Nuclear Medicine and Radiation Safety

How are radioisotopes used in medicine, and how is radiation dose measured and kept safe?

IntermediateNuclear ChemistryLast reviewed 5 October 2026

What is it?

Radiation from radioactive nuclei is ionizing: it has enough energy to knock electrons out of atoms and molecules in the body. This can break bonds in DNA directly, or split water into reactive free radicals that damage cells. The same property makes radiation dangerous and useful: it can harm healthy tissue, but it can also be aimed at cancer cells, and the gamma rays that escape the body can be detected to make images.

Three quantities describe radiation and its effects:

QuantityUnitMeaning
Activitybecquerel, Bq1 Bq = 1 decay per second
Absorbed dosegray, Gy1 Gy = 1 J of energy absorbed per kg of tissue
Equivalent dosesievert, Svabsorbed dose × radiation weighting factor

The radiation weighting factor is 1 for gamma rays, X-rays and beta particles, but 20 for alpha particles, which deposit their energy in a very short track and do much more damage per joule.

Key idea

Medicine uses the right radiation for the job. Imaging needs gamma rays that escape the body, from an isotope with a short half-life. Therapy needs radiation that deposits its energy locally (beta or alpha particles), or a carefully aimed external beam.

Why does it matter?

  • Diagnosis. Millions of scans a year use technetium-99m to image bones, the heart and other organs, and fluorine-18 PET scans find cancers and study the brain.
  • Treatment. Iodine-131 treats overactive and cancerous thyroid glands; external gamma and X-ray beams and implanted sources treat many cancers.
  • Safety. Patients, staff and the public must be protected, so doses are measured, limited and kept as low as reasonably achievable (ALARA).

How does it work?

1. Choosing a tracer for imaging

A good imaging isotope:

  • emits gamma rays (they pass out of the body to the camera; alpha and beta would be absorbed and only cause damage);
  • has a short half-life (hours), long enough for the scan but short enough to limit the dose;
  • can be attached to a molecule that collects in the organ of interest.

Technetium-99m (t1/2t_{1/2} = 6.01 h, 140 keV gamma rays) fits all three. Iodine-123 (t1/2t_{1/2} = 13.2 h) is taken up by the thyroid, which uses iodine.

2. PET scans

In positron emission tomography, a glucose-like molecule labelled with fluorine-18 (t1/2t_{1/2} = 110 min) collects in very active tissue such as tumours. Each fluorine-18 nucleus emits a positron, which meets an electron within a millimetre or so. The two annihilate, and their mass becomes two 511 keV gamma rays that fly apart in opposite directions. Detectors around the patient record both, and a computer traces back where they came from.

3. Radiotherapy

  • Iodine-131 (beta and gamma emitter, t1/2t_{1/2} = 8.02 d) concentrates in the thyroid, where its beta particles destroy the tissue.
  • External beams of high-energy X-rays or gamma rays are aimed at a tumour from several directions, so the tumour gets a high dose while surrounding tissue gets less.
  • Brachytherapy places small sealed sources inside or next to the tumour. Some newer treatments use alpha emitters, such as radium-223 (t1/2t_{1/2} = 11.4 d) for cancer that has spread to bone.

4. Protection: time, distance, shielding

  • Time: dose = dose rate × time, so spend as little time near a source as possible.
  • Distance: the dose rate from a small source falls with the square of the distance (inverse-square law): rate2rate1=(d1d2)2\dfrac{\text{rate}_2}{\text{rate}_1} = \left(\dfrac{d_1}{d_2}\right)^2.
  • Shielding: lead, concrete or water absorbs radiation. Each half-value thickness of a shield halves the intensity.

Workers wear dosimeters that record their dose, and dose limits are set well below the levels known to cause harm.

Dose (equivalent)Example
about 2.4 mSv per yearworldwide average natural background (radon, rocks, cosmic rays, food)
1 mSv per yearlimit for the public from artificial sources (excluding medical)
20 mSv per yearlimit for radiation workers (averaged over 5 years)

Think of it like this

Standing near a radioactive source is like standing near a campfire. You warm up less if you stay a shorter time (time), step back (distance: much cooler two steps away), or stand behind a wall (shielding). The fire itself (the activity) stays the same; what changes is how much reaches you.

More precisely

The effective dose (also in sieverts) goes one step further: it weights each organ by its sensitivity to radiation, so that doses to different parts of the body can be compared and added. In a patient, the activity of a tracer falls both by radioactive decay and by the body excreting it; the combined rate is described by an effective half-life, which is shorter than the physical half-life. Radiation effects are of two kinds: high doses cause predictable tissue damage (such as burns or radiation sickness), while low doses slightly raise the long-term probability of cancer, which is why every unnecessary dose is avoided.

Visualise it

A PET scanner seen end-on: a ring of detector blocks around a patient. Inside the patient, a fluorine-18 nucleus emits a positron (e+), which travels a short way and meets an electron (e−). They annihilate, and two gamma rays of 511 keV fly apart in opposite directions to detectors on opposite sides of the ring. Positron and electron annihilate: their mass becomes two gamma rays.
Two gamma rays at 180° let the scanner trace each decay back to its source.
The inverse-square law. Radiation spreads out from a small source. At 1 m it covers 1 square and the dose rate is 40.0 µSv/h; at 2 m it covers 4 squares and the rate is 10.0 µSv/h; at 3 m it covers 9 squares and the rate is 4.44 µSv/h. Dose rate is proportional to 1 divided by distance squared. Also: less time near the source, and more shielding.
Doubling the distance cuts the dose rate to a quarter.

Worked example

Worked example: Absorbed dose and equivalent dose

Question: A 70.0 kg person absorbs 0.0150 J from gamma rays. Find the absorbed dose and the equivalent dose. What would the equivalent dose be if the same energy came from alpha particles?

  1. Absorbed dose =0.0150 J70.0 kg=2.14×10−4 J/kg=2.14×10−4 Gy= \dfrac{0.0150\ \text{J}}{70.0\ \text{kg}} = 2.14 \times 10^{-4}\ \text{J/kg} = 2.14 \times 10^{-4}\ \text{Gy}
  2. Gamma (weighting factor 1): 2.14×10−4 Gy×1=2.14×10−4 Sv=2.14 \times 10^{-4}\ \text{Gy} \times 1 = 2.14 \times 10^{-4}\ \text{Sv} = 0.214 mSv
  3. Alpha (weighting factor 20): 2.14×10−4 Gy×20=4.29×10−3 Sv=2.14 \times 10^{-4}\ \text{Gy} \times 20 = 4.29 \times 10^{-3}\ \text{Sv} = 4.29 mSv, twenty times more harmful for the same energy.

Worked example: Distance and the inverse-square law

Question: The dose rate 1.00 m from a source is 40.0 µSv/h. What is it at 2.00 m and at 3.00 m?

  1. rate2=rate1×(d1d2)2\text{rate}_2 = \text{rate}_1 \times \left(\dfrac{d_1}{d_2}\right)^2
  2. At 2.00 m: 40.0 µSv/h×(1.00 m2.00 m)2=40.0\ \text{µSv/h} \times \left(\dfrac{1.00\ \text{m}}{2.00\ \text{m}}\right)^2 = 10.0 µSv/h
  3. At 3.00 m: 40.0 µSv/h×(1.00 m3.00 m)2=40.0\ \text{µSv/h} \times \left(\dfrac{1.00\ \text{m}}{3.00\ \text{m}}\right)^2 = 4.44 µSv/h

Worked example: How much tracer is left?

Question: A patient receives 740 MBq of technetium-99m (t1/2t_{1/2} = 6.01 h). What activity remains 18.0 h later (ignoring excretion)?

  1. n=18.0 h6.01 h=2.995n = \dfrac{18.0\ \text{h}}{6.01\ \text{h}} = 2.995 half-lives
  2. A=740 MBq×(12)2.995=A = 740\ \text{MBq} \times \left(\tfrac{1}{2}\right)^{2.995} = 92.8 MBq, about one-eighth of the starting activity.

Common mistake

Common mistake: Confusing activity with dose

A source with a high activity (Bq) does not necessarily give a high dose: the dose depends on the type and energy of the radiation, how much is absorbed, the distance and the time. Activity describes the source; dose describes what the body receives.

Common mistake: Halving the dose rate when the distance doubles

The inverse-square law means doubling the distance divides the dose rate by 4, not 2; tripling it divides by 9.

Common mistake: Thinking an imaged patient becomes permanently radioactive

Tracers have short half-lives and are also excreted, so the activity falls quickly. After a technetium-99m scan, the activity is about one-sixteenth of the starting value within a day. External X-ray and gamma beams leave no radioactivity in the body at all.

Notation note

  • Prefixes: 1 mSv = 10⁻³ Sv; 1 µSv = 10⁻⁶ Sv; 1 MBq = 10⁶ Bq.
  • Older units: 1 curie (Ci) = 3.7 × 10¹⁰ Bq; 1 rad = 0.01 Gy; 1 rem = 0.01 Sv.
  • keV and MeV are units of energy for single photons or particles (1 keV = 1.602 × 10⁻¹⁶ J).

Remember this

Remember this

  • Activity in Bq (decays per second); absorbed dose in Gy (J/kg); equivalent dose in Sv = Gy × weighting factor (1 for γ, X, β; 20 for α).
  • Imaging: gamma emitters with short half-lives (Tc-99m, 6.01 h). PET: F-18 positrons give two 511 keV gamma rays.
  • Therapy: beta or alpha emitters placed in the tumour (I-131, Ra-223), or aimed external beams.
  • Protection: less time, more distance (rate ∝ 1/d²), more shielding; keep doses as low as reasonably achievable.

Test yourself

Check your understanding before moving on.

Flashcards

Nuclear Medicine and Radiation Safety: Flashcards

10 cards

  1. Question
    What is ionizing radiation, and how does it harm cells?
    Answer

    Radiation that knocks electrons out of atoms. It breaks bonds in DNA directly or forms damaging free radicals from water.

  2. Question
    Becquerel, gray, sievert?
    Answer

    Bq = decays per second (activity). Gy = J absorbed per kg (absorbed dose). Sv = Gy × weighting factor (equivalent dose).

  3. Question
    Radiation weighting factors?
    Answer

    1 for gamma, X-rays and beta; 20 for alpha.

  4. Question
    What makes a good imaging isotope?
    Answer

    Gamma emitter, short half-life (hours), can be targeted to an organ. Example: technetium-99m (6.01 h).

  5. Question
    Why are alpha emitters not used for imaging?
    Answer

    Alpha particles do not leave the body, so they cannot be detected outside, and they give a high local dose.

  6. Question
    How does a PET scan detect fluorine-18?
    Answer

    Its positron annihilates with an electron, giving two 511 keV gamma rays in opposite directions, recorded by a detector ring.

  7. Question
    Name a radioisotope used to treat the thyroid.
    Answer

    Iodine-131 (beta and gamma, 8.02 d): the thyroid takes up iodine, and the beta particles destroy the tissue.

  8. Question
    Three ways to reduce radiation dose?
    Answer

    Less time, more distance, more shielding (and keep doses as low as reasonably achievable).

  9. Question
    Inverse-square law?
    Answer

    Dose rate ∝ 1 ÷ distance²: doubling the distance cuts the rate to one-quarter.

  10. Question
    Average natural background dose?
    Answer

    About 2.4 mSv per year worldwide, mostly from radon, rocks, cosmic rays and food.

Quiz

Nuclear Medicine and Radiation Safety: Quiz

7 questions

  1. Question 1EasyWhich unit measures the energy absorbed per kilogram of tissue?
    Show answer

    Answer: gray

    1 Gy = 1 J/kg. The becquerel measures activity; the sievert adds the radiation weighting factor.

  2. Question 2EasyWhich isotope is best suited to imaging an organ?
    Show answer

    Answer: a gamma emitter with a half-life of 6 hours

    Gamma rays leave the body to reach the camera, and a short half-life limits the dose. This describes technetium-99m.

  3. Question 3MediumAn absorbed dose of 0.010 Gy from alpha particles gives an equivalent dose of:
    Show answer

    Answer: 0.20 Sv

    Equivalent dose = 0.010 Gy × 20 = 0.20 Sv.

  4. Question 4MediumThe dose rate at 1.0 m is 36 µSv/h. What is it at 3.0 m?
    Show answer

    Answer: 4.0 µSv/h

    36 µSv/h × (1.0 m ÷ 3.0 m)² = 36 µSv/h ÷ 9 = 4.0 µSv/h.

  5. Question 5MediumIn a PET scan, what is detected?
    Show answer

    Answer: two gamma rays travelling in opposite directions

    Positrons annihilate with electrons inside the body, producing pairs of 511 keV gamma rays at 180°.

  6. Question 6HardA shield has a half-value thickness of 2.0 mm. What fraction of the gamma rays passes through 6.0 mm?
    Show answer

    Answer: 1/8

    6.0 mm ÷ 2.0 mm = 3 half-value thicknesses, so (½)³ = 1/8 gets through.

  7. Question 7MediumWhy is iodine-131 used to treat the thyroid?
    Show answer

    Answer: the thyroid takes up iodine, and its beta particles destroy nearby tissue

    Chemistry delivers the isotope to the right organ; the short-range beta particles deposit their energy there.

Notes and downloads

References

  1. Brown, T. L.; LeMay, H. E., Jr.; Bursten, B. E.; Murphy, C. J.; Woodward, P. M.; Stoltzfus, M. W. Chemistry: The Central Science, 15th ed.; Pearson, 2022.

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