How Radioactivity Developed into Modern Nuclear Medicine

Imagine trying to find out where a substance travels inside the human body without opening the body. You need a way to follow it from the outside. What if you could replace a tiny fraction of its atoms with radioactive versions that behave almost exactly the same chemically, but emit radiation that a detector can measure?

That idea became the foundation of nuclear medicine. Radioactive isotopes could be used to trace substances through living systems, produce images of internal organs, and even deliver energy to particular tissues.

However, these applications did not come from one discovery. Scientists had to solve several different problems: discovering natural radioactivity, developing radioactive tracers, learning to manufacture radioactive isotopes, and finally demonstrating that they could be used clinically.

The story involves some particularly important experiments. Each one added a different piece of evidence.

01Henri Becquerel: discovering natural radioactivity (1896)

In 1895, Wilhelm Röntgen discovered X-rays. Scientists were suddenly interested in other substances that might produce invisible, penetrating radiation.

Henri Becquerel had been investigating phosphorescence, the process by which certain materials continue glowing after exposure to light.

He wondered whether fluorescent or phosphorescent uranium compounds might also produce penetrating radiation after being exposed to sunlight.

There was a straightforward way to test this.

The first experiment

Becquerel placed a photographic plate beneath crystals of a uranium-containing salt. The plate was protected from ordinary light using opaque paper.

He also placed a coin between the crystals and the plate.

His prediction was that sunlight would cause the uranium salt to emit radiation capable of reaching the photographic plate. The coin would block some of that radiation, leaving its outline on the developed plate.

When he developed the plate, he found an image of the crystals and the coin.

This supported his initial explanation.

But Becquerel needed to rule out another possibility.

Perhaps chemicals from the uranium salt were affecting the photographic plate rather than radiation passing through the paper.

He repeated the experiment with a thin sheet of paper between the crystals and the plate. The photographic effect remained.

This made chemical contamination an unlikely explanation.

The observation that changed everything

Becquerel intended to repeat the experiment after exposing the uranium salts to sunlight.

Unfortunately, the weather in Paris was cloudy.

He put the experimental arrangement away and waited for better conditions. When the sunlight failed to appear, he eventually decided to develop the photographic plates anyway.

He expected a weak image because the uranium salts had not received much sunlight.

Instead, the plates were strongly blackened.

This observation challenged his original explanation.

If sunlight was necessary to activate phosphorescence, why had the photographic plate been exposed while the uranium salts were kept in darkness?

Becquerel investigated further. He placed uranium crystals against photographic plates and stored them in darkness for extended periods.

The plates continued to darken.

He also tested materials placed between the uranium and the photographic plate, including glass and aluminium. The radiation could pass through these materials, although its intensity was affected.

The important conclusion was that uranium produced invisible radiation without needing to be exposed to sunlight.

Becquerel had discovered natural radioactivity.

There is an important distinction between what he observed and what scientists understand today.

• Observation: Uranium-containing materials exposed photographic plates even when kept in darkness.
• Initial interpretation: Uranium produced a previously unknown form of penetrating radiation.
• Modern explanation: Unstable atomic nuclei undergo spontaneous radioactive decay, releasing radiation.

Becquerel had not yet discovered the detailed nuclear processes responsible for the radiation. His experiments established the phenomenon that later scientists would investigate.

Marie and Pierre Curie: identifying new radioactive elements

Becquerel’s discovery raised another question.

Was uranium the only element capable of producing this radiation?

Marie Curie investigated the activity of different substances using sensitive electrical measurements.

She noticed something unexpected. Pitchblende, a uranium-containing ore, was more radioactive than could be explained by its uranium content alone.

She proposed that the ore must contain other, previously unknown radioactive substances.

Working with Pierre Curie, she investigated the ore through repeated chemical separations. They measured the radioactivity of the resulting fractions to identify where the strongest activity was concentrated.

Their work led to the identification of two new elements:

• Polonium, announced in July 1898.
• Radium, announced in December 1898.

Marie Curie also introduced the term radioactivity to describe the phenomenon.

Her work strengthened the idea that radioactivity was a property of atoms rather than simply a consequence of chemical reactions.

This was a major conceptual step. Radioactivity was not something that uranium needed to be switched on by sunlight. It originated within the atoms themselves.

The discoveries of Becquerel and the Curies established that natural radioactive materials could act as sources of detectable radiation.

But could scientists use that radiation to follow the movement of individual elements through chemical reactions?

That was the next problem.

02George de Hevesy: developing the radioactive tracer (1911-1913)

Imagine trying to measure a very small amount of lead moving through a chemical process.

Ordinary chemical analysis might struggle to detect it, particularly when it is mixed with a much larger quantity of other material.

Radioactivity offered a clever solution: give a tiny proportion of the lead atoms a detectable radioactive label.

To understand why this works, we need to understand isotopes.

Why radioactive isotopes can act as tracers

Isotopes are atoms of the same element that contain different numbers of neutrons.

They have:

• The same number of protons.
• The same electron arrangement when neutral.
• Different numbers of neutrons.
• Potentially different nuclear stability.

Because most chemical behaviour depends on the electron arrangement, isotopes of the same element generally behave almost identically in chemical reactions.

For example, lead-210 and stable lead isotopes are all chemically lead.

The radioactive isotope can therefore travel through chemical processes alongside ordinary lead, while its radiation provides a way to detect it.

Think of mixing a few specially marked beads into a large container of identical beads. If the marked beads behave just like the others, tracking them tells you where the whole collection has travelled.

The analogy has one limitation: radioactive labels are not permanent. Their nuclei decay, so the detectable signal decreases with time.

The failed separation experiment

In 1911, Hungarian chemist George de Hevesy was working in Ernest Rutherford’s laboratory in Manchester.

Rutherford challenged him to separate a radioactive substance called radium D from ordinary lead.

Radium D was later identified as lead-210 (\(^{210}\mathrm{Pb}\)).

Hevesy tried numerous chemical methods to separate the two substances.

He failed.

This was not simply a lack of experimental skill. The two substances were isotopes of the same element, so their chemical properties were almost identical.

Chemical reactions could not effectively distinguish the radioactive lead atoms from the stable lead atoms.

Hevesy recognised that this apparent failure could be turned into an advantage.

If radium D could not be separated from ordinary lead, perhaps it could be used to follow ordinary lead instead.

Turning radioactivity into a measurement tool

Hevesy proposed adding a very small amount of radioactive lead to a larger quantity of ordinary lead.

The mixture could then undergo chemical reactions, precipitation, filtration, or other processes.

At each stage, the radioactive material could be detected by measuring its radiation.

The amount of radioactivity in a sample would indicate how much labelled lead was present.

In 1913, Hevesy and chemist Fritz Paneth published work using radioactive isotopes as indicators in investigations of the solubility of lead compounds.

An electroscope was used to measure the activity of very small samples.

There is a historical detail worth getting right: although Hevesy originally considered radium D, his first published solubility experiments used another radioactive lead isotope, thorium B, as the indicator.

The underlying principle was powerful:

If the radioactive isotope behaves chemically like the stable isotope, measuring its activity allows scientists to infer the movement and distribution of the element itself.

Hevesy subsequently applied radioactive tracers to investigations involving plants and animals.

His work established an important foundation for nuclear medicine: radioactive substances could reveal processes occurring inside living systems.

He received the 1943 Nobel Prize in Chemistry for his work involving isotopic indicators.

Notice that the tracer technique existed before scientists could routinely manufacture radioactive isotopes. Early researchers relied on radioactive materials occurring naturally.

The next breakthrough would make the supply of useful radioactive isotopes much greater.

03Irène and Frédéric Joliot-Curie: creating artificial radioactivity (1934)

Natural radioactive materials were useful, but they created a practical limitation.

Scientists could not simply choose any element and obtain whichever radioactive isotope they wanted.

They needed a way to create radioactive nuclei in the laboratory.

In 1934, Irène and Frédéric Joliot-Curie demonstrated that this was possible.

Their experiment changed the future of nuclear physics and medical isotope production.

The aluminium bombardment experiment

The Joliot-Curies used a radioactive polonium source that emitted alpha particles.

They directed these alpha particles towards a thin sheet of aluminium.

A detector was positioned to measure radiation coming from the aluminium.

The prediction was that the alpha particles would interact with aluminium nuclei. If a nuclear reaction occurred, new particles might be emitted.

During bombardment, the detector registered radiation.

But the really interesting observation came when they removed the polonium source.

The detector continued registering emissions from the aluminium.

The signal did not disappear immediately. Instead, it decreased over time in a pattern characteristic of radioactive decay.

Why was this important?

If the detector were simply measuring alpha particles coming from the polonium source, removing that source should stop the signal.

But the aluminium continued emitting radiation.

The aluminium must therefore have been transformed into something radioactive.

Understanding the nuclear reaction

An alpha particle is a helium-4 nucleus:

\[ {}^{4}_{2}\mathrm{He} \]

It contains two protons and two neutrons.

When it strikes an aluminium-27 nucleus, a nuclear reaction can produce phosphorus-30 and a neutron.

The reaction is:

\[ {}^{27}_{13}\mathrm{Al} + {}^{4}_{2}\mathrm{He} \rightarrow {}^{30}_{15}\mathrm{P} + {}^{1}_{0}\mathrm{n} \]

Check conservation of mass number:

\[ 27+4=30+1=31 \]

Check conservation of atomic number:

\[ 13+2=15+0=15 \]

Both quantities are conserved.

The phosphorus-30 nucleus is unstable. It undergoes positron emission, transforming into silicon-30:

\[ {}^{30}_{15}\mathrm{P} \rightarrow {}^{30}_{14}\mathrm{Si} + {}^{0}_{+1}\mathrm{e} + \nu_e \]

Here, \({}^{0}_{+1}\mathrm{e}\) represents a positron, and \(\nu_e\) represents an electron neutrino.

During positron emission, a proton in the nucleus changes into a neutron.

The atomic number therefore decreases by one, while the mass number remains unchanged.

The important point is that the aluminium bombardment created an unstable phosphorus isotope that continued decaying after the alpha-particle source was removed.

The Joliot-Curies had demonstrated artificial radioactivity.

They had created a radioactive isotope that was not simply being extracted from a naturally radioactive material.

Their discovery earned them the 1935 Nobel Prize in Chemistry.

Why artificial radioactivity mattered for medicine

Scientists could now investigate methods of producing particular radioactive isotopes through nuclear reactions.

Cyclotrons and, later, nuclear reactors became important sources of artificially produced radioisotopes.

This opened up possibilities that natural radioactive materials alone could not provide.

A medical researcher could now investigate whether a particular radioactive isotope might:

• Follow a substance through the body.
• Produce radiation that could be detected externally.
• Deliver energy to a particular organ or tissue.

The discovery of artificial radioactivity did not immediately produce modern nuclear medicine. It provided a much more flexible supply of the radioactive materials needed to develop it.

04Saul Hertz and Arthur Roberts: radioiodine becomes a medical treatment (1937-1941)

Consider the thyroid gland, a small organ in the neck.

The thyroid needs iodine to produce thyroid hormones. It therefore takes up iodine from the bloodstream.

This creates an interesting possibility.

If radioactive iodine behaves chemically like ordinary iodine, would the thyroid take it up as well?

Yes.

The thyroid does not distinguish between stable and radioactive iodine isotopes through ordinary chemical processes.

That means radioactive iodine can follow the same biological pathway as ordinary iodine.

This was the central idea behind the development of radioiodine treatment.

The early experiments

Saul Hertz was a physician and director of the Thyroid Clinic at Massachusetts General Hospital.

He worked with physicist Arthur Roberts at the Massachusetts Institute of Technology.

Beginning in 1937, they investigated whether radioactive iodine could be used to study the behaviour of iodine in the body.

Their early experiments used iodine-128 in rabbits.

The results showed that the thyroid rapidly accumulated radioactive iodine.

This was important evidence. It demonstrated that radioactive iodine could be used to investigate the behaviour of iodine inside living organisms.

However, iodine-128 had a half-life of approximately 25 minutes.

That was too short for the therapeutic application Hertz was considering.

The researchers needed radioactive iodine that would remain active for longer.

The development of cyclotrons made other iodine isotopes available, including:

• Iodine-130, with a half-life of approximately 12 hours.
• Iodine-131, with a half-life of approximately 8 days.

These isotopes offered a more practical opportunity to investigate radioiodine treatment.

The first clinical application

On 31 March 1941, Hertz administered the first radioiodine treatment to a patient with thyroid disease at Massachusetts General Hospital.

The treatment used a mixture of iodine-130 and iodine-131.

Roberts had contributed to the experimental and physics work that made this application possible.

The initial target was hyperthyroidism, a condition in which the thyroid produces excessive amounts of thyroid hormones.

Why could radioactive iodine help?

The answer lies in the radiation emitted by iodine-131.

Iodine-131 emits both beta particles and gamma radiation.

The two emissions have different roles.

Beta particles: These are energetic electrons emitted from the nucleus during radioactive decay. They transfer energy to nearby tissue, causing ionisation and damaging cells. Because beta particles have a relatively limited range in tissue, much of their energy is deposited close to where the iodine has accumulated.

Gamma radiation: These are high-energy electromagnetic photons. They can travel out of the body and be detected by external instruments.

The beta radiation provides the main therapeutic effect, while gamma radiation can provide information about the distribution of the radioactive material.

This is a particularly important development in the history of radio-medicine.

Scientists were no longer simply detecting radioactive substances. They were using the biological behaviour of an element to direct radioactive energy towards particular tissue.

The first treatment was a clinical milestone, not an instant demonstration that every patient could be cured. Subsequent clinical experience helped establish and refine the application.

The work of Hertz and Roberts helped establish radioiodine as an important part of nuclear medicine.

05Technetium-99m: making diagnostic imaging more practical (1957 onwards)

Radioiodine demonstrated that radioactive isotopes could be used in medical treatment.

But another application was developing: diagnostic imaging.

In treatment, the aim is to deposit energy in a target.

In diagnostic imaging, the aim is to detect radiation coming from inside the body and use it to produce an image.

This requires an isotope with suitable radioactive properties and a way of detecting its emissions.

One of the most important developments was technetium-99m.

The development of the technetium-99m generator

Technetium was identified as an element in the 1930s.

However, a major step towards its medical use came in 1957, when Walter Tucker and Margaret Greene at Brookhaven National Laboratory developed the first technetium-99m generator.

They did not discover technetium itself. Their contribution was developing a practical way to make technetium-99m available for medical applications.

The generator uses molybdenum-99 as a parent isotope.

As molybdenum-99 decays, it produces technetium-99m.

The technetium-99m can then be separated from the parent material and used to prepare radioactive compounds for imaging.

This made it possible to obtain fresh technetium-99m without requiring a cyclotron at every hospital.

Powell Richards at Brookhaven strongly promoted its medical potential.

In 1961, the laboratory received its first order for a technetium-99m generator from Argonne Cancer Research Hospital, where it was used for blood-flow measurements.

Paul Harper and colleagues also investigated its usefulness for imaging structures such as the thyroid and brain.

Why technetium-99m is useful for imaging

The symbol Tc-99m contains an important detail.

The letter \(m\) stands for metastable. It indicates that the technetium-99 nucleus is in an excited nuclear state.

Technetium-99m has two particularly useful properties:

1. Its half-life is approximately 6.02 hours.
2. It emits gamma photons with an energy of approximately 140.5 keV.

These photons can escape from the body and be detected by a gamma camera.

The relatively short half-life also means its activity decreases over time rather than remaining high for an extended period.

The isotope can be chemically attached to different compounds. These compounds travel through the body according to their chemical and biological properties.

A gamma camera detects the photons emitted from the body and converts the measurements into an image.

The image therefore shows the distribution of the radioactive tracer.

It is not a photograph of the organ. It is a representation of where the detectable radioactive emissions originate.

This was a major development in diagnostic medicine.

Radioactive tracers could now be used to investigate the structure and function of organs without requiring invasive exploration.

Technetium-99m became one of the most widely used medical radioisotopes.

Activity is not the same as the number of radioactive atoms

There is one more idea worth understanding before comparing different radioisotopes.

Activity is the number of radioactive decays occurring per second.

It is related to the number of undecayed nuclei by:

\[ A=\lambda N \]

Where:

• \(A\) is activity.
• \(\lambda\) is the decay constant.
• \(N\) is the number of undecayed radioactive nuclei.

The decay constant is related to half-life:

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

This gives us a subtle but important result.

Two radioactive samples can have the same activity but contain different numbers of radioactive atoms.

A sample with a longer half-life has a smaller decay constant. It therefore needs more radioactive nuclei to produce the same activity.

Keep this distinction in mind. It becomes important when comparing the radiation emitted by different isotopes.

06The historical developments compared

The development of radio-medicine involved several separate contributions. It is worth keeping them distinct.

One detail is particularly important: tracer development occurred before artificial radioactivity.

Hevesy was already using naturally occurring radioactive isotopes to follow chemical processes in 1913.

The Joliot-Curies’ discovery in 1934 subsequently expanded the possibilities by showing that scientists could create radioactive isotopes artificially.

Together, these developments made modern nuclear medicine possible.

07Historical sources and further reading

The following sources provide additional historical detail and support the scientific explanations in this article.

• Institute of Physics: Henri Becquerel discovers radioactivity – a detailed reconstruction of Becquerel’s experimental sequence.
• Nobel Prize: Marie and Pierre Curie and the discovery of polonium and radium – the identification of new radioactive elements.
• George de Hevesy: Some Applications of Isotopic Indicators – Hevesy’s own Nobel lecture explaining the development of radioactive indicators.
• Nobel Prize: Frédéric Joliot – the discovery of artificial radioactivity.
• Sawin and Becker: Radioiodine and the treatment of hyperthyroidism – a historical review of the early development of radioiodine treatment.
• Massachusetts General Hospital: Honouring the legacy of Saul Hertz – the clinical development of radioiodine.
• Brookhaven National Laboratory: Sharing the story of technetium-99m – the development of the technetium-99m generator and its medical applications.

08Questions and solutions

Question 1

A radioactive iodine-131 sample has an initial activity of 80 MBq.

Its half-life is 8.0 days.

Calculate its activity after 16 days.

Solution 1

The remaining activity is 20 MBq.

First, calculate the number of half-lives:

\[ n=\frac{16}{8.0}=2 \]

The activity halves twice:

\[ 80\rightarrow40\rightarrow20 \]

Alternatively:

\[ A=A_0\left(\frac12\right)^n \]

\[ A=80\left(\frac12\right)^2 \]

\[ \boxed{A=20\ \mathrm{MBq}} \]

The sample retains one-quarter of its initial activity.

Remember that the half-life describes the time taken for the activity, or number of undecayed nuclei, to decrease to half its initial value.

Question 2

Becquerel places uranium salt crystals above a photographic plate protected from ordinary light.

He performs three experiments:

• Experiment A: The uranium is exposed to sunlight before the photographic plate is developed.
• Experiment B: The uranium and plate are kept in darkness, but the plate is still strongly blackened.
• Experiment C: A sheet of aluminium is placed between the uranium and the photographic plate. The blackening becomes weaker.

(a) Which observation most strongly challenges the idea that sunlight activates the uranium to produce radiation?

(b) Explain why Experiment B provides evidence against the idea that the photographic plate is being affected by ordinary light.

(c) What does the result of Experiment C suggest about the radiation?

(d) Explain why these experiments do not, by themselves, establish that the radiation is X-rays.

Solution 2

Experiment B most strongly challenges the sunlight-activation explanation because the uranium produces the photographic effect even when it has not been exposed to sunlight.

(a) In Experiment B, the uranium is kept in darkness but still produces strong photographic blackening.

If sunlight were necessary to activate the uranium, the plate should show little or no blackening.

The observation contradicts that prediction.

(b) The photographic plate is protected from ordinary light by opaque material.

Therefore, ordinary visible light cannot explain the blackening.

The radiation must be capable of passing through the protective covering and reaching the photographic plate.

This is evidence for an invisible penetrating emission.

(c) The reduced blackening in Experiment C suggests that aluminium absorbs or attenuates some of the radiation.

The radiation is therefore not behaving like ordinary visible light.

The result also shows that the intensity reaching the plate depends on the material placed between the source and detector.

(d) The experiments establish that uranium emits penetrating radiation, but they do not identify its exact nature.

Photographic blackening is evidence that radiation reaches the plate. It does not establish whether the radiation consists of charged particles or electromagnetic photons.

Further experiments would be needed to investigate its properties, such as testing whether the radiation is deflected by electric or magnetic fields.

This distinction is important in experimental physics: an observation can strongly support one conclusion without proving every part of a proposed explanation.

Question 3

In an experiment similar to that performed by the Joliot-Curies, alpha particles bombard a thin aluminium-27 target.

After the alpha-particle source is removed, the aluminium continues emitting positrons.

The activity of the aluminium decreases to 12.5% of its initial value after 7.5 minutes.

(a) Write the nuclear equation for the production of phosphorus-30 from aluminium-27.

(b) Write the nuclear equation for the positron decay of phosphorus-30.

(c) Calculate the half-life of phosphorus-30 using the experimental data.

(d) Explain why the continued emission after removal of the alpha-particle source provides evidence for artificial radioactivity.

Solution 3

The experiment produces radioactive phosphorus-30, which has a half-life of 2.5 minutes according to the supplied measurements.

(a) The alpha particle is a helium-4 nucleus.

The reaction is:

\[ {}^{27}_{13}\mathrm{Al} + {}^{4}_{2}\mathrm{He} \rightarrow {}^{30}_{15}\mathrm{P} + {}^{1}_{0}\mathrm{n} \]

The mass numbers balance:

\[ 27+4=30+1 \]

The atomic numbers also balance:

\[ 13+2=15+0 \]

(b) Phosphorus-30 undergoes positron emission:

\[ {}^{30}_{15}\mathrm{P} \rightarrow {}^{30}_{14}\mathrm{Si} + {}^{0}_{+1}\mathrm{e} + \nu_e \]

The atomic number decreases from 15 to 14 because a proton changes into a neutron.

The mass number remains 30.

(c) The remaining activity is 12.5% of the initial activity.

Convert this to a fraction:

\[ \frac{12.5}{100}=\frac18 \]

Since:

\[ \frac18=\left(\frac12\right)^3 \]

Three half-lives have passed.

Therefore:

\[ T_{1/2}=\frac{7.5}{3} \]

\[ \boxed{T_{1/2}=2.5\ \mathrm{min}} \]

(d) The continued emission after the source has been removed demonstrates that the aluminium target has become radioactive.

The original polonium source was producing the alpha particles, but it was not responsible for the continued emission from the target.

The nuclear reaction had created unstable phosphorus-30 nuclei inside the aluminium.

These nuclei continued undergoing spontaneous radioactive decay after the bombardment ended.

The crucial evidence is the persistence of the radiation and its subsequent exponential decrease.

If the signal had stopped immediately when the polonium source was removed, there would have been no evidence that the target itself had become radioactive.

Question 4

A chemist is investigating the solubility of a lead compound using the radioactive tracer technique.

A sample contains 0.80 mg of lead. A negligible mass of radioactive lead isotope is added, and the radioactive isotope is assumed to mix uniformly with the stable lead atoms.

After a chemical separation, 12.5% of the total measured radioactive activity is found in the filtrate.

Assume:

• The radioactive isotope behaves chemically like stable lead.
• The tracer remains uniformly distributed among the lead atoms.
• There is negligible radioactive decay during the experiment.
• The detector responds equally to equal amounts of radioactive material in both samples.

(a) Calculate the mass of lead present in the filtrate.

(b) Explain why measuring the radioactive activity allows the chemist to estimate the amount of stable lead in the filtrate.

(c) Explain why the radioactive isotope cannot be separated from stable lead using ordinary chemical methods.

Solution 4

The filtrate contains 0.10 mg of lead.

(a) The radioactive activity is proportional to the amount of radioactive lead present.

Because the radioactive isotope is uniformly mixed with stable lead, the fraction of total activity in the filtrate represents the fraction of total lead in the filtrate.

Therefore:

\[ \frac{m_{\mathrm{filtrate}}}{m_{\mathrm{total}}} = \frac{12.5}{100} \]

Substitute the total lead mass:

\[ m_{\mathrm{filtrate}}=0.125\times0.80 \]

\[ \boxed{m_{\mathrm{filtrate}}=0.10\ \mathrm{mg}} \]

(b) The radioactive isotope acts as an indicator for the stable lead.

Both isotopes are distributed through the chemical mixture in approximately the same proportions because they have almost identical chemical properties.

Measuring the activity of the filtrate therefore provides information about the amount of lead present.

This is particularly useful when the amount of material is too small to measure easily using conventional chemical methods.

(c) The radioactive and stable atoms are isotopes of the same element.

They have the same number of protons and, when neutral, the same electron arrangement.

Ordinary chemical reactions depend mainly on electron interactions, so they cannot effectively distinguish between these isotopes.

The radioactive isotope can be detected through its nuclear emissions, but its chemical behaviour remains almost identical to that of stable lead.

This is precisely why it makes such a useful tracer.

Question 5

A research team compares two hypothetical radioactive isotopes of iodine.

Both isotopes have identical chemical behaviour and are attached to the same tracer compound.

After administration, the same fraction of each tracer enters the thyroid. Both samples initially produce the same activity, \(A_0\), inside the thyroid.

The isotopes have the following properties:

Assume:

• No radioactive tracer leaves the thyroid during the 24-hour observation period.
• Each decay produces one detectable gamma photon.
• Detector geometry and efficiency are identical.
• All beta-particle energy is absorbed within the thyroid.
• The quoted beta energies represent the average energy deposited per decay.

(a) Calculate the ratio of the gamma count rates produced by isotope B and isotope A after 24 hours.

(b) A researcher argues that isotope B must have been taken up approximately 15 times more strongly by the thyroid because its gamma count rate is much higher after 24 hours.

Evaluate this argument.

(c) The researcher also predicts that isotope B must have deposited more total beta energy in the thyroid during the 24-hour period because its remaining activity is higher.

Determine whether this prediction is correct.

Solution 5

Isotope B produces approximately 14.7 times the gamma count rate of isotope A after 24 hours, but isotope A releases more total beta energy during that period.

This result exposes an important distinction between instantaneous activity, the number of radioactive nuclei, and total energy deposited.

(a) Calculate the gamma count-rate ratio.

Because each decay produces one detectable gamma photon, and the detector conditions are identical, the gamma count rate is proportional to the activity.

For isotope A:

\[ n_A=\frac{24}{6}=4 \]

Its remaining activity is:

\[ A_A=A_0\left(\frac12\right)^4 \]

\[ A_A=0.0625A_0 \]

For isotope B, convert its half-life into hours:

\[ 8.0\times24=192\ \mathrm{h} \]

Its remaining activity is:

\[ A_B=A_0\left(\frac12\right)^{24/192} \]

\[ A_B=A_0(2^{-1/8}) \]

\[ A_B\approx0.917A_0 \]

The ratio is:

\[ \frac{A_B}{A_A} = \frac{0.917}{0.0625} \]

\[ \boxed{\frac{A_B}{A_A}\approx14.7} \]

Isotope B therefore produces approximately 14.7 times the gamma count rate after 24 hours.

(b) Evaluate the claim about thyroid uptake.

The researcher is incorrect.

The question states that both isotopes have identical chemical behaviour and that the same fraction of each tracer enters the thyroid.

Therefore, there is no difference in their biological uptake.

The difference in measured count rate comes from their different half-lives.

Isotope A has undergone four half-lives, while isotope B has undergone only one-eighth of a half-life.

Consequently, isotope A’s activity has decreased much more rapidly.

There is another subtlety.

Equal initial activity does not mean equal numbers of radioactive nuclei.

Recall:

\[ A=\lambda N \]

For equal initial activities:

\[ \lambda_A N_A=\lambda_B N_B \]

Rearranging:

\[ \frac{N_B}{N_A} = \frac{\lambda_A}{\lambda_B} \]

Since the decay constant is inversely proportional to half-life:

\[ \frac{N_B}{N_A} = \frac{T_B}{T_A} \]

Therefore:

\[ \frac{N_B}{N_A} = \frac{192}{6} \]

\[ \boxed{\frac{N_B}{N_A}=32} \]

Initially, isotope B actually contains 32 times as many radioactive nuclei as isotope A to produce the same activity.

This is because each individual B nucleus has a much lower probability of decaying per unit time.

A higher measured activity at a later time does not automatically mean stronger biological uptake.

(c) Compare the total beta energy released.

The researcher’s prediction is incorrect.

The total energy released depends on two quantities:

1. The total number of radioactive decays during the observation period.
2. The average beta energy released in each decay.

Looking only at the final activity ignores how many decays occurred throughout the preceding 24 hours.

Let \(D\) represent the number of decays during the interval.

The number of decays is the initial number of nuclei minus the number remaining:

\[ D=N_0-N_t \]

Using the radioactive decay equation:

\[ D=N_0\left(1-2^{-t/T_{1/2}}\right) \]

We also know:

\[ A_0=\lambda N_0 \]

and:

\[ \lambda=\frac{\ln2}{T_{1/2}} \]

Therefore:

\[ D= \frac{A_0T_{1/2}}{\ln2} \left(1-2^{-t/T_{1/2}}\right) \]

We can compare the number of decays for both isotopes. Their initial activities are equal, so the common factors cancel.

For isotope B compared with isotope A:

\[ \frac{D_B}{D_A} = \frac{ 192(1-2^{-24/192}) }{ 6(1-2^{-24/6}) } \]

\[ \boxed{\frac{D_B}{D_A}\approx2.83} \]

Isotope B undergoes approximately 2.83 times as many decays during the 24-hour period.

However, each B decay releases only 0.020 MeV of beta energy, compared with 0.10 MeV for each A decay.

The ratio of total beta energies is:

\[ \frac{E_B}{E_A} = \frac{D_B}{D_A} \times \frac{0.020}{0.10} \]

\[ E_B/E_A=2.83\times0.20 \]

\[ \boxed{\frac{E_B}{E_A}\approx0.57} \]

Isotope B releases only approximately 57% of the total beta energy released by isotope A.

Equivalently, isotope A releases approximately 1.76 times as much beta energy over the 24-hour period.

The result is surprising if you focus only on the final activity.

Isotope B has a much higher gamma count rate at the end of the experiment, yet isotope A has released more total beta energy.

The reason is that instantaneous activity is not the same as cumulative energy deposition. The energy released per decay and the total number of decays both matter.

This is exactly why radioactive imaging measurements and therapeutic effects must be interpreted differently.

09Where the physics goes next

The history of radio-medicine connects several important ideas: nuclear stability, radioactive decay, isotope chemistry, radiation detection, and energy transfer.

The next useful step is understanding how scientists select an appropriate radioisotope for a particular application.

That decision depends on its half-life, emitted radiation, energy, and the biological behaviour of the compound carrying it.

PET imaging takes the story further by using positron-emitting isotopes and detecting the photons produced when positrons annihilate with electrons. It uses a different detection process from the gamma-camera imaging discussed here.

Understanding the discoveries in this article gives you the foundation to understand why these different medical technologies work.