Medical Physics Tutorial 8 - Imaging with Ionising Radiation 2
Before you attempt this tutorial, you may find it helpful to revise the tutorials on Nuclear Physics, especially Tutorial 3 which covers metastable nuclei. You should be aware of the concept of half-life (Tutorial 5). A good understanding of all of these will be assumed throughout.
Many imaging techniques use the the idea of tracers. A radioisotope can be injected as a salt, or a radioactive isotope that is attached to:
a biological molecule;
a pharmaceutical molecule (a drug);
or a carrier molecule.
The molecule is taken to a particular target organ. Then the doctors can see the way the way that molecule is processed by that organ.
The radioisotope is a gamma emitter. An alpha emitter would not only be useless, but also highly dangerous.
Why is an alpha emitter no good for tracers?
A beta emitter is less ionising. In air, the range is less than a metre, and in body tissues it's typically a few millimetres. While such a range can be used for treatment of tumours, it is no good for imaging.
Gamma rays can pass out of the body easily, so these are used for imaging. We will look at the radioisotopes that are most commonly used.
Iodine is a component of thyroxin which made by the thyroid gland and is important in maintenance of the body's metabolism. The radioisotope is administered as radioactive potassium iodide, which is taken up by the thyroid gland.
Iodine-131 is a radioactive isotope of iodine, made in a research reactor. It is made from tellurium-130 (which is radioactive by beta minus decay with half-life of 8.2 × 1020 years) which absorbs a neutron:
This isotope of tellurium decays by beta minus decay:
The half life is about 25 minutes. (This contrasts with Te-128, which has a half life of 2 × 1024 years, which is quite a long time, 1011 life times of the Universe.)
Iodine 131 that is used widely for imaging. It is also used for radiotherapy, which is the treatment of cancerous tumours by radioactive emission. It has a half-life of 8.02 days (= 6.93 × 105 s). The radioisotope decays to xenon with a decay energy of 970 keV by beta minus decay. 90 % of the iodine decays by beta-minus decay in two steps like this:
The xenon nucleus (shown by Xe*) is excited and almost immediately loses its energy as a gamma-ray photon:
The gamma photon has an energy of 364 keV.
For the other 10 %, the decay pattern is like this:
The gamma photon has an energy of 636 keV:
The two pathways are shown in the diagram below:
Calculate the speed at which an electron of energy 334 keV travels. Comment on your answer.
Electronic charge = 1.60 × 10-19 C
Mass of electron = 9.11 × 10-31 kg
Calculate the wavelength of 364 keV gamma ray photons that form 90 % of the emissions.
Planck's Constant, h = 6.63 × 10-34 J s
The beta particles have a tissue penetration of about 2 mm. The image comes from the gamma rays. The picture below shows the accumulation of I-131 in the thyroid:
Image by Drahreg01, Wikimedia Commons
Iodine-131 is also a fall-out product of nuclear fission. Uncontrolled exposure can lead to cancers of the thyroid. Iodine-131 is also used as an industrial tracer.
The nuclear mass of I-131 is 130.9061246 u
The nuclear mass of Xe-131 is 130.9050824 u
Show that the total energy of the decay from I-131 to Xe-131 is about 970 keV.
1 u = 1.6605 × 10-27 kg
Patients who are treated with I-131 have a high radioactive burden for quite a time. After a month the radioactivity is about 7 % of the original. Therefore the patients may have to keep away from their children or grandchildren for several days. (My late mother had some iodine treatment, and she was told that she should not see any of her grandchildren for about two weeks at least.)
This is one of the most commonly used radioisotopes used in medicine nowadays. Almost 85 % of diagnoses that use radioactive imaging use Technetium as a radioactive tracer. There are a large number of pharmaceutical molecules that act as a vehicle. Tissues that are investigated in this way include:
Myocardium (muscles of the heart);
Technetium-99 m is made by bombarding moybdenum-98 with neutrons to give Mo-99:
This decays to technetium-99 by beta minus emission:
The half life for the decay is 66 hours (2.38 × 105 s).
The technetium nucleus is metastable, which means that its remains in an excited state for an extended period of time. Most excited nuclei lose their excess energy almost immediately (about 1.0 × 1016 s). The half life of the metastable state is 6.03 hours (21700 s). As the energy level of the nucleus falls, a gamma photon of energy 140 keV is emitted. The low energy technetium decays to ruthenium by beta minus decay, with a half life of 211 000 years (6.65 × 1012 s). The energy loss events of the metastable nucleus form an exponential decay, just like any other radioactive decay.
Write down the decay equation for technetium to ruthenium.
The energy of the gamma photons is low, so the radiation is less ionising, so it safer to use than other radionuclides. Additionally the half life of the metastable state is short. The low energy technetium has a low rate of decay, and is excreted from the body in the urine.
Technetium has to be prepared on site for immediate use, as the metastable state has a half life of just 6 hours. It is supplied as Mo-99 in the form of molybdate ions (MoO42-), and placed into a technetium generator, sometimes light-heartedly called a moly-cow. Aluminium oxide in a column bonds with the molybdenum but does not bond with the technetium. Saline is passed down the column in a process called column chromatography. The technetium in the form of pertechnate ions (TcO4-) is flushed out with a saline solution. The process is sometimes called a moly-milk. The idea is shown in this simplified diagram:
The pertechnate ions may then be bonded to a pharmaceutical. Then they are injected into the patient. Some technetium generators are shown below:
Image by Kieran Maher, Wikimedia Commons
The technetium generator and the syringes used for injection are shielded to reduce exposure to radiation. Since the molybdenum-99 has a half life of 66 hours, hospitals need to buy the generators on a regular and staggered basis.
Indium is produced by bombarding cadmium (Cd) nuclei with protons in a cyclotron. The event is shown in the equation below:
The product formed does NOT decay by beta minus decay, but by electron capture to make Cd-111:
Draw a Feynman diagram to illustrate this event. Show that it is possible.
The daughter cadmium nucleus is left in an excited state. It loses the excess energy by emitting a gamma photon. There are two energies for the gamma photons, 171.3 keV and 245.4 keV. The half life for the electron capture is 2.80 days (2.42 × 105 s).
The vehicles used to carry indium-111 include:
oxine, an organic compound that allows the indium to be used as a tracer for cell components of blood;
peptides and proteins that are taken up by tumours.
PET stands for Positron Emission Tomography. A PET scanner is shown in the picture below:
Image: Jens Maus - Wikimedia Commons
A biologically active molecule acts as a vehicle for a positron emitter and is injected into the patient. The most widely used molecule is flourodexoyyglucose, which is taken up by organs like the brain.
The tracer used is fluorine-18, which has 9 protons and 9 neutrons. Is is made by bombarding water that carries the stable isotope oxygen-18 with high speed protons (18 MeV) from a cyclotron.
(In all the sources I have seen, they simply say that oxygen-18 takes in a proton to form a nucleus of fluorine-18. No mention of the neutron is made, but it is needed to make the equation balanced.)
Fluorine-18 is not stable and decays by beta plus emission to stable oxygen-18. This is because there are too few neutrons for stability. The half life of the beta plus decay is about 110 minutes (6586 s). The energy of the positrons is 633.5 keV. This type of decay accounts for 97 %. 3 % of the decay is by electron capture.
Write down the equation that shows the positron decay of fluorine-18.
The positrons travel about 1 mm before meeting an electron. A positronium particle is formed (a positron and electron).
Explain what happens next.
Notice that the gamma photons move at an angle to the path of the positron and the electron, rather than perpendicularly (as seen in Particle Physics Tutorial 6). This is because the positron is moving fast, while the electron will be moving much slower. Momentum has to be conserved.
The gamma photons are picked up in an array of sensors that are mounted in a circular ring about the patient.
The gamma photons move off in a straight line in opposite directions. Each pair strikes sensors around the sensor array. When each pair of sensors is activated, data are sent to the computer. The combination of sensors activated enables the computer to locate the source with precision. The PET scan does not make an image that is particularly easy to interpret on its own. The resolution is not good. At best it's about 0.5 mm, but can be as much as 6 mm. Therefore the PET scan is often used in combination with a CT scan to produce high quality images to allow doctors to determine the shape and location of the tumour. The picture below shows a 3-dimensional image from a PET and CT Scan.
Image: Jens Maus - Wikimedia Commons
In this case, the tracer has collected in the brain, kidneys, bladder, and liver. Also there are faint images of tumours in the abdominal cavity. This shows that metastasis has occurred, which means that the cancerous cells have broken away from the original tumour and spread about the body.
In Nuclear Physics Tutorial 5 we came across the equation for exponential decay:
– no of nuclei;
N0 – original number of nuclei;
– exponential number, 2.718…;
l - decay constant (s-1);
– time (s)]
We often use the activity rather than the number of nuclei, so the equation becomes:
– activity (Bq);
A0 – original
– exponential number, 2.718…;
l - decay constant (s-1);
– time (s)]
Technetium in its metastable state has a half life of 21700 s.
(a) Calculate the decay constant.
(b) A patient is given a dose of 300 MBq. Calculate the activity after 24 hours.
(c) What proportion of the original dose has decayed?
In reality, the patient will not be emitting that number of gamma photons per second. This is because the body is eliminating the radiopharmaceutical through normal methods of waste disposal (i.e. going to the lavatory). Therefore there are two factors in the effective half-life of a radiopharmaceutical:
The physical half life, which is a property of the radionuclide itself and cannot be changed at all;
The biological half life, the time in the body.
There are a number of factors that determine the length of the biological half life. For example, if the tracer is targeted at bones, it can hang around there for quite some time. Other tracers are eliminated from the body quite quickly through the kidneys.
As well as the physical decay constant of the radionuclide, there is the biological decay constant. So we can write:
Effective decay constant (s-1) = physical decay constant (s-1) + biological decay constant (s-1)
In Physics code, we write:
We can rewrite the equation as:
Note that doctors often measure the half-life in days, not seconds, as this is more meaningful to the patients.
Phosphorus-32 has a physical half-life of 14.3 days, and a biological half-life of 1155 days. What is the effective half-life?
tE-1 = tP-1 + tB-1 = 14.3 d-1 + 1155 d-1 = 0.0708 d-1
tE = 14.1 days
In the case of phosphorus-32, the tracer will have decayed before it is eliminated from the body.
The biological half life of techetium-99m is 86400 s and the physical half life is 21700 s.
(a) Calculate the effective half life in s.
(b) Calculate the effective decay constant.
(c) Calculate the activity of the technetium after 3 days, if the initial activity is 3.0 × 106 Bq.
Some crystals can emit flashes of light (or scintillate) when exposed to gamma radiation. The gamma camera uses this property. It is constructed like this:
The gamma rays from the patient spread out radially. In the diagram above, they are shown coming from one source. In reality the gamma rays would come from several sources. This could cause a very unclear image, with several scintillation events from each source. Therefore a collimator is used. A collimator is a set of narrow tubes made from lead that selects only the photons that are coming up vertically. If the photons come at an angle, they are absorbed by the lead; the idea is shown below:
There are far fewer photons that strike the scintillator than there would otherwise. However gamma rays (and X-rays) cannot be refracted by materials, so it's not possible to focus them with a lens.
Once the photons pass the collimator, they arrive at the scintillator, often a crystal of sodium iodide. Electrons are excited by the gamma photon and lose the extra energy as a visible light photon. This is picked up by the photomultiplier to be sent out as a signal to the computer.
This is an indirect process. The gamma photons have to set off a scintillation to be observed. The electron avalanche spreads out across the whole photomultiplier, which reduces the resolution. You can see how the electrons are spread out, which results in a lower resolution. The resolution of a gamma camera of this kind is about 10 mm.
One problem with sodium iodide is that it is hygroscopic (not hydroscopic) which means that it absorbs humidity from the air. Eventually they dissolve in the water they absorb, a process called deliquescence. So the sodium iodide crystal has to be kept away from the air.
More recently the resolution can be improved by direct detection of the gamma photons by semiconductors, which can be made very small. The picture below show the idea:
The avalanche of electrons is large and in a narrower spread, leading to better resolution. This type of gamma camera has a resolution is about 4 mm (which isn't that brilliant).
High Energy X-rays
These are often referred to as hard X-rays. The X-ray photons carry an energy of between 100 keV to 1000 MeV. Their energies are comparable to gamma rays. (This is consistent with the idea of the electromagnetic spectrum being continuous. Soft X-rays can be considered to be hard UV.) They can be used to penetrate deep into matter, which makes them suitable for radiotherapy.
Image by Michael Goodyear, Wikimedia Commons
In a diagnostic machine, the electrons are accelerated by a large voltage between a filament (like a light bulb) at 0 volts and a target at the anode. The potential difference is 50 kV.
The radiotherapy machine needs a much higher energy, so it uses a linear accelerator (LINAC) to accelerate electrons to hit a heavy metal target. The way a linear accelerator works is shown in the animation (kindly donated by Stephen Lucas). In the animation, an alpha particle is used, but the radiotherapy machine uses electrons. The linear accelerator alternates at microwave frequencies (about 1 × 1010 Hz). The accelerated electrons are bent by magnetic fields to strike a heavy metal target, like tungsten. The process is inefficient, and only 1 % of the accelerated electrons actually produce an X-ray photon. The rest of the energy is lost as heat. The X-rays can shaped by collimators to form the shape of the cancer. Even so, most of the X-ray photons that are produced do not leave the head, but are absorbed by the casing. This is because they go off in random directions. The proportion that leave the head to irradiate the patient is quite low, approximately 10 % of the photons produced.
The idea of the machine is to target high energy X-rays to destroy the DNA of the cancerous cells, which will kill them. The head is movable, so it can be made to rotate about the patient. A static head would cause considerable damage to healthy cells near to the tumour, which is undesirable. Therefore the head is rotated so that while the cancerous cells are bombarded, the healthy cells get a much lower dose. While there is risk of secondary cancers to the healthy cells, they generally repair themselves to remain healthy. The cancerous cells cannot repair themselves, so they die.
A radiotherapy machine gives out X-ray photons of 150 keV. A patient needs 1.0 J of energy to treat the cancer cells of a tumour.
(a) Calculate the number of photons required to deliver this dose.
(b) The machine uses a current of 5.0 mA to produce the beam. Assuming that the efficiency of photon production is 1.0 %, and that 5 % of the photons produced are used to irradiate the patient, calculate that time taken for the machine to be switched on.
The X-ray dose is released in short bursts so the collateral damage to healthy cells is reduced as low as possible. The machine head is then moved to another position, so that another short burst can be given.
In some machines, a linear accelerator is not used. Laser light is shone onto the target to release the X-rays. These use titanium in a matrix of aluminium oxide to produce intense red light. With a sufficient intensity (in the order of 1 × 1012 W m-2), X-rays can be produced.
Gamma rays are also used in a similar way for radiotherapy. The source of these is usually cobalt-60. In this case, the head does not need a power source. (The use of gamma rays for radiotherapy is not on the AQA syllabus.)
Sometimes it is necessary to carry out internal radiotherapy, which involves placing a radioactive source next to a cancer. The cancers most commonly treated in this way include:
Prostate cancer (the prostate is a gland found in men that produces a white oily secretion that carries sperm cells).
Other cancers can be treated in this way as well.
The most common treatment using a radioactive implant is called brachytherapy. The implant is simply a capsule that carries a radionuclide. It is inserted by a surgical procedure to place the capsules around the cancerous tissue.
Sometimes bare wires made of radioactive material are used. They are called radiotherapy wires.
It is possible to use gamma rays or beta minus particles. Common beta minus emitters include:
Half Life / d
Energy / eV
3.8 × 105
3.54 × 106
The beta particles penetrate into the cancerous cells and the ionising radiation does damage to the DNA. This is much more likely to kill cancerous cells than it is to kill healthy cells. A large dose can be applied to a small area, and does not require the use of large and expensive equipment to deliver the radiotherapy from outside. Often the patient can undergo the treatment at home, once the implants have been placed. The radiation risks to other members of the family are low, since the tissue penetration is only a few millimetres.
There can be short time side effects as with any procedure, such as bruising and discomfort in the implanted region. Longer term side effects could result from the irradiation of the area about the tumour, such as scarring, or even secondary cancers (which defeats the object of the treatment).
Liver cancer is treated using selective internal radiation therapy (SIRT). In this relatively new procedure, very tiny beads are injected into a blood vessel leading to the liver. They block the tiny blood vessels that feed the cancer. So the cancerous cells are not only irradiated, they are starved of nutrition.
Initial tests to make a diagnosis are carried out by the general practitioner (GP). The kinds of tests carried out by the GP include:
Visual tests of the mouth, ears, and nose;
Stethoscope, an instrument that allows the doctor to listen to noises going on in the body;
Measurements of temperature, and blood pressure;
Taking blood and urine tests.
The imaging techniques described in this and previous tutorials are only carried out after the GP has exhausted the first line tests. They require specialist equipment that is expensive, and not always easy to interpret. The equipment is expensive to buy and to run. The personnel that use this equipment have to be well-trained. They are often called radiographers. The specialist doctors that make diagnoses from the radiology equipment are called radiologists. They work in the radiology department of a hospital.
We have seen how some diagnostic imaging equipment does not use ionising radiation, for example, the ultrasound scan and MRI scan. However many scans use X-rays. The amount that these are used has to be carefully monitored, to ensure that the doses given do not put a patient at risk. A normal X-ray gives a dose that is a very small fraction of the overall safe dose that a patient can receive. The main principle that governs the use of radiation is "as low as reasonably achievable" (ALARA).
The use of radioactive tracers is more risky, and the radiology department would not wish to undertake such procedures unless the potential benefits out weight the risks. Clearly, if there is a condition that is life-threatening, or likely to be so, then use of diagnostic techniques that use ionising radiation is going to be essential. Doses of radiopharmaceuticals are very carefully calculated for the patient. Not all cancer treatment is, of course, radiotherapy. Chemotherapy is often used. This can be effective, but can have many unpleasant side-effects (they are highly toxic chemicals). They can make the patient feel very ill, and lose their hair.
In the exam, you may be asked to compare imaging techniques for:
The question is most likely to be a "six-pointer", in which you need not only to describe the imaging techniques correctly, but also write coherently in good English.
A patient who has had a cough for a number of months has seen his doctor. The doctor has done a variety of tests, and sent him to the hospital to have a chest X-ray. She has now received the results back from the hospital and she thinks he has a respiratory condition that needs further investigation. There are three imaging techniques that have been suggested:
(b) CT scan;
(c) PET scan.
Compare these techniques for convenience, safety, and resolution. Decide which technique should be used in the first instance.
Dosimetry (Welsh Board and Eduqas)
The average activity of a radioactive material is defined as:
the number of disintegrations every second.
Over a short period of time, this statement can be summed up using the simple equation:
A - activity (Bq);
N - number of nuclei disintegrating;
t - time (s)
A sample of radioactive material measured to have an activity of 1200 counts per minute. If the background radiation is 120 counts per minute, calculate:
(a) the activity of the material in Bq;
(b) the number of disintegrations over a 15 hour period.
The beta and gamma radiation will penetrate the body (alpha is stopped by skin). Even if the chance of an interaction was quite low, your answer to Question 11 will show you that over a period of time, there would be a significant number of interactions, each of which could cause damage. So we need to have a way of measuring the absorbed dose of radiation.
The absorbed dose is defined as:
energy absorbed per unit mass
In physics code, this is written:
D - absorbed dose (units are Gray (Gy))
E - energy (J);
m - mass (kg).
1 Gy is the equivalent of 1 J kg-1.
What is the absorbed dose if a cancerous tissue of mass 150 g is exposed to 12 J of energy?
That is not the whole story, though. The risk of damage depends not only on the absorbed dose, but also on:
The type of radiation (alpha, beta, gamma, or neutrons);
The type of tissue that is exposed to the radiation. Some tissues are more sensitive to damage than others.
So we give a weighting factor for each kind of radiation:
Weighting Factor Wr
A slow neutron is also known as a thermal neutron. It travels at about 14 km s-1 which may sound fast, but is quite slow as far as particles are concerned. The table shows that alpha particles are very damaging if they manage to penetrate the body.
The Weighting Factor is combined with the absorbed dose to give the equivalent dose, which is defined as:
the product between the weighting factor and the absorbed dose
It has the physics code, H, and the units are Sieverts (Sv). The equation is:
What is the equivalent dose on the tissue in Question 13, assuming that neutrons are used for the treatment?
The time of exposure needs to be taken into account as well. A dose of, say, 15 millisieverts (mSv) will have more impact on the body if it is received in 1 day than if it is received in a year. This is because the body has very good repair mechanisms. So we need another quantity called the equivalent dose rate, which is the rate at which the equivalent dose is received. As an equation, this is written as:
H-dot - the equivalent dose rate (Sv s-1);
H - equivalent dose (Sv);
t - time (s).
You don't have to use seconds for time. You can use hours, or minutes, as long as you are consistent. Use the time units given in the question.
A technician works with radiation in the form of beta radiation and thermal (slow) neutrons. His monthly dose from the beta radiation is 20 mGy and the thermal neutrons is 50 mGy.
(a) Calculate the total equivalent dose received;
(b) Calculate the daily equivalent dose rate.