lives

If the isotope has:

t₁/₂ = 4 hours

then:

t = 3 × 4

t = 12 hours

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A Common Background-Radiation Trap

Suppose:

initial measured rate = 500 counts/min

background = 100 counts/min

The corrected initial rate is:

400 counts/min

After one half-life, the source contribution should be:

200 counts/min

But the detector will measure:

source + background

= 200 + 100

= 300 counts/min

Therefore, the raw detector reading does not simply halve from 500 to 250.

Background must be accounted for.

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Interpreting Decay Graphs

A decay graph commonly plots:

x-axis → time

y-axis → number of nuclei, mass of radioactive parent isotope, activity, or corrected count rate

The graph should show:

• rapid decrease initially
• progressively slower decrease
• a smooth exponential curve
• equal half-life intervals for successive halvings

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Finding Half-Life from a Graph

Suppose a graph shows:

800 Bq at 2 hours

400 Bq at 8 hours

The activity has halved.

Therefore:

t₁/₂ = 8 − 2

t₁/₂ = 6 hours

Notice that you do not have to begin at t = 0.

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Checking a Half-Life on a Graph

Once you find one halving interval, check another.

For example:

800 Bq at 2 h

400 Bq at 8 h

200 Bq at 14 h

Each halving takes:

6 hours

This confirms:

t₁/₂ = 6 hours

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Comparing Decay Curves

Suppose two isotopes begin with the same amount.

Isotope A

Half-life = 2 hours

Isotope B

Half-life = 10 hours

Isotope A's graph falls more steeply.

Isotope B's graph falls more gradually.

Therefore:

steeper decay curve → shorter half-life

shallower decay curve → longer half-life

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Example: Comparing Two Isotopes

Both isotopes begin with:

6400 nuclei.

Isotope A:

t₁/₂ = 2 h

Isotope B:

t₁/₂ = 8 h

After 8 hours:

Isotope A

n = 8/2 = 4

N = 6400(1/2)⁴

N = 400

Isotope B

n = 8/8 = 1

N = 6400(1/2)

N = 3200

Therefore, isotope A decays much more rapidly.

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Decay Calculations in Radiometric Dating

Decay calculations can be used to estimate how much time has passed since a radioactive system began evolving under the assumptions of a particular dating method.

For example, carbon-14 has a half-life of approximately:

5730 years

If an appropriate sample contains:

1/4

of the expected original carbon-14 proportion, then:

1 → 1/2 → 1/4

This represents:

2 half-lives

Estimated elapsed time:

2 × 5730

= 11,460 years

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Example: Carbon-14

A suitable sample contains:

12.5%

of its expected original carbon-14 proportion.

Convert:

12.5% = 1/8

Then:

1 → 1/2 → 1/4 → 1/8

Therefore:

3 half-lives

Elapsed time:

3 × 5730

= 17,190 years

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Choosing the Best Method

For simple whole half-lives:

repeated halving is fastest

Example:

800 → 400 → 200 → 100

For a known number of half-lives:

N = N₀(1/2)ⁿ

For elapsed time:

n = t/t₁/₂

For fractional half-lives:

N = N₀(1/2)^(t/t₁/₂)

For general exponential calculations:

N = N₀e^(−λt)

Choose the simplest method that fits the information provided.

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A Full Problem-Solving Strategy

When solving a radioactive decay problem:

Identify what is given

Look for:

• initial amount
• final amount
• half-life
• elapsed time
• activity
• count rate
• background radiation

Check units

Make sure time units match.

Correct for background if necessary

Use:

corrected count = measured count − background

Find the number of half-lives

If possible:

n = t/t₁/₂

Choose the calculation

For whole half-lives:

halve repeatedly

or:

remaining = initial × (1/2)ⁿ

Work backwards if necessary

To find the original amount:

original = remaining × 2ⁿ

Check your answer

For ordinary decay without production of the parent isotope:

• remaining quantity should not exceed the initial quantity
• quantity should decrease with time
• units should be correct
• longer elapsed time should normally mean less parent isotope remains

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Worked Mixed Problem

A source has:

measured initial count rate = 1640 counts/min

background = 40 counts/min

half-life = 3 hours

What measured count rate would be expected after:

9 hours?

First correct the initial count:

1640 − 40 = 1600 counts/min

Number of half-lives:

n = 9/3

n = 3

Source count after three half-lives:

1600 → 800 → 400 → 200

Corrected source rate:

200 counts/min

Now add the background back to predict the detector reading:

200 + 40 = 240 counts/min

Therefore, the expected measured count rate is:

240 counts/min

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Worked Mixed Problem: Original Amount

A radioactive isotope has:

half-life = 12 years

remaining mass = 15 mg

elapsed time = 48 years

Number of half-lives:

n = 48/12

n = 4

Original mass:

m₀ = 15 × 2⁴

m₀ = 15 × 16

m₀ = 240 mg

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Worked Mixed Problem: Finding Half-Life

A source has an initial corrected activity of:

7200 Bq

After:

15 days

its activity is:

900 Bq

Halve:

7200 → 3600 → 1800 → 900

Therefore:

3 half-lives

Then:

t₁/₂ = 15/3

t₁/₂ = 5 days

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Common Mistakes

Mistake: Subtracting half of the original amount repeatedly.
Each half-life halves the amount currently remaining.

Mistake: Using n = half-life/elapsed time.
Correct:

n = elapsed time/half-life

Mistake: Mixing time units.
Convert days, hours, minutes or years before calculating.

Mistake: Multiplying by 1/2 only once when several half-lives have passed.
Use:

(1/2)ⁿ

Mistake: Halving when working backwards.
To find the original quantity, double for each previous half-life.

Mistake: Assuming radioactive decay is linear.
It is exponential.

Mistake: Forgetting background radiation in count-rate problems.
Correct the measured rate before applying half-life calculations.

Mistake: Confusing percentage remaining with percentage decayed.
They must add to:

100%

Mistake: Assuming the amount becomes exactly zero after several half-lives.
The exponential model approaches zero rather than suddenly reaching it.

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Key Terms

Half-life (t₁/₂) — time required for the expected amount or activity of radioactive parent material to fall to half its previous value.

Elapsed time (t) — amount of time that has passed.

Number of half-lives — number of half-life intervals that have elapsed.

Initial quantity — amount of radioactive parent material at the beginning of the calculation.

Remaining quantity — amount of radioactive parent material still undecayed.

Activity — number of nuclear decays occurring each second.

Becquerel (Bq) — SI unit of activity; 1 Bq = 1 decay/s.

Count rate — number of radiation events detected per unit time.

Corrected count rate — measured count rate after subtracting background radiation.

Decay constant (λ) — measure of decay probability per unit time.

Exponential decay — decrease in which the rate of change depends on the quantity remaining.

Background radiation — radiation detected from environmental sources rather than only the source being investigated.

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Key Takeaways

• Radioactive quantities halve during each half-life.
• The number of elapsed half-lives is n = t/t₁/₂.
• After n half-lives, the fraction remaining is (1/2)ⁿ.
• The remaining quantity is N = N₀(1/2)ⁿ.
• The original quantity can be found using N₀ = N × 2ⁿ.
• Elapsed time can be calculated using t = n × t₁/₂.
• Half-life can be calculated using t₁/₂ = t/n.
• After 1 half-life, 50% remains.
• After 2, 25% remains.
• After 3, 12.5% remains.
• After 4, 6.25% remains.
• Each half-life halves the remaining quantity, not the original quantity.
• Radioactive decay is exponential rather than linear.
• The same fraction decays during equal half-life intervals.
• Fractional numbers of half-lives can be handled using N = N₀(1/2)^(t/t₁/₂).
• The general exponential equation is N = N₀e^(−λt).
• Half-life and decay constant are related by t₁/₂ = 0.693/λ.
• Count-rate calculations should normally use background-corrected data.
• Decay graphs can be used to determine half-life by measuring the time between successive halvings.
• A shorter half-life produces a steeper decay curve.

Check Your Understanding

1. An isotope has a half-life of 5 hours. How many half-lives occur in 25 hours?
2. An isotope has a half-life of 8 days. How many half-lives occur in 40 days?
3. A sample begins with 6400 radioactive nuclei. How many remain after three half-lives?
4. A radioactive isotope initially has a mass of 320 mg. How much remains after four half-lives?
5. A source initially has an activity of 12,800 Bq. What is its activity after five half-lives?
6. What fraction of a radioactive sample remains after six half-lives?
7. What percentage remains after three half-lives?
8. What percentage has decayed after three half-lives?
9. A sample has a half-life of 4 days and an initial mass of 800 mg. How much remains after 16 days?
10. A source decreases from 6400 Bq to 400 Bq. How many half-lives have elapsed?
11. If the source in Question 10 has a half-life of 3 hours, how much time has elapsed?
12. A sample decreases from 960 mg to 120 mg in 18 years. Determine its half-life.
13. After four half-lives, 25 mg of a radioactive isotope remains. Determine the original mass.
14. A source has an activity of 125 Bq after five half-lives. Determine its original activity.
15. A detector records 850 counts/min and the background is 50 counts/min. What is the corrected count rate?
16. The corrected count rate in Question 15 falls to 100 counts/min. How many half-lives have elapsed?
17. An isotope has a half-life of 6 hours and an initial activity of 2000 Bq. Calculate its expected activity after 15 hours.
18. Explain why a radioactive-decay graph is exponential rather than linear.
19. A carbon-14 sample contains 1/8 of its expected original carbon-14 proportion. Using a half-life of 5730 years, estimate the elapsed time.
20. A detector initially reads 1640 counts/min with a background of 40 counts/min. The isotope has a half-life of 4 hours. Calculate the expected measured detector reading after 12 hours.