Counting half-lives

Definition

Half-life

The half-life of a radioactive isotope is the time taken for half the undecayed nuclei in a sample to decay, or for the activity of a source to fall by half.

Definition

Undecayed nuclei

Undecayed nuclei are the nuclei in a sample that have not yet decayed at a given moment.

Every half-life calculation turns on one number: how many half-lives have passed.
The number of half-lives is $\dfrac{t}{t_{1/2}}$ , where $t$ is the total elapsed time and $t_{1/2}$ is the half-life.
Both times must be in the same unit before dividing, so a time given in hours and a half-life given in minutes must be converted first.
Each half-life multiplies whatever is left by $12\dfrac{1}{2}$ , so after $n$ half-lives the fraction remaining is $(12)n=12n\left(\dfrac{1}{2}\right)^{n} = \dfrac{1}{2^{n}}$ .
It is worth knowing the first few values by sight: $12\dfrac{1}{2}$ , $14\dfrac{1}{4}$ , $18\dfrac{1}{8}$ , $116\dfrac{1}{16}$ , $132\dfrac{1}{32}$ , $164\dfrac{1}{64}$ for $n = 1$ to $6$ .
The same fraction applies whether the quantity being tracked is the activity, the count rate or the number of undecayed nuclei, because all three fall together in proportion.
For activity, this gives $A_{0} \times \left(\dfrac{1}{2}\right)^{n}$ , where $A_{0}$ is the starting activity and $A$ is the activity after $n$ half-lives.
For nuclei, the same relationship is $N_{0} \times \left(\dfrac{1}{2}\right)^{n}$ .
The fraction that has decayed is whatever is left over, so it is $\dfrac{1}{2^{n}}$ ; after $3$ half-lives, $18\dfrac{1}{8}$ remains and $78\dfrac{7}{8}$ has decayed.
Reading the question carefully matters here, because it may ask for the amount remaining, the amount decayed, or the ratio between them, and these are three different answers.

The four question types

Type one: find the amount left after a given time. Work out $n$ , then halve the starting value $n$ times, or multiply it by $12n\dfrac{1}{2^{n}}$ .
Type two: find the time for a given fall. Count how many halvings take the starting value to the final value, then multiply that count by the half-life, since $\times t_{1/2}$ .
Type three: find the half-life. Count the halvings between the two given values to get $n$ , then divide the elapsed time by it, since $t1/2=tnt_{1/2} = \dfrac{t}{n}$ .
Type four: express the fall as a ratio. Give the answer in the form final to initial, such as $1 : 8$ , or as the fraction $18\dfrac{1}{8}$ , whichever the question asks for.
When $n$ is not a whole number, halving repeatedly will not reach the answer exactly, and the value must be read from a graph instead.
Working in whole halvings is the quickest and safest route whenever the numbers allow it, because it needs no power key on the calculator.

Example

Finding the activity after a given time

A source has a starting activity of $Bq6400\ \text{Bq}$ and a half-life of $15$ minutes, and the activity after $1$ hour is required.
Converting to a common unit gives $t = 60$ minutes and $t_{1/2} = 15$ minutes.
The number of half-lives is $\dfrac{60}{15} = 4$ .
The fraction remaining is $124=116\dfrac{1}{2^{4}} = \dfrac{1}{16}$ .
Substituting gives $\times \dfrac{1}{16} = 400\ \text{Bq}$ .
Halving four times checks this: $\rightarrow 3200 \rightarrow 1600 \rightarrow 800 \rightarrow 400\ \text{Bq}$ .

Example

Finding the half-life from two readings

A count rate falls from $960$ counts per second to $60$ counts per second in $12$ hours, and the half-life is required.
Halving from the start gives $\rightarrow 480 \rightarrow 240 \rightarrow 120 \rightarrow 60$ counts per second.
That is four halvings, so $n = 4$ .
Substituting into $t1/2=tnt_{1/2} = \dfrac{t}{n}$ gives $t1/2=124=3t_{1/2} = \dfrac{12}{4} = 3$ hours.
Checking the fraction confirms it: $60960=116=124\dfrac{60}{960} = \dfrac{1}{16} = \dfrac{1}{2^{4}}$ , which needs exactly four half-lives.

Example

Finding the time for a given fall

An isotope has a half-life of $8$ days, and the time for its activity to fall to one eighth of its starting value is required.
Writing the fraction as a power of a half gives $18=123\dfrac{1}{8} = \dfrac{1}{2^{3}}$ , so $n = 3$ .
Substituting into $\times t_{1/2}$ gives $\times 8 = 24$ days.
Stated as a ratio, the activity has changed from $8$ to $1$ , so the final to initial ratio is $1 : 8$ .
The fraction that has decayed is $\dfrac{1}{8} = \dfrac{7}{8}$ of the original nuclei.

Example

Working with a number of nuclei

A sample starts with $4.8 \times 10^{12}$ undecayed nuclei of an isotope whose half-life is $25$ years, and the number left after a century is required.
The number of half-lives is $\dfrac{100}{25} = 4$ .
The fraction remaining is $124=116\dfrac{1}{2^{4}} = \dfrac{1}{16}$ .
Substituting gives $\times 10^{12} \times \dfrac{1}{16} = 3.0 \times 10^{11}$ .
The number that has decayed is $4.8 \times 10^{12} - 3.0 \times 10^{11} = 4.5 \times 10^{12}$ .
Keeping the answer in standard form avoids a power of ten error, which is the commonest slip in questions of this kind.

Reading a decay curve

Definition

Decay curve

A decay curve is a graph of activity, count rate or number of undecayed nuclei against time for a radioactive source.

A decay curve plots activity, count rate or number of undecayed nuclei on the vertical axis against time on the horizontal axis.
Start by reading the value on the vertical axis at $t = 0$ , which is the starting value $A_{0}$ .
Halve that value, find the halved value on the vertical axis, and draw a horizontal construction line across to the curve.
From that meeting point, drop a vertical construction line down to the time axis and read off the time, which is the half-life.
Repeat the same steps starting from a different point on the curve, for instance from the half value down to the quarter value, and check that the two times agree.
Averaging two or three such readings gives a better value than one, because the curve is drawn through scattered points.
Leave the construction lines on the graph, since they show the method and are what a marker looks for.
Where the vertical axis is a count rate, subtract any background count rate the question supplies before halving, otherwise the halving is applied to the wrong starting value.
To read the value at a time between plotted points, go up from the time axis to the curve and across, rather than trying to calculate it.

Example

Taking a half-life from a graph

A decay curve of count rate against time starts at $80$ counts per second at $t = 0$ .
Half of $80$ is $40$ , so a horizontal line is drawn from $40$ on the vertical axis across to the curve.
Dropping a vertical line from that point meets the time axis at $6$ minutes, giving a first half-life value of $6$ minutes.
For a second reading, half of $40$ is $20$ , and the same construction gives a time of $12$ minutes.
The interval between the two readings is $12 - 6 = 6$ minutes, which agrees with the first value.
Since both readings give $6$ minutes, the half-life is $6$ minutes.

Exam technique

Setting out a half-life calculation

Show the number of half-lives as a separate line of working, because $n$ is usually worth a mark of its own.
Print the substitution with the numbers in it, such as $\times \dfrac{1}{16}$ , rather than only the rearranged formula.
Convert both times into the same unit before dividing, and state the unit you converted to.
Give the unit with the final answer, since a bare number can be marked down where the unit carries a mark.
Write out the successive halvings when the numbers are friendly, because each halving is visible working that can earn credit if the final value slips.
Answer the question that was asked: remaining, decayed, or a ratio are three different answers from the same value of $n$ .
For a graph question, draw and leave the construction lines, and take two readings from different parts of the curve.
Keep large or small numbers in standard form throughout, because a power of ten error can cost every mark after it.

Common Mistake

Do not multiply the starting value by $n$ or divide it by $2 n$ , because each half-life multiplies what is left by $12\dfrac{1}{2}$ and the effect compounds.
Do not assume the activity is zero once several half-lives have passed, because a fraction always remains.
Do not divide the elapsed time by the half-life and then treat the result as the answer, because that quotient is $n$ and not the activity.
Do not give the fraction remaining when the question asks for the fraction that has decayed, since the two add to $1$ .
Do not read a half-life from a graph without halving the starting value first, because the interval must run between a value and its half.
Do not forget to subtract a stated background count rate before halving, or the starting value will be too high.
Do not mix units, because a half-life in minutes with a time in hours gives an answer wrong by a factor of $60$ .

Self review

Write the equation linking the number of half-lives, the elapsed time and the half-life.
What fraction of the original nuclei remains after $5$ half-lives?
A source of activity $Bq1200\ \text{Bq}$ has a half-life of $20$ minutes. What is its activity after $1$ hour?
A count rate falls from $640$ to $40$ counts per second in $18$ hours. What is the half-life?
An isotope has a half-life of $5$ days. How long does its activity take to fall to one sixteenth of its starting value?
Describe how to find a half-life from a graph of count rate against time.
After $4$ half-lives, what fraction of the original nuclei has decayed?

Counting half-lives

Definition

Half-life

The half-life of a radioactive isotope is the time taken for half the undecayed nuclei in a sample to decay, or for the activity of a source to fall by half.

Definition

Undecayed nuclei

Undecayed nuclei are the nuclei in a sample that have not yet decayed at a given moment.

Every half-life calculation turns on one number: how many half-lives have passed.
The number of half-lives is $\dfrac{t}{t_{1/2}}$ , where $t$ is the total elapsed time and $t_{1/2}$ is the half-life.
Both times must be in the same unit before dividing, so a time given in hours and a half-life given in minutes must be converted first.
Each half-life multiplies whatever is left by $12\dfrac{1}{2}$ , so after $n$ half-lives the fraction remaining is $(12)n=12n\left(\dfrac{1}{2}\right)^{n} = \dfrac{1}{2^{n}}$ .
It is worth knowing the first few values by sight: $12\dfrac{1}{2}$ , $14\dfrac{1}{4}$ , $18\dfrac{1}{8}$ , $116\dfrac{1}{16}$ , $132\dfrac{1}{32}$ , $164\dfrac{1}{64}$ for $n = 1$ to $6$ .
The same fraction applies whether the quantity being tracked is the activity, the count rate or the number of undecayed nuclei, because all three fall together in proportion.
For activity, this gives $A_{0} \times \left(\dfrac{1}{2}\right)^{n}$ , where $A_{0}$ is the starting activity and $A$ is the activity after $n$ half-lives.
For nuclei, the same relationship is $N_{0} \times \left(\dfrac{1}{2}\right)^{n}$ .
The fraction that has decayed is whatever is left over, so it is $\dfrac{1}{2^{n}}$ ; after $3$ half-lives, $18\dfrac{1}{8}$ remains and $78\dfrac{7}{8}$ has decayed.
Reading the question carefully matters here, because it may ask for the amount remaining, the amount decayed, or the ratio between them, and these are three different answers.

The four question types

Type one: find the amount left after a given time. Work out $n$ , then halve the starting value $n$ times, or multiply it by $12n\dfrac{1}{2^{n}}$ .
Type two: find the time for a given fall. Count how many halvings take the starting value to the final value, then multiply that count by the half-life, since $\times t_{1/2}$ .
Type three: find the half-life. Count the halvings between the two given values to get $n$ , then divide the elapsed time by it, since $t1/2=tnt_{1/2} = \dfrac{t}{n}$ .
Type four: express the fall as a ratio. Give the answer in the form final to initial, such as $1 : 8$ , or as the fraction $18\dfrac{1}{8}$ , whichever the question asks for.
When $n$ is not a whole number, halving repeatedly will not reach the answer exactly, and the value must be read from a graph instead.
Working in whole halvings is the quickest and safest route whenever the numbers allow it, because it needs no power key on the calculator.

Example

Finding the activity after a given time

A source has a starting activity of $Bq6400\ \text{Bq}$ and a half-life of $15$ minutes, and the activity after $1$ hour is required.
Converting to a common unit gives $t = 60$ minutes and $t_{1/2} = 15$ minutes.
The number of half-lives is $\dfrac{60}{15} = 4$ .
The fraction remaining is $124=116\dfrac{1}{2^{4}} = \dfrac{1}{16}$ .
Substituting gives $\times \dfrac{1}{16} = 400\ \text{Bq}$ .
Halving four times checks this: $\rightarrow 3200 \rightarrow 1600 \rightarrow 800 \rightarrow 400\ \text{Bq}$ .

Example

Finding the half-life from two readings

A count rate falls from $960$ counts per second to $60$ counts per second in $12$ hours, and the half-life is required.
Halving from the start gives $\rightarrow 480 \rightarrow 240 \rightarrow 120 \rightarrow 60$ counts per second.
That is four halvings, so $n = 4$ .
Substituting into $t1/2=tnt_{1/2} = \dfrac{t}{n}$ gives $t1/2=124=3t_{1/2} = \dfrac{12}{4} = 3$ hours.
Checking the fraction confirms it: $60960=116=124\dfrac{60}{960} = \dfrac{1}{16} = \dfrac{1}{2^{4}}$ , which needs exactly four half-lives.

Example

Finding the time for a given fall

An isotope has a half-life of $8$ days, and the time for its activity to fall to one eighth of its starting value is required.
Writing the fraction as a power of a half gives $18=123\dfrac{1}{8} = \dfrac{1}{2^{3}}$ , so $n = 3$ .
Substituting into $\times t_{1/2}$ gives $\times 8 = 24$ days.
Stated as a ratio, the activity has changed from $8$ to $1$ , so the final to initial ratio is $1 : 8$ .
The fraction that has decayed is $\dfrac{1}{8} = \dfrac{7}{8}$ of the original nuclei.

Example

Working with a number of nuclei

A sample starts with $4.8 \times 10^{12}$ undecayed nuclei of an isotope whose half-life is $25$ years, and the number left after a century is required.
The number of half-lives is $\dfrac{100}{25} = 4$ .
The fraction remaining is $124=116\dfrac{1}{2^{4}} = \dfrac{1}{16}$ .
Substituting gives $\times 10^{12} \times \dfrac{1}{16} = 3.0 \times 10^{11}$ .
The number that has decayed is $4.8 \times 10^{12} - 3.0 \times 10^{11} = 4.5 \times 10^{12}$ .
Keeping the answer in standard form avoids a power of ten error, which is the commonest slip in questions of this kind.

Reading a decay curve

Definition

Decay curve

A decay curve is a graph of activity, count rate or number of undecayed nuclei against time for a radioactive source.

A decay curve plots activity, count rate or number of undecayed nuclei on the vertical axis against time on the horizontal axis.
Start by reading the value on the vertical axis at $t = 0$ , which is the starting value $A_{0}$ .
Halve that value, find the halved value on the vertical axis, and draw a horizontal construction line across to the curve.
From that meeting point, drop a vertical construction line down to the time axis and read off the time, which is the half-life.
Repeat the same steps starting from a different point on the curve, for instance from the half value down to the quarter value, and check that the two times agree.
Averaging two or three such readings gives a better value than one, because the curve is drawn through scattered points.
Leave the construction lines on the graph, since they show the method and are what a marker looks for.
Where the vertical axis is a count rate, subtract any background count rate the question supplies before halving, otherwise the halving is applied to the wrong starting value.
To read the value at a time between plotted points, go up from the time axis to the curve and across, rather than trying to calculate it.

Example

Taking a half-life from a graph

A decay curve of count rate against time starts at $80$ counts per second at $t = 0$ .
Half of $80$ is $40$ , so a horizontal line is drawn from $40$ on the vertical axis across to the curve.
Dropping a vertical line from that point meets the time axis at $6$ minutes, giving a first half-life value of $6$ minutes.
For a second reading, half of $40$ is $20$ , and the same construction gives a time of $12$ minutes.
The interval between the two readings is $12 - 6 = 6$ minutes, which agrees with the first value.
Since both readings give $6$ minutes, the half-life is $6$ minutes.

Exam technique

Setting out a half-life calculation

Show the number of half-lives as a separate line of working, because $n$ is usually worth a mark of its own.
Print the substitution with the numbers in it, such as $\times \dfrac{1}{16}$ , rather than only the rearranged formula.
Convert both times into the same unit before dividing, and state the unit you converted to.
Give the unit with the final answer, since a bare number can be marked down where the unit carries a mark.
Write out the successive halvings when the numbers are friendly, because each halving is visible working that can earn credit if the final value slips.
Answer the question that was asked: remaining, decayed, or a ratio are three different answers from the same value of $n$ .
For a graph question, draw and leave the construction lines, and take two readings from different parts of the curve.
Keep large or small numbers in standard form throughout, because a power of ten error can cost every mark after it.

Common Mistake

Do not multiply the starting value by $n$ or divide it by $2 n$ , because each half-life multiplies what is left by $12\dfrac{1}{2}$ and the effect compounds.
Do not assume the activity is zero once several half-lives have passed, because a fraction always remains.
Do not divide the elapsed time by the half-life and then treat the result as the answer, because that quotient is $n$ and not the activity.
Do not give the fraction remaining when the question asks for the fraction that has decayed, since the two add to $1$ .
Do not read a half-life from a graph without halving the starting value first, because the interval must run between a value and its half.
Do not forget to subtract a stated background count rate before halving, or the starting value will be too high.
Do not mix units, because a half-life in minutes with a time in hours gives an answer wrong by a factor of $60$ .

Self review

Write the equation linking the number of half-lives, the elapsed time and the half-life.
What fraction of the original nuclei remains after $5$ half-lives?
A source of activity $Bq1200\ \text{Bq}$ has a half-life of $20$ minutes. What is its activity after $1$ hour?
A count rate falls from $640$ to $40$ counts per second in $18$ hours. What is the half-life?
An isotope has a half-life of $5$ days. How long does its activity take to fall to one sixteenth of its starting value?
Describe how to find a half-life from a graph of count rate against time.
After $4$ half-lives, what fraction of the original nuclei has decayed?

6.4.3 Half-life calculations

Counting half-lives

The four question types

Finding the activity after a given time

Finding the half-life from two readings

Finding the time for a given fall

Working with a number of nuclei

Reading a decay curve

Taking a half-life from a graph

Setting out a half-life calculation

Review 6.4.3 Half-life calculations by teaching Genie

6.4.3 Half-life calculations

Counting half-lives

The four question types

Finding the activity after a given time

Finding the half-life from two readings

Finding the time for a given fall

Working with a number of nuclei

Reading a decay curve

Taking a half-life from a graph

Setting out a half-life calculation

Review 6.4.3 Half-life calculations by teaching Genie

6.4.3 Half-life calculations

Counting half-lives

The four question types

Finding the activity after a given time

Finding the half-life from two readings

Finding the time for a given fall

Working with a number of nuclei

Reading a decay curve

Taking a half-life from a graph

Setting out a half-life calculation

Review 6.4.3 Half-life calculations by teaching Genie

6.4.3 Half-life calculations Revision Guide

6.4.3 Half-life calculations

Counting half-lives

The four question types

Finding the activity after a given time

Finding the half-life from two readings

Finding the time for a given fall

Working with a number of nuclei

Reading a decay curve

Taking a half-life from a graph

Setting out a half-life calculation

Review 6.4.3 Half-life calculations by teaching Genie

6.4.3 Half-life calculations Revision Guide