Why You Must Subtract Background Counts Before Analysing Decay

A Geiger-Müller tube records 150 counts in a 10-second interval. Two minutes later, it records only 90 counts. Has the radioactive sample’s count rate fallen by 40%?

Not necessarily. Suppose the detector records 30 background counts every 10 seconds, even when the sample is removed. The sample itself contributed 120 counts initially and 60 counts two minutes later.

The sample’s count rate has actually halved.

This is why you must subtract the background count before analysing radioactive decay. The detector measures radiation from the sample and its surroundings together, but only the sample-related signal follows the sample’s decay law. If you leave the background in your data, you can calculate the wrong half-life, decay constant, and activity.

01What is background radiation, and why does the detector count it?

Imagine placing a radioactive source in front of a Geiger-Müller (GM) tube.

The source emits ionising radiation. When that radiation enters the tube, it causes ionisation of the gas inside. This produces an electrical pulse, which travels to an electronic counter. The counter records each detected event as one count.

The important detail is that the detector cannot identify where each pulse originated.

Some radiation comes from the radioactive sample. Other radiation comes from the surrounding environment.

This environmental radiation is called background radiation. It can include:

• Cosmic radiation arriving from space.
• Radiation from naturally occurring radioactive materials in rocks, soil, and building materials.
• Radiation from other radioactive sources in the surrounding area.

Even with the sample removed, the detector will usually continue recording counts.

The two contributions to a measured count

The total number of counts recorded by the detector consists of two contributions:

1. Counts caused by radiation from the radioactive sample.
2. Counts caused by background radiation.

Therefore:

\[ N_{\mathrm{observed}}= N_{\mathrm{sample}}+N_{\mathrm{background}} \]

Rearranging gives:

\[ \boxed{ N_{\mathrm{sample}}= N_{\mathrm{observed}}-N_{\mathrm{background}} } \]

This is the background correction.

If both measurements are taken over the same time interval, you can subtract the counts directly.

However, if the measurements have different durations, you must first convert them into count rates.

Count rate is calculated using:

\[ R=\frac{N}{t} \]

where:

• \(R\) is the count rate, measured in counts per second or counts per minute.
• \(N\) is the number of recorded counts.
• \(t\) is the measurement time.

The corrected count rate is therefore:

\[ \boxed{ R_{\mathrm{sample}}= R_{\mathrm{observed}}-R_{\mathrm{background}} } \]

This is the quantity you should use when analysing the sample’s radioactive decay.

02Why background subtraction matters when calculating half-life

Now consider what happens when you measure a radioactive sample repeatedly.

You expect its activity to decrease exponentially. As unstable nuclei decay, fewer remain to undergo further decay.

For a detector operating under unchanged conditions, the sample-related count rate follows the same exponential pattern as the sample’s activity.

The difficulty is that background radiation does not decrease as the sample decays.

Assuming a constant background count rate:

\[ R_{\mathrm{observed}}(t) = R_0e^{-\lambda t}+R_{\mathrm{background}} \]

where:

• \(R_{\mathrm{observed}}(t)\) is the total measured count rate at time \(t\).
• \(R_0\) is the sample-related count rate at the beginning of the experiment.
• \(\lambda\) is the decay constant, measured in inverse units of time.
• \(t\) is the elapsed time.
• \(R_{\mathrm{background}}\) is the constant background count rate.

The first term decreases exponentially. The second term remains approximately constant.

This creates a problem if you analyse the uncorrected readings.

What happens if you do not subtract background?

Imagine a sample whose corrected count rate halves every 2 minutes.

Its background count rate is 30 counts per 10 seconds.

The following table contains illustrative measurements:

The corrected count rate halves every 2 minutes:

\[ 120\rightarrow60\rightarrow30\rightarrow15 \]

Therefore:

\[ \boxed{t_{1/2}=2\ \mathrm{min}} \]

But look at the observed readings:

\[ 150\rightarrow90\rightarrow60\rightarrow45 \]

They do not halve at regular intervals.

Why?

As the sample becomes less active, background radiation makes up an increasingly large fraction of the total reading.

Eventually, the sample contributes very few counts, and the detector’s readings approach the background level of 30 counts per 10 seconds.

A common mistake is to assume that the observed count rate should eventually reach zero.

It should not.

Even if every unstable nucleus in the sample has decayed, the detector will continue recording background radiation.

This is why the background-corrected count rate must be used when determining half-life.

03How should background radiation be measured experimentally?

Understanding the correction is one thing. Collecting reliable measurements is another.

A good experiment needs to separate the sample’s contribution from the radiation already present in the environment.

Consider how you would set up the investigation.

First, position the GM tube and electronic counter. Keep the detector at a fixed location, with a consistent operating voltage.

Second, remove the radioactive sample from the detector’s vicinity. Record the number of background counts over a measured interval.

Third, place the radioactive sample in its holder at a fixed distance from the detector. Record the total count over the same time interval.

Fourth, subtract the background count from the observed count to estimate the sample’s contribution.

Finally, repeat the measurements at regular time intervals. Use the corrected readings to analyse the decay.

Why should the background measurement be repeated?

Background radiation is not perfectly constant.

It can fluctuate because of changes in the surrounding environment, nearby radioactive sources, and the random nature of radiation detection.

A background reading taken at the beginning of an experiment might not perfectly represent the background several hours later.

A sensible experimental approach is to:

• Measure background radiation before starting the experiment.
• Repeat the background measurement after the main experiment.
• Compare the two readings to check whether the background has remained reasonably stable.
• If the background changes gradually, estimate the background at each measurement time using an appropriate interpolation.
• If it changes substantially or suddenly, investigate the cause rather than assuming a single background value applies throughout.

For a more reliable estimate, background measurements can also be taken over longer intervals or repeated several times.

What if the corrected count is negative?

Suppose the detector records 18 counts in a sample measurement, while the estimated background is 21 counts over the same interval.

The subtraction gives:

\[ N_{\mathrm{sample}}=18-21=-3 \]

Does this mean the sample has negative activity?

Of course not.

Radioactive counting is random. The measured background is an estimate, not an exact number that must appear in every interval.

A negative corrected result indicates that the sample’s contribution is too small to distinguish reliably from the background in that particular measurement.

The sensible response is to repeat the measurement, increase the counting time, or improve the experimental conditions.

Do not interpret a negative corrected count as physically negative radiation.

04Count rate is not the same as activity

There is another distinction worth understanding before moving on.

A corrected count rate tells you how frequently the detector records radiation attributable to the sample.

Activity tells you how frequently nuclei actually decay.

Activity is measured in becquerels (Bq):

\[ 1\ \mathrm{Bq}=1\ \mathrm{decay/s} \]

The two quantities are related, but they are not necessarily equal.

A detector does not record every decay. Some radiation travels away from the detector, some is absorbed, and some may not produce a detectable pulse.

For a fixed experimental arrangement, we can write:

\[ R_{\mathrm{sample}}=\varepsilon A_{\mathrm{act}} \]

where:

• \(R_{\mathrm{sample}}\) is the corrected count rate in counts per second.
• \(\varepsilon\) is the effective detection efficiency, expressed as a fraction.
• \(A_{\mathrm{act}}\) is the sample’s activity in becquerels.

For example, if a detector has an effective efficiency of 20%, it records approximately 20% of the relevant decay events.

If the corrected count rate is 40 counts per second:

\[ A_{\mathrm{act}}=\frac{40}{0.20} \]

\[ \boxed{A_{\mathrm{act}}=200\ \mathrm{Bq}} \]

This assumes that each decay produces one relevant emission and that the stated efficiency includes the relevant detection and geometric effects.

The distinction matters because changing the distance between the sample and detector can change the detection efficiency.

You could therefore record fewer counts without the sample’s actual activity having changed.

For a half-life experiment, keeping the geometry and detector settings fixed allows the corrected count rate to track the activity reliably.

05A separate calculation: What energy does a nuclear decay release?

Background subtraction tells us how much of the detector reading comes from the sample.

It does not tell us how much energy each individual decay releases.

To investigate that, we need to look at the nuclear reaction itself.

This introduces three different ideas:

1. Conservation of nuclear charge and nucleon number.
2. Nuclear mass compared with atomic mass.
3. Binding energy compared with energy released in a reaction.

These are separate calculations, so do not confuse them with background correction.

Conserving charge and nucleon number

Consider alpha decay of uranium-238:

\[ {}^{238}_{92}\mathrm{U} \rightarrow {}^{234}_{90}\mathrm{Th} + {}^{4}_{2}\mathrm{He} \]

An alpha particle is a helium nucleus containing two protons and two neutrons.

Two quantities must be conserved in a nuclear equation.

Nucleon number, \(A\)

The mass number counts the total number of protons and neutrons.

For this reaction:

\[ 238=234+4 \]

Nuclear charge, \(Z\)

The atomic number counts the number of protons and determines the nuclear charge.

For this reaction:

\[ 92=90+2 \]

Both quantities balance.

The same principle applies to beta-minus decay.

For example:

\[ {}^{14}_{6}\mathrm{C} \rightarrow {}^{14}_{7}\mathrm{N} + {}^{0}_{-1}\mathrm{e} + \bar{\nu}_e \]

Here, a neutron changes into a proton, releasing an electron and an electron antineutrino.

Check nucleon number:

\[ 14=14+0+0 \]

Check nuclear charge:

\[ 6=7-1+0 \]

Both are conserved.

Remember that the electron has a nucleon number of zero. It contributes negative electric charge, but it is not a nucleon.

Nuclear mass versus atomic mass

Before calculating nuclear energy, you need to know exactly which mass you have been given.

These two quantities are not interchangeable.

An atomic mass is slightly greater than the corresponding nuclear mass because it includes the electrons.

The difference is small, but it matters in nuclear energy calculations.

For example, if you are calculating the mass defect using nuclear masses, the appropriate expression is:

\[ \Delta m= Zm_p+Nm_n-m_{\mathrm{nucleus}} \]

Here:

• \(Z\) is the number of protons.
• \(N\) is the number of neutrons.
• \(m_p\) is the mass of a proton.
• \(m_n\) is the mass of a neutron.
• \(m_{\mathrm{nucleus}}\) is the measured nuclear mass.

If you are given neutral atomic masses instead, you must account for the electrons.

A useful method for binding-energy calculations is to use the mass of a neutral hydrogen atom in place of the bare proton mass:

\[ \Delta m= Zm_{\mathrm H}+Nm_n-m_{\mathrm{atom}} \]

This works because the hydrogen atoms and the neutral atom being compared contain matching numbers of electrons. Small differences in electronic binding energies are neglected.

The key rule is simple: make sure both sides of your mass calculation account for the same particles.

Binding energy versus energy released

These quantities are related, but they answer different questions.

Nuclear binding energy is the energy required to separate a nucleus completely into its individual protons and neutrons.

A bound nucleus has less mass than its separate constituent nucleons. This difference is called the mass defect.

Einstein’s mass-energy relationship connects the mass defect to binding energy:

\[ E=mc^2 \]

If the mass difference is measured in atomic mass units:

\[ \boxed{ B(\mathrm{MeV})= \Delta m(\mathrm{u})\times931.5 } \]

Here, \(B\) represents binding energy, and \(931.5\ \mathrm{MeV/u}\) is the approximate conversion factor.

Energy released in a nuclear reaction, usually represented by \(Q\), is different.

It compares the total initial mass with the total final mass:

\[ \boxed{Q=(m_i-m_f)c^2} \]

where:

• \(m_i\) is the total initial mass.
• \(m_f\) is the total final mass.
• \(Q\) is the energy released or absorbed.

If the masses are in atomic mass units:

\[ Q=(m_i-m_f)\times931.5 \]

Our sign convention is that a positive \(Q\) means energy is released, while a negative \(Q\) means energy must be supplied.

The difference is the comparison being made:

A reaction’s released energy is not automatically equal to the binding energy of its final nucleus. The initial nuclei were already bound, so the reaction releases the net increase in total binding energy.

For a more detailed treatment of mass defects and binding energy per nucleon, see How to Calculate Binding Energy per Nucleon from Nuclear Mass.

06Questions and solutions

The following questions test background correction first, then extend the reasoning to decay measurements and nuclear energy.

Question 1

A GM tube records 86 counts in one minute when a radioactive sample is present.

When the sample is removed, the detector records 14 counts in one minute.

Calculate the corrected count rate due to the sample.

Solution 1

The corrected count rate is 72 counts per minute.

The sample-present measurement includes both the sample radiation and background radiation.

Since both measurements were taken over the same time interval, subtract the counts directly:

\[ \begin{aligned} N_{\mathrm{sample}} &=N_{\mathrm{observed}}-N_{\mathrm{background}}\\ &=86-14\\ &=72 \end{aligned} \]

The measurement interval is one minute, so:

\[ \boxed{R_{\mathrm{sample}}=72\ \mathrm{counts/min}} \]

The detector recorded 86 counts, but only 72 are attributed to the radioactive sample.

Question 2

A student measures background radiation for 3 minutes and records 54 counts.

They then measure a radioactive sample for 2 minutes and record 300 counts.

The student calculates the corrected count rate using:

\[ R=\frac{300-54}{2} \]

Explain why this method is incorrect, and calculate the correct sample count rate.

Solution 2

The correct sample count rate is 132 counts per minute.

The student’s mistake is subtracting counts recorded over different time intervals.

The background measurement lasted 3 minutes, while the sample measurement lasted only 2 minutes. Therefore, the two raw counts cannot be subtracted directly.

Step 1: Calculate the background count rate.

\[ R_{\mathrm{background}}= \frac{54}{3} =18\ \mathrm{counts/min} \]

Step 2: Calculate the observed count rate.

\[ R_{\mathrm{observed}}= \frac{300}{2} =150\ \mathrm{counts/min} \]

Step 3: Subtract the background rate.

\[ \begin{aligned} R_{\mathrm{sample}} &=150-18\\ &=132\ \mathrm{counts/min} \end{aligned} \]

Therefore:

\[ \boxed{R_{\mathrm{sample}}=132\ \mathrm{counts/min}} \]

An alternative method is to calculate how many background counts would be expected during the 2-minute sample measurement:

\[ N_{\mathrm{background}}=18\times2=36 \]

Then:

\[ R_{\mathrm{sample}}= \frac{300-36}{2} =132\ \mathrm{counts/min} \]

The student’s original method gives 123 counts per minute because it incorrectly treats a 3-minute background count as though it were collected in 2 minutes.

Question 3

A radioactive sample is monitored using a GM tube.

The background count is measured as 20 counts per 15-second interval.

The following illustrative observations are recorded:

(a) Calculate the corrected count at each time.

(b) Determine the half-life of the sample.

(c) Explain why calculating half-life directly from the observed counts would produce an unreliable result.

Solution 3

The corrected counts are 96, 48, 24, and 12 counts per 15 seconds, giving a half-life of 3 minutes.

Part (a): Correct the measurements.

Subtract 20 counts from every observed reading:

Part (b): Determine the half-life.

The corrected count decreases from 96 to 48 after 3 minutes.

It then decreases from 48 to 24 over another 3 minutes.

Therefore:

\[ \boxed{t_{1/2}=3\ \mathrm{min}} \]

The same result is confirmed by the final decrease from 24 to 12.

Part (c): Explain the problem with the observed readings.

The observed counts are:

\[ 116,\ 68,\ 44,\ 32 \]

These values do not halve over equal time intervals.

This happens because the background contribution remains approximately constant while the sample’s contribution decreases exponentially.

The observed count therefore approaches 20 counts per 15 seconds rather than zero.

Using the uncorrected measurements would make the sample appear to decay more slowly than it actually does.

The half-life is a property of the radioactive sample, not the combined radiation detected from the sample and its surroundings.

Question 4

Polonium-210 undergoes alpha decay:

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

The following illustrative rounded neutral atomic masses are provided:

Use \(1\ \mathrm{u}\,c^2=931.5\ \mathrm{MeV}\).

(a) Verify that the nuclear reaction conserves nucleon number and nuclear charge.

(b) Calculate the energy released by the reaction.

(c) Explain why the energy calculated in part (b) is not the total binding energy of the lead-206 nucleus.

Solution 4

The reaction conserves both nucleon number and nuclear charge, and releases approximately 5.41 MeV per decay.

Part (a): Check conservation.

For nucleon number:

\[ 210=206+4 \]

For nuclear charge:

\[ 84=82+2 \]

Both quantities are conserved.

Part (b): Calculate the released energy.

The initial mass is:

\[ m_i=209.982874\ \mathrm{u} \]

The final mass is:

\[ \begin{aligned} m_f&=205.974465+4.002603\\ &=209.977068\ \mathrm{u} \end{aligned} \]

The mass difference is:

\[ \begin{aligned} \Delta m&=m_i-m_f\\ &=209.982874-209.977068\\ &=0.005806\ \mathrm{u} \end{aligned} \]

Convert this into energy:

\[ \begin{aligned} Q&=0.005806\times931.5\\ &=5.4083\ \mathrm{MeV} \end{aligned} \]

Therefore:

\[ \boxed{Q=5.41\ \mathrm{MeV}} \]

The positive mass difference indicates that energy is released.

The use of neutral atomic masses is valid here because the electron contributions balance: the parent polonium atom has 84 electrons, while the daughter lead atom and neutral helium bookkeeping mass contain 82 and 2 electrons respectively.

Part (c): Distinguish binding energy from released energy.

The 5.41 MeV is the net energy released when polonium-210 transforms into lead-206 and an alpha particle.

The binding energy of lead-206 answers a different question: how much energy would be required to separate its 206 nucleons into 82 free protons and 124 free neutrons?

The starting polonium nucleus was already bound, so the reaction releases only the net increase in total binding energy between the initial and final nuclear systems.

It would therefore be incorrect to call the 5.41 MeV the total binding energy of lead-206.

Question 5

Two radioactive alpha sources, A and B, are investigated using separate GM detector arrangements.

Each decay produces exactly one alpha particle, and each detected alpha particle produces one recorded count.

The following illustrative measurements are obtained:

Assume:

• Both detectors have constant effective efficiencies throughout the measurements.
• The efficiency values include the effects of geometry and detector response.
• Detector dead-time losses are negligible.
• Each source undergoes only the stated alpha decay.
• All released nuclear energy is included in the supplied energy per decay.

(a) Calculate the corrected count rate for each source.

(b) Determine the activity of each source in becquerels.

(c) Calculate the energy released per second by each source in watts.

Use:

\[ 1\ \mathrm{MeV}=1.602\times10^{-13}\ \mathrm{J} \]

(d) Which source releases more energy per second? Explain why comparing the observed count rates alone would lead to an incorrect conclusion.

Solution 5

Source B releases more energy per second, approximately \(1.16\times10^{-11}\ \mathrm{W}\), compared with \(7.80\times10^{-12}\ \mathrm{W}\) for source A.

The key is to correct the background first, account for detector efficiency second, and only then calculate the energy released per second.

Part (a): Calculate the corrected count rates.

For source A:

\[ \begin{aligned} R_A&=150-30\\ &=120\ \mathrm{counts/min} \end{aligned} \]

For source B:

\[ \begin{aligned} R_B&=100-20\\ &=80\ \mathrm{counts/min} \end{aligned} \]

Therefore:

\[ \boxed{R_A=120\ \mathrm{counts/min}} \]

\[ \boxed{R_B=80\ \mathrm{counts/min}} \]

Source A has the greater corrected count rate.

However, this does not necessarily mean it has greater activity.

Part (b): Calculate the activities.

The relationship between corrected count rate and activity is:

\[ R_{\mathrm{sample}}=\varepsilon A_{\mathrm{act}} \]

Rearranging:

\[ A_{\mathrm{act}}=\frac{R_{\mathrm{sample}}}{\varepsilon} \]

For source A:

\[ \begin{aligned} A_A&=\frac{120}{0.20}\\ &=600\ \mathrm{decays/min} \end{aligned} \]

Convert to decays per second:

\[ A_A=\frac{600}{60}=10\ \mathrm{Bq} \]

For source B:

\[ \begin{aligned} A_B&=\frac{80}{0.10}\\ &=800\ \mathrm{decays/min} \end{aligned} \]

Convert to decays per second:

\[ A_B=\frac{800}{60} =13.33\ \mathrm{Bq} \]

Therefore:

\[ \boxed{A_A=10.0\ \mathrm{Bq}} \]

\[ \boxed{A_B=13.33\ \mathrm{Bq}} \]

Despite producing fewer corrected counts per minute, source B has the greater activity.

This is because its detector has a lower efficiency. A smaller fraction of its actual decays is being recorded.

Part (c): Calculate the energy released per second.

The energy released per second is the number of decays per second multiplied by the energy released per decay.

Therefore:

\[ P=A_{\mathrm{act}}\times E_{\mathrm{decay}} \]

The result is initially obtained in MeV/s. We then convert to J/s, which is equivalent to watts.

For source A:

\[ \begin{aligned} P_A&=10.0\times4.87\\ &=48.7\ \mathrm{MeV/s} \end{aligned} \]

Convert into joules per second:

\[ \begin{aligned} P_A&=48.7\times1.602\times10^{-13}\\ &=7.80\times10^{-12}\ \mathrm{W} \end{aligned} \]

For source B:

\[ \begin{aligned} P_B&=13.33\times5.41\\ &\approx72.13\ \mathrm{MeV/s} \end{aligned} \]

Convert into joules per second:

\[ \begin{aligned} P_B&=72.13\times1.602\times10^{-13}\\ &\approx1.16\times10^{-11}\ \mathrm{W} \end{aligned} \]

Therefore:

\[ \boxed{P_A=7.80\times10^{-12}\ \mathrm{W}} \]

\[ \boxed{P_B=1.16\times10^{-11}\ \mathrm{W}} \]

Part (d): Explain the result.

Source B releases more energy per second, even though its observed and background-corrected count rates are both lower.

There are two reasons.

First, its detector efficiency is lower. Once we account for this, source B actually has a greater activity:

\[ 13.33\ \mathrm{Bq}>10.0\ \mathrm{Bq} \]

Second, each decay of source B releases more energy:

\[ 5.41\ \mathrm{MeV}>4.87\ \mathrm{MeV} \]

Both factors increase its total energy release per second.

The tempting approach is to compare the observed counts and conclude that source A must be more active. Even comparing the corrected count rates alone would suggest the same thing.

But neither comparison accounts for the different detector efficiencies.

This is the important distinction: background correction isolates the sample’s detected signal, while detection efficiency determines how that signal relates to the actual number of nuclear decays.

07What to do next

Once you have a reliable background-corrected count-rate series, you can use it to determine the sample’s half-life and decay constant.

The exponential decay model applies to the sample’s corrected count rate, not the raw detector readings.

For the next step, see Decay constant and half-life: HSC Physics explained.