How to Find Half-Life from Two Activity Measurements

A detector measures the activity of a radioactive source at 2:00 pm and again at 10:00 pm. The first reading is 800 Bq, and the second is 200 Bq. What is the half-life?

Before calculating, predict how many half-lives have passed. The activity has fallen to one-quarter of its original value. That means two half-lives have passed in 8.0 hours.

Therefore,

\[ T_{1/2}=\frac{8.0}{2}=\boxed{4.0\text{ h}} \]

This is straightforward when one reading is an exact fraction of the other. But experimental measurements rarely produce such convenient numbers. You need a method that works for any pair of activity measurements, including readings taken at times other than the start of an experiment.

01The method: compare the two activity measurements

For a single radioactive isotope, activity decreases exponentially. Each nucleus has a constant probability of decaying per unit time, so the activity falls by the same fraction over equal time intervals.

Suppose you measure:

• Initial activity \(A_1\) at time \(t_1\).
• Later activity \(A_2\) at time \(t_2\).

The elapsed time between measurements is:

\[ \Delta t=t_2-t_1 \]

The fraction of the original activity remaining is:

\[ \frac{A_2}{A_1} = \left(\frac{1}{2}\right)^{\Delta t/T_{1/2}} \]

Rearranging gives the equation you need:

\[ \boxed{ T_{1/2}= \frac{\Delta t\ln 2} {\ln(A_1/A_2)} } \]

Here:

• \(T_{1/2}\) is the half-life.
• \(\Delta t\) is the time between measurements.
• \(A_1\) is the earlier activity.
• \(A_2\) is the later activity.
• \(\ln\) is the natural logarithm.

Activity is measured in becquerels (Bq), where \(1\text{ Bq}\) represents one nuclear decay per second. The half-life has the same time units as \(\Delta t\).

An important detail: The first measurement does not need to be taken at \(t=0\). You only need the time difference between the two readings.

Why does the logarithm appear?

Imagine the activity falls from 800 Bq to 400 Bq, then to 200 Bq.

Each halving represents one half-life.

However, suppose the activity falls from 800 Bq to 300 Bq. The ratio is no longer a whole power of two, so you cannot simply count complete halvings.

The logarithm allows you to calculate the fractional number of half-lives:

\[ n=\frac{\ln(A_1/A_2)}{\ln 2} \]

Once you know \(n\), divide the elapsed time by it:

\[ T_{1/2}=\frac{\Delta t}{n} \]

Both approaches give exactly the same result.

A quick decision rule

When you see two activity measurements in an HSC question:

1. Identify the earlier and later readings.
2. Calculate the elapsed time between them.
3. Check whether the readings are actual activities or detector count rates.
4. If they are genuine activities, calculate \(A_1/A_2\).
5. Substitute into the half-life equation.
6. Check that the answer makes physical sense.

If the activity has fallen by a factor of four, two half-lives have passed. If it has fallen by a factor of eight, three have passed.

For other ratios, use logarithms rather than trying to estimate the number of halvings.

02How do scientists obtain the two measurements?

There is an experimental detail worth understanding before using detector readings in a calculation.

A radioactive source does not normally come with a display showing its activity. In a typical school experiment, a Geiger-Müller (GM) tube detects radiation from the source and records electrical pulses.

The detector measures a count rate, not necessarily the actual activity.

The measurement process follows a simple sequence:

Step 1: Measure the background radiation.

With the radioactive source removed, record the detector’s count rate over a fixed time interval.

Background radiation comes from the surroundings, including naturally occurring radioactive materials and cosmic radiation.

Step 2: Position the radioactive source.

Place the source at a fixed distance from the GM tube. Keep its orientation and the detector arrangement unchanged.

Step 3: Record the first count rate.

Count the detector pulses over a known interval, such as 60 seconds.

Step 4: Wait for a known period.

Allow the radioactive sample to decay for a measured time.

Step 5: Record the second count rate.

Repeat the measurement using the same detector, distance, orientation, and counting interval.

What should happen if the sample contains one radioactive isotope?

The background-corrected count rate should decrease exponentially. If the elapsed time is one half-life, the sample’s contribution to the detector reading should halve.

Count rate is not quite the same as activity

This is an important distinction.

Activity measures the number of nuclear decays occurring each second. Count rate measures the number of radiation events registered by the detector each second.

Not every decay necessarily produces a recorded count. Some radiation travels in directions that miss the detector, and the detector may not register every particle that reaches it.

For a constant detector arrangement, the relationship can be written as:

\[ C_{\text{sample}}=\varepsilon A \]

where:

• \(C_{\text{sample}}\) is the sample’s measured count rate.
• \(\varepsilon\) is the detector’s effective counting efficiency.
• \(A\) is the actual activity.

If the detector arrangement stays unchanged, \(\varepsilon\) stays approximately constant.

Therefore, when comparing two readings:

\[ \frac{C_1}{C_2} = \frac{\varepsilon A_1}{\varepsilon A_2} = \frac{A_1}{A_2} \]

The detector efficiency cancels.

This is why you can calculate a half-life from two count rates, provided the experimental conditions remain consistent and background radiation has been removed.

03Correcting for background radiation

A detector records radiation from both the radioactive source and its surroundings.

The total measured count rate is:

\[ C_{\text{measured}} = C_{\text{sample}}+C_{\text{background}} \]

Rearranging:

\[ \boxed{ C_{\text{sample}} = C_{\text{measured}}-C_{\text{background}} } \]

You must subtract the background before calculating the ratio.

Why? Background radiation does not decrease according to the half-life of your radioactive sample. It continues contributing to the detector reading.

Consider a detector that initially records 500 counts/min, including 50 counts/min of background radiation.

The radioactive sample itself contributes:

\[ 500-50=450\text{ counts/min} \]

If the sample’s activity halves, its contribution becomes:

\[ \frac{450}{2}=225\text{ counts/min} \]

The detector will therefore record:

\[ 225+50=275\text{ counts/min} \]

Notice that the total detector reading has not halved. It has decreased from 500 to 275 counts/min.

This is why applying the half-life equation directly to uncorrected detector readings can produce an incorrect answer.

04What can make two measurements unreliable?

Even if you use the correct equation, the result can be wrong if the measurements are not comparable.

Three experimental conditions deserve particular attention.

1. The detector must remain in the same position

Suppose the source is moved further away before the second measurement.

The detector may record fewer particles simply because a smaller fraction of the emitted radiation reaches it.

That does not mean the radioactive isotope has decayed faster.

The solution is to use a fixed holder and keep the source-to-detector distance constant.

2. Background radiation must be consistent

If the background changes between measurements, subtracting an old background reading may introduce an error.

Ideally, measure the background under the same conditions as the sample measurements. If necessary, repeat the background measurement before and after the experiment.

3. Counting results fluctuate

Radioactive decay is random. Even if the source has a perfectly constant activity, two measurements taken under identical conditions will not necessarily produce exactly the same count rate.

Longer counting intervals and repeated measurements help reduce the influence of random fluctuations.

For an ordinary HSC calculation, you will generally be given readings that can be treated as exact. In an actual experiment, the calculated half-life is an estimate with some uncertainty.

06Don’t confuse activity with nuclear mass and energy

There is another distinction worth keeping clear in nuclear physics questions.

Two activity measurements tell you how quickly a radioactive sample is decaying. They do not directly tell you how much energy is released in each decay.

These quantities describe different properties.

Charge and nucleon number are conserved

Consider alpha decay of polonium-210:

\[ {}^{210}_{84}\mathrm{Po} \rightarrow {}^{206}_{82}\mathrm{Pb} + {}^{4}_{2}\mathrm{He} \]

Check the nucleon numbers:

\[ 210=206+4 \]

Check the proton numbers, which determine nuclear charge:

\[ 84=82+2 \]

Both quantities are conserved.

This matters when identifying the products of a nuclear decay and deciding which masses belong in an energy calculation.

Nuclear mass and atomic mass are different

A nuclear mass includes only the protons and neutrons in the nucleus.

An atomic mass includes the nucleus and all the electrons in a neutral atom.

For example, a neutral polonium-210 atom contains 84 electrons, whereas its nucleus contains none.

When calculating nuclear energy changes, use a consistent mass convention. You can use nuclear masses for the parent and all nuclear products, or compatible atomic masses when the electron numbers balance. Do not mix a nuclear mass with an atomic mass without making the appropriate electron correction.

Binding energy is not the same as released energy

Binding energy is the energy required to separate a nucleus into its individual nucleons.

It is related to the mass defect:

\[ E_b=\Delta mc^2 \]

where:

• \(E_b\) is the total nuclear binding energy.
• \(\Delta m\) is the mass defect.
• \(c\) is the speed of light.

Released energy, usually represented by \(Q\), is calculated using the mass difference between the initial nucleus and all final products:

\[ Q=(m_{\text{initial}}-m_{\text{final}})c^2 \]

Here, \(Q>0\) means energy is released.

For alpha decay, this is the mass difference between the parent nucleus and the combined masses of the daughter nucleus and alpha particle.

The activity measurements alone cannot give you this energy. You need the relevant mass data.

For a more detailed treatment of these calculations, see How to Calculate Binding Energy per Nucleon from Nuclear Mass.

07Questions and solutions

The following questions move from a direct half-life calculation to experimental corrections and more demanding problems involving background radiation and nuclear decay.

Question 1

A calibrated instrument measures the activity of a radioactive isotope as 960 Bq at 1.0 hour and 240 Bq at 13.0 hours.

Calculate the half-life of the isotope.

Solution 1

The half-life is 6.0 hours. The activity has decreased by a factor of four, meaning two half-lives have passed during the 12-hour interval.

Step 1: Calculate the elapsed time.

\[ \Delta t=13.0-1.0=12.0\text{ h} \]

Step 2: Calculate the activity ratio.

\[ \frac{A_1}{A_2}=\frac{960}{240}=4 \]

Since \(4=2^2\), two half-lives have passed.

Step 3: Calculate the half-life.

\[ T_{1/2}=\frac{12.0}{2} \]

\[ \boxed{T_{1/2}=6.0\text{ h}} \]

The important point is that you use the time between measurements, not the time of the second measurement measured from zero.

Question 2

A student measures a radioactive sample using a GM tube.

The following data are recorded:

The background count rate is 10 counts/min.

Calculate the half-life of the isotope.

Explain why using the ratio of the two raw count rates would give an incorrect result.

Solution 2

The half-life is approximately 4.53 hours. The background contribution must be removed before calculating the ratio.

Step 1: Convert the background reading into a count rate.

The background is already given as:

\[ C_{\text{background}}=10\text{ counts/min} \]

Step 2: Calculate the gross count rates.

First measurement:

\[ C_1=\frac{540}{3.0} =180\text{ counts/min} \]

Second measurement:

\[ C_2=\frac{180}{3.0} =60\text{ counts/min} \]

Step 3: Calculate the net count rates.

\[ C_{\text{net},1}=180-10=170 \]

\[ C_{\text{net},2}=60-10=50 \]

Both are measured in counts/min.

Step 4: Apply the half-life equation.

\[ T_{1/2} = \frac{8.0\times0.693} {\ln(170/50)} \]

\[ T_{1/2} = \frac{5.544}{1.224} \]

\[ \boxed{T_{1/2}=4.53\text{ h}} \]

Using the raw count rates would give:

\[ T_{1/2} = \frac{8.0\times0.693}{\ln(180/60)} \]

\[ T_{1/2}=5.05\text{ h} \]

This is incorrect because the background contribution remains approximately constant while the radioactive sample decays.

Question 3

A radioactive source is measured using the same GM tube at two different times.

The background count rate is 15 counts/min.

The detector has been moved between measurements. Its efficiency at the second position is 40% of its original efficiency.

Calculate the actual half-life of the isotope.

Explain why simply subtracting background radiation and taking the ratio of the resulting count rates is insufficient.

Solution 3

The actual half-life is 6.0 hours. The second detector arrangement has a lower efficiency, so the observed reduction in count rate is greater than the reduction in actual activity.

Step 1: Subtract the background count rate.

For the first measurement:

\[ C_1=415-15=400\text{ counts/min} \]

For the second measurement:

\[ C_2=95-15=80\text{ counts/min} \]

Step 2: Account for the change in detector efficiency.

The first reading is:

\[ 400=\varepsilon A_1 \]

The second reading is:

\[ 80=0.40\varepsilon A_2 \]

Rearranging:

\[ A_1=\frac{400}{\varepsilon} \]

\[ A_2=\frac{80}{0.40\varepsilon} =\frac{200}{\varepsilon} \]

Therefore:

\[ \frac{A_1}{A_2}=\frac{400}{200}=2 \]

The actual activity has halved.

Step 3: Calculate the half-life.

\[ T_{1/2} = \frac{6.0\times0.693}{\ln 2} \]

\[ \boxed{T_{1/2}=6.0\text{ h}} \]

The tempting approach is to calculate \(400/80=5\), giving an apparent half-life of approximately 2.58 hours.

That approach assumes the detector efficiency is unchanged. It is not.

The observed count-rate reduction is caused by two separate effects: radioactive decay and the reduction in detector efficiency. You must separate them before finding the true activity ratio.

Question 4

A pure sample of polonium-210 undergoes alpha decay according to:

\[ {}^{210}_{84}\mathrm{Po} \rightarrow {}^{206}_{82}\mathrm{Pb} + {}^{4}_{2}\mathrm{He} \]

A calibrated instrument measures the activity as 960 Bq. Ten hours later, it measures 240 Bq.

A student makes the following claims:

• The half-life is 5.0 hours.
• Three-quarters of the original polonium nuclei have decayed.
• Three-quarters of the original nuclear mass has been converted into energy.
• The energy released can be calculated directly from the two activity readings.

Determine which claims are correct. Check the conservation of charge and nucleon number, and explain what additional information is needed to calculate the energy released.

Solution 4

The first two claims are correct, but the final two are incorrect. The activity measurements determine the half-life and the fraction of parent nuclei remaining, not the energy released per decay.

Step 1: Calculate the half-life.

The activity ratio is:

\[ \frac{A_1}{A_2}=\frac{960}{240}=4 \]

Two half-lives have passed.

Therefore:

\[ T_{1/2}=\frac{10}{2} \]

\[ \boxed{T_{1/2}=5.0\text{ h}} \]

Step 2: Determine the fraction of parent nuclei remaining.

For a single isotope:

\[ A=\lambda N \]

Here, \(\lambda\) is the decay constant and \(N\) is the number of undecayed parent nuclei.

Because the isotope is unchanged, its decay constant remains constant. Therefore, the ratio of activities equals the ratio of undecayed nuclei:

\[ \frac{N_2}{N_1}=\frac{A_2}{A_1} \]

\[ \frac{N_2}{N_1}=\frac{240}{960} =\frac14 \]

One-quarter of the original polonium nuclei remain.

Therefore, the fraction that has decayed is:

\[ 1-\frac14=\frac34 \]

So 75% of the original parent nuclei have decayed.

Step 3: Check nucleon number conservation.

For the alpha decay:

\[ 210=206+4 \]

There are 210 nucleons before the decay and 210 afterwards.

Step 4: Check charge conservation.

\[ 84=82+2 \]

The total proton number is also conserved.

Step 5: Explain why the mass-energy claim is incorrect.

The decay does not convert the entire mass of each polonium nucleus into energy.

Instead, the original polonium nucleus becomes a lead nucleus and an alpha particle. Both products retain almost all the original mass.

Only the small difference between the initial and final total masses becomes released energy.

To calculate the energy released, you need the mass of the parent polonium nucleus, the mass of the daughter lead nucleus, and the mass of the alpha particle.

Using nuclear masses:

\[ Q= (m_{\mathrm{Po}}-m_{\mathrm{Pb}}-m_{\alpha})c^2 \]

The activity measurements do not provide these masses.

Also, this released energy is not simply the binding energy of the polonium nucleus. Binding energy describes the energy required to separate a nucleus into its individual nucleons.

The two calculations describe related but different physical quantities.

Question 5

A radioactive source is measured using a GM tube at three equally spaced times.

The detector remains in exactly the same position, and the source contains only one radioactive isotope.

The following gross count rates are recorded:

Assume that:

• The background count rate is constant but unknown.
• The detector efficiency remains constant.
• The measurements are idealised exact values.
• No radioactive decay products contribute to the detector readings.

A student uses the first and last measurements to calculate a half-life of approximately 4.70 hours.

The student argues that the middle reading is unnecessary because two measurements are enough to determine half-life.

Determine the actual background count rate and the isotope’s half-life. Then explain whether the student’s method can uniquely determine the half-life using only the first and last gross readings.

Solution 5

The background count rate is 20 counts/min, and the actual half-life is 4.0 hours. The student’s calculation using only the first and last gross readings is not sufficient to determine the true half-life because the background is unknown.

This question tests an important limitation of the two-reading method: two measurements determine a half-life only when the measured quantities are proportional to the actual activity.

Step 1: Define the unknown background count rate.

Let \(B\) represent the constant background count rate, in counts/min.

Subtracting this background from each measurement gives the actual sample contributions:

\[ C_0=260-B \]

\[ C_1=140-B \]

\[ C_2=80-B \]

Step 2: Use the equal time intervals.

The measurements are separated by equal intervals of 4.0 hours.

For a single radioactive isotope, the same fraction of activity remains during each equal interval.

Therefore, the three background-corrected readings form a geometric sequence.

This means:

\[ \frac{C_1}{C_0} = \frac{C_2}{C_1} \]

Rearranging:

\[ C_1^2=C_0C_2 \]

Substitute the expressions containing the unknown background:

\[ (140-B)^2=(260-B)(80-B) \]

Step 3: Solve for the background count rate.

Expand both sides:

\[ 19600-280B+B^2 \]

\[ =20800-340B+B^2 \]

Cancel the \(B^2\) terms and rearrange:

\[ 60B=1200 \]

\[ \boxed{B=20\text{ counts/min}} \]

Step 4: Calculate the actual sample count rates.

Subtract the background from each gross measurement:

All rates are in counts/min.

Step 5: Calculate the half-life.

The net count rate falls from 240 to 120 counts/min in 4.0 hours.

That is exactly one halving.

Therefore:

\[ \boxed{T_{1/2}=4.0\text{ h}} \]

The second interval confirms the result:

\[ 120\rightarrow60\text{ counts/min} \]

Again, one half-life has passed.

Step 6: Explain why the student’s original method is insufficient.

Using only the first and last gross readings gives:

\[ T_{\text{apparent}} = \frac{8.0\times0.693}{\ln(260/80)} \]

\[ T_{\text{apparent}}\approx4.70\text{ h} \]

This assumes that the entire gross count rate comes from the radioactive sample.

But the background is constant, not decaying. Without knowing its value, you cannot separate the sample’s contribution from the total detector reading.

For example, assuming zero background gives an apparent half-life of approximately 4.70 hours. Assuming a background of 20 counts/min gives a half-life of 4.0 hours. Both assumptions fit the first and last readings, but only the second also fits the middle reading under the single-isotope model.

The third measurement supplies the extra constraint needed to determine the unknown background.

This is the important distinction: two genuine activity measurements are enough to calculate a half-life, but two uncorrected detector readings are not necessarily enough when the background is unknown.

Once you can calculate half-life from two measurements, the next useful step is connecting it to the decay constant:

\[ \boxed{\lambda=\frac{\ln 2}{T_{1/2}}} \]

The decay constant gives you another way to describe how quickly a radioactive isotope decays. See Decay constant and half-life: HSC Physics explained.

Remember the distinction when approaching an exam question: use activity measurements to determine how quickly nuclei decay, and use mass differences to calculate the energy released by a nuclear reaction.