Free FRCR Physics questions
40 practice questions in the same format as the First FRCR Physics exam: a stem and five items, each one true or false. Answer them, check, and read why every answer is right. No sign-up.
How the exam marks these
The RCR paper has 40 questions like these, 200 true/false items in two hours. Each item is marked on its own and there is no negative marking, so answer every one. Read each item as a complete sentence: one wrong word, such as “always” or a unit off by a factor of ten, makes it false.
For how the exam works and how to plan your revision, read what FRCR Physics is and the true/false traps examiners set.
X-ray physics questions
Question 1
Regarding the anode (target) of a diagnostic X-ray tube:
ATungsten is used mainly because of its high atomic number and high melting point.
True
Bremsstrahlung efficiency rises with target atomic number, and tungsten (Z = 74) also has a melting point of about 3,400 °C, so it survives the intense local heating at the focal spot.BMost of the energy of the electrons striking the target is converted into X-rays.
False
At diagnostic energies only about 1% of the electron energy becomes X-rays. The other ~99% is deposited as heat, which is why tube rating and anode cooling matter so much.CRhenium is alloyed with tungsten in the target to reduce surface cracking.
True
A tungsten–rhenium alloy (typically 5–10% Re) is more ductile than pure tungsten, so the focal track resists the surface roughening and cracking caused by repeated heating and cooling.DA rotating anode allows a higher tube rating than a stationary anode with the same focal spot size.
True
Rotation spreads the heat over a whole focal track instead of one small spot, so much more power can be applied for a short exposure without melting the target.EDuring an exposure, the anode is held at a negative potential relative to the filament.
False
The anode is positive relative to the cathode (filament). That potential difference accelerates the electrons emitted from the filament across the tube onto the target.
Question 2
Concerning the X-ray spectrum from a tungsten target:
AThe maximum photon energy in keV is numerically equal to the tube voltage in kVp.
True
An electron accelerated through V kilovolts gains V keV of kinetic energy. The most energetic photon possible is produced when all of that energy goes into a single bremsstrahlung photon.BTungsten K-characteristic radiation is produced at a tube voltage of 50 kVp.
False
The tungsten K-shell binding energy is about 69.5 keV. Below roughly 70 kVp, electrons cannot eject K-shell electrons, so no K lines appear in the spectrum.CBremsstrahlung output increases with the atomic number of the target.
True
Bremsstrahlung is produced when electrons are decelerated by the nuclear field, and the yield is roughly proportional to Z. Higher-Z targets give more X-rays for the same tube current.DThe energies of the characteristic lines rise as the tube voltage increases.
False
Characteristic line energies are fixed by the electron shell binding energies of the target element. Raising kVp increases how many characteristic photons are produced, not their energy.EBremsstrahlung produces a continuous spectrum of photon energies.
True
An electron can lose any fraction of its energy in a single deceleration, so bremsstrahlung photons span a continuum from very low energies up to the kVp.
Question 3
Increasing the tube voltage (kVp), with all other settings unchanged:
AIncreases both the maximum and the mean photon energy of the beam.
True
The spectrum's upper limit rises with kVp, and the whole distribution shifts towards higher energies, so the mean energy increases too.BIncreases the X-ray output approximately in proportion to kVp squared.
True
Tube output (air kerma per mAs) rises roughly as kVp². This is why a small kVp increase allows a large reduction in mAs for the same detector exposure.CReduces subject contrast between soft tissue and bone.
True
At higher energies the photoelectric effect, which depends strongly on atomic number, contributes less. Attenuation differences between tissues shrink, so subject contrast falls.DHas no effect on the amount of scatter reaching the detector.
False
Higher-energy photons are more likely to undergo Compton scatter and the scattered photons are more forward-directed and penetrating, so more scatter reaches the detector.EIncreases the proportion of photoelectric interactions in soft tissue.
False
Photoelectric probability falls steeply with energy (roughly 1/E³). Raising kVp makes Compton interactions an even larger share of the total in soft tissue.
Question 4
Concerning Compton scattering:
AIt involves an outer-shell, loosely bound electron.
True
Compton scatter occurs when the photon energy is much greater than the electron's binding energy, so the electron behaves as if it were free. In tissue this means outer-shell electrons.BIts probability per electron is largely independent of atomic number.
True
Compton probability depends on the number of electrons per unit mass (electron density), which is similar for most tissues. It does not have the strong Z dependence of the photoelectric effect.CThe scattered photon has the same energy as the incident photon.
False
Part of the photon's energy is given to the recoil electron, so the scattered photon always has less energy. Scatter with no energy loss is coherent (Rayleigh) scattering.DIt is the dominant interaction in soft tissue above about 30 keV.
True
In soft tissue the photoelectric and Compton contributions are equal at around 25–30 keV. Above that, Compton dominates across most of the diagnostic range.EBack-scattered photons carry more energy than forward-scattered photons.
False
The energy lost increases with scatter angle. A photon scattered through 180° loses the most energy, so back-scatter is lower in energy than forward scatter.
Question 5
Regarding attenuation and the half-value layer (HVL):
AAfter two HVLs, 25% of a narrow monoenergetic beam remains.
True
Each HVL halves the intensity of a monoenergetic beam: one HVL leaves 50%, two leave 25%, three leave 12.5%.BThe HVL equals 0.693 divided by the linear attenuation coefficient.
True
From I = I₀e^(−μx), setting I = I₀/2 gives HVL = ln 2 / μ = 0.693 / μ.CThe linear attenuation coefficient of water does not depend on photon energy.
False
Attenuation coefficients fall as photon energy rises across the diagnostic range, which is why higher-energy beams penetrate further.DFor a polyenergetic beam, the second HVL is thicker than the first.
True
The first HVL removes mostly low-energy photons, so the beam that remains is harder (more penetrating). It therefore takes more material to halve it again.EBroad-beam geometry gives a smaller measured HVL than narrow-beam geometry.
False
In broad-beam conditions scattered photons also reach the detector, so the measured transmission is higher and the measured HVL is larger. HVL is defined under narrow-beam conditions.
Question 6
Regarding the focal spot and geometric unsharpness:
AGeometric unsharpness equals the focal spot size multiplied by the object-to-image distance divided by the source-to-object distance.
True
By similar triangles, the penumbra is Ug = f × OID / SOD. A bigger focal spot, a bigger object-to-image gap or a shorter source-to-object distance all increase blur.BA steeper (smaller) target angle makes the focal spot look smaller from the patient’s side.
True
This is the line focus principle: the effective focal spot is roughly the actual focal track length × sin(anode angle), so a steeper (smaller) angle gives a smaller projected spot.CIncreasing the source-to-image distance increases geometric unsharpness.
False
A longer source distance makes the beam less divergent at the object, so geometric unsharpness decreases (and magnification falls) for the same object-to-image distance.DA smaller anode angle increases the field size the beam can cover.
False
A small anode angle limits field coverage at a given distance and makes the heel effect worse. That trade-off is why anode angles are typically around 7–16°.EBecause of the heel effect, beam intensity is higher on the anode side of the field.
False
X-rays leaving towards the anode side travel through more target material and are attenuated more, so intensity is higher on the cathode side.
More on this topic: X-ray physics question bank
CT questions
Question 7
Regarding CT numbers and display:
AWater is 0 HU and air is approximately −1000 HU.
True
The Hounsfield scale is defined as HU = 1000 × (μ − μwater) / μwater, which sets water at 0 and air (μ ≈ 0) at about −1000.BThe CT number of every tissue is independent of the tube voltage.
False
Water is fixed at 0 HU by definition, but materials whose attenuation varies differently with energy, such as bone and iodine, give CT numbers that change with kVp.CFat measures around −100 HU.
True
Fat attenuates slightly less than water, so it has a negative CT number of roughly −50 to −120 HU.DIncreasing the window width increases the displayed contrast.
False
A wider window spreads the grey scale over more HU, so each HU step changes brightness less and displayed contrast falls. Narrow windows (e.g. brain) increase contrast.EThe window level sets the CT number shown at the middle of the grey scale.
True
The window level (centre) is the HU value displayed as mid-grey. The window width sets how many HU above and below it are spread across black to white.
Question 8
Regarding CT dose descriptors:
ACTDIvol takes helical pitch into account.
True
CTDIvol = CTDIw / pitch. A higher pitch spreads the same tube output over a longer length, lowering the average dose in the scanned volume.BDLP equals CTDIvol multiplied by the scan length.
True
Dose-length product (mGy·cm) = CTDIvol × scan length. It reflects the total radiation output of the scan.CCTDIvol is the absorbed dose received by the individual patient.
False
CTDIvol is measured in a standard acrylic phantom (16 or 32 cm). It is a scanner output index for comparing protocols, not a patient dose.DDoubling the scan length, all else equal, doubles the DLP but not the CTDIvol.
True
CTDIvol describes dose within the scanned volume and does not depend on its length. DLP multiplies by length, so it doubles.EAutomatic tube current modulation raises the dose to thinner parts of the body.
False
Tube current modulation lowers the mA through thinner or less attenuating regions and raises it only where needed, keeping image noise more uniform at lower overall dose.
Question 9
Regarding noise in CT images:
AQuantum noise is proportional to 1/√mAs.
True
The number of detected photons is proportional to mAs, and Poisson noise scales as 1/√N. Quadrupling the mAs halves the noise.BHalving the reconstructed slice thickness increases noise by about √2.
True
A thinner slice collects half as many photons per voxel, so noise rises by √2 (about 41%) if nothing else changes.CA sharp (bone) reconstruction kernel reduces image noise.
False
Sharp kernels boost high spatial frequencies, which improves edge detail but increases noise. Smooth (soft tissue) kernels reduce noise at the cost of resolution.DIterative reconstruction can reduce noise for a given dose.
True
Iterative methods model the noise statistics and the scanner geometry, so they can produce lower-noise images than filtered back projection from the same data, or allow dose reduction.EImage noise does not depend on patient size.
False
A larger patient attenuates more of the beam, so fewer photons reach the detectors and noise increases unless the tube output is raised.
Question 10
Regarding CT artefacts:
ABeam hardening can cause cupping in images of a uniform phantom.
True
Rays through the centre travel further and the beam hardens more, so the centre appears less attenuating than the edges. Water phantoms show a cupped profile without correction.BA faulty detector element on a third-generation scanner produces a ring artefact.
True
In third-generation geometry each detector element always measures rays at the same distance from the rotation centre, so a miscalibrated element creates a ring in the image.CPartial volume artefact is reduced by using thinner slices.
True
Partial volume averaging happens when a voxel contains more than one tissue. Thinner slices make that less likely, especially at the skull base and at small structures.DMotion artefact is reduced by using a longer rotation time.
False
Motion artefact comes from the patient moving during data collection, so a shorter rotation time reduces it. A longer rotation gives more time for movement.EPhoton starvation streaks are worst along the shortest path through the patient.
False
Photon starvation happens where attenuation is greatest, along the longest, densest paths such as laterally through the shoulders or pelvis, where very few photons reach the detectors.
Question 11
Regarding helical multislice CT:
APitch is the table travel per rotation divided by the total collimated beam width.
True
This is the IEC definition of pitch. A pitch of 1 means the table moves exactly one beam width per rotation; below 1 the helices overlap.BIncreasing the pitch with the mAs unchanged increases patient dose.
False
A higher pitch covers the same length with fewer rotations, so dose falls: CTDIvol is inversely proportional to pitch.CIsotropic voxels allow images to be reformatted in any plane without loss of resolution.
True
When voxels are equal in all three dimensions, spatial resolution is the same in every direction, so coronal, sagittal and oblique reformats are as sharp as the axial images.DThe reconstructed slice thickness is fixed by the detector row width and cannot be changed after the scan.
False
The thinnest slice is limited by the detector row width, but thicker slices can be reconstructed from the same raw data at any time afterwards.EOverbeaming is proportionally worse for narrow total beam collimations.
True
The beam is widened slightly so that the outer detector rows receive the full beam. That extra penumbra is a fixed width, so it is a larger fraction of a narrow beam, wasting more dose.
More on this topic: CT question bank
Ultrasound questions
Question 12
Regarding the propagation of ultrasound:
AScanners assume a speed of sound of 1540 m/s in soft tissue.
True
Depth is calculated from the echo return time using an assumed 1540 m/s. Tissues with a different speed cause small errors in depth and size.BThe speed of sound in tissue depends on the ultrasound frequency.
False
Speed depends on the medium's stiffness and density (c = √(K/ρ)), not on frequency. Changing frequency changes the wavelength instead.CThe wavelength of 5 MHz ultrasound in soft tissue is about 0.3 mm.
True
λ = c / f = 1540 / 5,000,000 ≈ 0.31 mm.DSound travels faster in air than in soft tissue.
False
Sound travels at only about 330 m/s in air, compared with about 1540 m/s in soft tissue and around 3–4,000 m/s in bone.EAcoustic impedance is the product of density and the speed of sound.
True
Z = ρc, measured in rayls (kg m⁻² s⁻¹). Differences in impedance between tissues determine how much of the beam is reflected at a boundary.
Question 13
Regarding reflection and attenuation of ultrasound:
AThe fraction reflected at an interface depends on the difference in acoustic impedance across it.
True
The intensity reflection coefficient is ((Z₂ − Z₁)/(Z₂ + Z₁))². The bigger the impedance mismatch, the stronger the echo.BAlmost all of the beam is reflected at a soft tissue–air interface.
True
Air has a tiny impedance compared with tissue, so about 99.9% of the intensity is reflected. That is why coupling gel is needed and why gas blocks the view.CAttenuation in soft tissue is roughly 0.5–1 dB per cm per MHz.
True
Attenuation rises roughly in proportion to frequency, at around 0.5–1 dB cm⁻¹ MHz⁻¹ in soft tissue, mainly through absorption.DHigher-frequency ultrasound penetrates deeper into tissue.
False
Attenuation increases with frequency, so higher frequencies penetrate less. They are used for superficial structures, where their better resolution matters most.ETime gain compensation corrects for refraction of the beam.
False
Time gain compensation amplifies later (deeper) echoes more to make up for attenuation with depth. It has no effect on refraction.
Question 14
Regarding spatial resolution in ultrasound:
AAxial resolution is about half the spatial pulse length.
True
Two reflectors along the beam can be separated if they are more than half the spatial pulse length apart, because the echo makes a round trip.BIncreasing the transmit frequency improves axial resolution.
True
Higher frequency means a shorter wavelength and so a shorter pulse for the same number of cycles, improving axial resolution.CLateral resolution is best in the focal zone.
True
Lateral resolution depends on beam width, which is narrowest at the focus. Above and below the focal zone the beam is wider and lateral resolution is poorer.DDamping the transducer element lengthens the pulse.
False
The backing (damping) layer stops the element ringing, which shortens the pulse and improves axial resolution.EElevation (slice-thickness) resolution is determined by the pulse length.
False
Elevation resolution depends on the beam thickness perpendicular to the scan plane, set by the transducer's acoustic lens or a multi-row array, not by pulse length.
Question 15
Regarding Doppler ultrasound:
AThe Doppler shift is proportional to the cosine of the angle between the beam and the direction of flow.
True
fD = 2 f₀ v cos θ / c. The shift is greatest when the beam is parallel to flow and falls as the angle increases.BIn pulsed Doppler, aliasing occurs when the Doppler shift exceeds half the pulse repetition frequency.
True
Pulsed Doppler samples the signal at the PRF, so by the Nyquist limit it can only measure shifts up to PRF / 2. Higher shifts wrap around and appear in the wrong direction.CIncreasing the transmitted frequency reduces aliasing.
False
The Doppler shift is proportional to the transmitted frequency, so a higher frequency gives a bigger shift and makes aliasing more likely. Lower frequency or higher PRF reduces it.DWith the beam at 90° to the flow, no Doppler shift is measured.
True
cos 90° = 0, so motion perpendicular to the beam produces no Doppler shift, and colour Doppler shows no flow there.EVelocity estimates are more accurate at an insonation angle of 80° than at 45°.
False
Near 90° the cosine changes rapidly with angle, so a small angle-correction error causes a large velocity error. Angles of 60° or less are recommended.
Question 16
Regarding ultrasound artefacts:
AAcoustic shadowing occurs behind strongly attenuating or reflecting structures such as gallstones.
True
Little sound passes through a gallstone or bone, so the region behind it returns few echoes and appears dark.BPosterior acoustic enhancement occurs behind fluid-filled structures.
True
Fluid attenuates less than the surrounding tissue, but time gain compensation still boosts echoes from that depth, so tissue behind a cyst appears brighter.CReverberation produces equally spaced repeated echoes.
True
Sound bouncing back and forth between two strong parallel reflectors returns echoes at regular time intervals, displayed as equally spaced lines deeper in the image.DMirror-image artefact places a duplicate structure on the transducer side of a strong reflector.
False
The duplicate appears deeper than the strong reflector (such as the diaphragm), because the extra path length is interpreted as extra depth.ESide-lobe artefacts are caused by setting the pulse repetition frequency too high.
False
Side lobes are weak off-axis beams from the transducer. Echoes they pick up are wrongly placed along the main beam. They are unrelated to PRF; too high a PRF causes range ambiguity.
More on this topic: Ultrasound question bank
MRI questions
Question 17
Regarding precession and resonance in MRI:
AThe Larmor frequency is proportional to the main magnetic field strength.
True
f = (γ/2π) × B₀. Doubling the field doubles the precession frequency.BThe Larmor frequency of hydrogen at 1.5 T is about 64 MHz.
True
For hydrogen, γ/2π ≈ 42.58 MHz/T, so 1.5 T gives about 64 MHz and 3 T about 128 MHz.CDoubling B₀ halves the Larmor frequency.
False
The relationship is directly proportional, so doubling B₀ doubles the Larmor frequency.DThe RF pulse must be at the Larmor frequency to tip the magnetisation.
True
Energy is only absorbed efficiently when the RF frequency matches the precession frequency, which is the condition for resonance.EAt equilibrium, the net magnetisation lies perpendicular to B₀.
False
At equilibrium the net magnetisation lies along B₀ (longitudinal). The RF pulse tips it into the transverse plane, where it induces a signal.
Question 18
Regarding relaxation:
AT1 describes the recovery of longitudinal magnetisation.
True
After an RF pulse, longitudinal magnetisation regrows as Mz = M₀(1 − e^(−t/T1)) as energy is given up to the surrounding lattice (spin–lattice relaxation).BIn tissue, T2 is never longer than T1.
True
Any process that restores longitudinal magnetisation also destroys transverse coherence, so T2 ≤ T1 in tissue.CT2* is longer than T2.
False
T2* includes dephasing from static field inhomogeneity as well as true spin–spin relaxation, so it is always shorter than T2.DT1 values of tissues generally increase with field strength.
True
Most tissue T1 values are longer at 3 T than at 1.5 T, which affects the choice of TR and inversion times at higher fields.EFat has a long T1 compared with water.
False
Fat has a short T1, so it recovers quickly and appears bright on T1-weighted images. Water and CSF have long T1 values.
Question 19
Regarding image weighting in spin echo MRI:
AT1 weighting uses a short TR and a short TE.
True
A short TR catches tissues at different stages of T1 recovery, and a short TE keeps T2 effects small, so contrast is dominated by T1.BT2 weighting uses a long TR and a long TE.
True
A long TR lets longitudinal magnetisation recover in most tissues, removing T1 differences, and a long TE lets T2 decay differences develop.CCSF is bright on T1-weighted images.
False
CSF has a very long T1, so it recovers little within a short TR and appears dark on T1-weighted images. It is bright on T2-weighted images.DFat appears bright on both T1-weighted and fast spin echo T2-weighted images.
True
Fat is bright on T1 because of its short T1, and it stays relatively bright on fast spin echo T2 images because the closely spaced refocusing pulses reduce J-coupling dephasing.EProton density weighting uses a short TR and a long TE.
False
Proton density weighting uses a long TR and a short TE, minimising both T1 and T2 effects so the signal mainly reflects the number of protons.
Question 20
Regarding MRI safety:
AOn a superconducting scanner, the static magnetic field is present even when no scan is running.
True
The superconducting magnet is always on. Projectile and implant risks apply whenever anyone enters the scan room, not just during scanning.BSpecific absorption rate (SAR) relates to heating of the patient by RF energy.
True
SAR (W/kg) measures RF power deposited in tissue. Scanners limit it to keep body temperature rise within safe limits.CThe loud acoustic noise comes mainly from the RF coils.
False
Noise is produced by the gradient coils. Rapidly switched currents in the strong static field create Lorentz forces that make the coils vibrate.DRapidly switched gradients can cause peripheral nerve stimulation.
True
A fast-changing gradient field (dB/dt) induces electric currents in tissue, which can stimulate peripheral nerves. Scanners limit dB/dt to prevent this.ETitanium implants are strongly attracted by the static magnetic field.
False
Titanium is not ferromagnetic, so it experiences little attraction or torque. It can still heat in the RF field, so each implant needs checking against its MR conditions.
More on this topic: MRI question bank
Nuclear medicine questions
Question 21
Regarding technetium-99m:
AIt emits a gamma photon of 140 keV.
True
The 140 keV photon is ideal for the gamma camera: penetrating enough to leave the patient, and low enough to be collimated and stopped efficiently by a NaI crystal.BIts physical half-life is about 66 hours.
False
Tc-99m has a half-life of about 6 hours. About 66 hours is the half-life of its parent, molybdenum-99.CIt is obtained from a molybdenum-99 generator.
True
Mo-99 bound to an alumina column decays to Tc-99m, which is washed off (eluted) with saline as sodium pertechnetate.DIts decay changes the atomic number of the nucleus.
False
Tc-99m decays by isomeric transition to Tc-99: the nucleus drops to a lower energy state and emits a gamma photon, with no change in atomic number or mass number. The ‘m’ stands for metastable.EMolybdenum breakthrough can be checked by measuring the eluate inside a lead shield.
True
The lead pot stops the 140 keV Tc-99m photons but not the higher-energy 740–780 keV Mo-99 photons, so any reading through the shield indicates Mo-99 contamination.
Question 22
Regarding the gamma camera:
AThe sodium iodide crystal converts gamma photons into light.
True
Thallium-doped NaI(Tl) is a scintillator: each absorbed gamma photon produces a flash of light in proportion to the energy deposited.BPhotomultiplier tubes convert the light into an electrical signal and amplify it.
True
Light frees electrons at the photocathode, which are multiplied at a series of dynodes into a measurable pulse. The pattern of signals across the tubes locates each event.CIncreasing the crystal thickness sharpens the intrinsic resolution of the camera.
False
A thicker crystal absorbs more photons (better sensitivity), but light spreads further before reaching the photomultipliers, so intrinsic resolution gets worse.DWith a parallel-hole collimator, spatial resolution worsens with distance from the collimator face.
True
Each hole accepts photons from a cone that widens with distance, so the patient should be placed as close to the collimator as possible.EA high-sensitivity collimator has longer, narrower holes than a high-resolution collimator.
False
High-sensitivity collimators have shorter, wider holes that accept more photons at the cost of resolution. High-resolution collimators have longer, narrower holes.
Question 23
Regarding positron emission tomography (PET):
AAnnihilation produces two 511 keV photons travelling in almost opposite directions.
True
A positron and an electron annihilate, converting their rest mass into two 511 keV photons emitted about 180° apart.BThe radionuclide in FDG decays with a half-life of a little under two hours.
True
FDG is labelled with fluorine-18, half-life about 110 minutes. That is long enough to make it at a regional cyclotron and deliver it to nearby hospitals.CCoincidence detection removes the need for a lead collimator.
True
Detecting both photons within a narrow timing window defines a line of response, so the direction is known without absorptive collimation. This is why PET is far more sensitive than SPECT.DThe fluorine-18 used for FDG is made by neutron irradiation in a reactor.
False
F-18 is made in a cyclotron, usually by proton bombardment of oxygen-18 enriched water.ERandom coincidences increase as the activity in the field of view falls.
False
Random coincidences rise roughly with the square of the activity, so they increase at higher count rates, not lower.
Question 24
Regarding radioactive decay and half-life:
AActivity is measured in becquerels, where 1 Bq is one decay per second.
True
The SI unit of activity is the becquerel. Clinical activities are usually in MBq (10⁶ Bq).BAfter four half-lives, about 6% of the original activity remains.
True
(½)⁴ = 1/16 = 6.25%.CThe effective half-life is longer than the physical half-life.
False
Biological clearance adds to physical decay, so the effective half-life is always shorter than either the physical or the biological half-life.DIf the physical and biological half-lives are both 6 hours, the effective half-life is 3 hours.
True
1/Teff = 1/Tphys + 1/Tbiol = 1/6 + 1/6 = 1/3, so Teff = 3 hours.EThe physical half-life of a radionuclide depends on its temperature and pressure.
False
Radioactive decay is a nuclear process and is essentially unaffected by temperature, pressure or chemical form.
More on this topic: Nuclear medicine question bank
Radiation protection questions
Question 25
Regarding radiation dose quantities:
AAbsorbed dose is measured in gray, where 1 Gy is 1 joule per kilogram.
True
Absorbed dose is the energy deposited per unit mass of material. Its SI unit is the gray (J kg⁻¹).BEquivalent dose is absorbed dose multiplied by the radiation weighting factor.
True
H = D × wR, measured in sieverts. It accounts for the different biological effectiveness of radiation types.CThe radiation weighting factor for X-rays is 20.
False
X-rays, gamma rays and electrons have wR = 1. A factor of 20 applies to alpha particles.DThe tissue weighting factors used to calculate effective dose add up to 1.
True
Effective dose is E = Σ wT × HT, and the ICRP tissue weighting factors sum to 1, so a uniform whole-body equivalent dose gives the same effective dose.EEffective dose can be measured directly with a dosemeter placed on the patient.
False
Effective dose is a calculated quantity, derived from organ doses and tissue weighting factors for a reference person. It cannot be measured directly.
Question 26
Under the Ionising Radiations Regulations 2017 (IRR17):
AThe annual effective dose limit for an employee aged 18 or over is 50 mSv.
False
The employee limit is 20 mSv per year. 50 mSv was the limit under older regulations.BThe annual equivalent dose limit for the lens of the eye of an employee is 20 mSv.
True
IRR17 cut the lens limit from 150 mSv to 20 mSv a year, following ICRP's evidence that cataracts occur at lower doses than previously thought.CThe annual effective dose limit for a member of the public is 1 mSv.
True
The public limit is 1 mSv per year of effective dose from work activities, excluding medical exposure.DThe annual equivalent dose limit for the skin of an employee is 150 mSv.
False
The employee skin limit is 500 mSv a year, averaged over any 1 cm² area. The same 500 mSv limit applies to the extremities.EEmployees likely to receive an effective dose above 6 mSv a year must be designated as classified persons.
True
Classification is required for those likely to exceed 6 mSv effective dose, or more than three-tenths of any equivalent dose limit (such as 15 mSv to the lens), in a year.
Question 27
Regarding staff protection in fluoroscopy:
ADoubling the distance from a point source reduces the dose rate to a quarter.
True
By the inverse square law, dose rate falls as 1/d², so twice the distance gives a quarter of the dose rate.BThe main source of staff dose is scatter from the patient.
True
Staff should never be in the primary beam, so their dose comes almost entirely from radiation scattered by the patient, highest close to where the beam enters.CWith a lateral C-arm projection, staff on the image detector side receive more dose than those on the tube side.
False
Scatter is strongest where the beam enters the patient, so the tube side has the higher dose rate. Stand on the detector side when possible.DLead aprons are designed to protect against the primary beam.
False
Aprons of about 0.25–0.5 mm lead equivalent are designed to attenuate scattered radiation. They do not give adequate protection in the primary beam.EAn under-couch tube gives a lower dose to the operator's head and eyes than an over-couch tube.
True
With the tube under the couch, the most intense scatter comes from the patient's underside and is directed downwards, so dose to the operator's upper body and eyes is lower.
Question 28
Under the Ionising Radiation (Medical Exposure) Regulations, IR(ME)R:
AThe referrer must supply the medical information needed to justify the exposure.
True
The referrer provides enough clinical information for the practitioner to decide whether the exposure is justified.BThe practitioner is responsible for justifying the exposure.
True
The IR(ME)R practitioner weighs the expected benefit against the radiation risk and justifies each exposure. Authorisation may be delegated to an operator following the practitioner's guidelines.CAn operator is a person entitled by the employer to carry out practical aspects of an exposure.
True
Operators perform practical aspects such as positioning, setting exposure factors and evaluating images, and must be entitled by the employer and adequately trained.DA repeat examination on the same patient does not need to be justified.
False
Every exposure must be justified, including repeats. A repeat must still offer a net benefit to the patient.EDiagnostic reference levels are dose limits that must not be exceeded for any patient.
False
DRLs are guidance levels for standard examinations on average-sized patients. They are not limits, and individual patients may legitimately exceed them; consistent excess should trigger a review.
More on this topic: Radiation protection question bank
Digital radiography questions
Question 29
Regarding flat panel detectors:
AIndirect conversion detectors use a scintillator such as caesium iodide.
True
In indirect detectors a scintillator (CsI or gadolinium oxysulphide) converts X-rays to light, which an amorphous silicon photodiode array converts to charge.BDirect conversion detectors use amorphous selenium to convert X-rays to charge.
True
Amorphous selenium is a photoconductor: absorbed X-rays create charge directly, which an applied electric field draws straight to the pixel electrodes.CColumnar caesium iodide crystals reduce lateral spread of light.
True
The needle-like CsI crystals act as light guides, channelling light towards the photodiodes. This allows a thicker, more absorbing layer without as much blur.DDirect conversion detectors suffer more from light spread than indirect detectors.
False
Direct detectors produce no light. Charge travels along the field lines with little sideways spread, which gives them excellent intrinsic resolution.EThe thin-film transistor array is read out by laser scanning.
False
The TFT array is read out electronically, row by row. Laser scanning is how computed radiography plates are read.
Question 30
Regarding digital image matrix and file size:
AA 2048 × 2048 image stored at 16 bits per pixel needs about 8 MB.
True
2048 × 2048 = about 4.2 million pixels, × 2 bytes (16 bits) = about 8.4 million bytes (8 MB) before compression.BThe limiting spatial frequency of a digital detector is 1 / (2 × pixel pitch).
True
This is the Nyquist frequency. A 0.1 mm pitch gives a limit of 5 line pairs per mm.CHalving the pixel pitch on a detector of the same size doubles the number of pixels.
False
Halving the pitch doubles the pixels along both axes, so the total number of pixels quadruples, as does the file size.DA 10-bit image can display 1024 grey levels.
True
The number of grey levels is 2ⁿ, so 10 bits gives 2¹⁰ = 1024.ELossless compression permanently discards some image data.
False
Lossless compression reconstructs the original data exactly. Only lossy compression discards information.
Question 31
Regarding PACS and DICOM:
AThe DICOM header carries patient and acquisition information with the image.
True
A DICOM object holds the pixel data and tagged metadata such as patient ID, study details and exposure factors, so they stay together.BHL7 is the messaging standard commonly used between the RIS and the hospital information system.
True
HL7 carries text messages such as orders, patient demographics and reports between hospital systems. DICOM handles the images.CA modality worklist reduces manual entry of patient details at the scanner.
True
The modality downloads scheduled patients and their details from the RIS, cutting typing errors and mismatched studies.DDICOM is a lossy image compression method.
False
DICOM is a standard for image format and communication. It can carry images compressed in several ways, lossless or lossy, but is not itself a compression method.EReporting monitors need no quality control once installed.
False
Display brightness and grey-scale response drift with age, so reporting monitors need regular quality control, including checks against the DICOM grey-scale display function.
Question 32
Regarding exposure in digital radiography:
ADigital detectors have a wider dynamic range than film-screen systems.
True
Digital detectors respond linearly over roughly four orders of magnitude of exposure, far wider than the useful range of film.BOverexposure can go unnoticed because image processing normalises brightness.
True
Automatic rescaling makes an overexposed image look normal, so doses can drift upwards unnoticed (dose creep). Exposure indices exist to catch this.CThe exposure index reflects the exposure reaching the detector, not the patient dose.
True
The exposure index indicates the air kerma at the detector. It helps check technique but does not measure patient dose.DUnderexposed digital images appear too light.
False
Processing still normalises brightness, so an underexposed image looks normal in brightness but noisier (more quantum mottle).EDoubling the mAs halves the quantum noise.
False
Relative quantum noise scales as 1/√N, so doubling the photons reduces noise by only √2 (about 30%). Halving it needs four times the mAs.
More on this topic: Digital radiography question bank
Image quality questions
Question 33
Regarding the modulation transfer function (MTF):
AThe MTF describes how well a system transfers contrast at each spatial frequency.
True
MTF is the ratio of output to input modulation as a function of spatial frequency, giving a full picture of a system's resolution.BThe MTF increases with spatial frequency.
False
MTF is 1 at zero frequency and falls as spatial frequency increases, because finer detail is blurred more.CAt a given spatial frequency, a system with a higher MTF reproduces that detail with more contrast.
True
A higher MTF at a frequency means more of the object's contrast at that scale survives into the image, so the detail is easier to see.DThe MTF of a complete imaging chain is the sum of its components' MTFs.
False
Component MTFs multiply: the system MTF at each frequency is the product of the focal spot, detector and display MTFs, so the weakest part limits the whole.ESpatial resolution is commonly expressed in line pairs per millimetre.
True
Limiting resolution is often quoted in lp/mm, the highest bar-pattern frequency that can still be distinguished, usually where the MTF has fallen to a few percent.
Question 34
Regarding noise in X-ray images:
AQuantum noise follows Poisson statistics.
True
X-ray photons arrive randomly, so the count in each pixel varies with a standard deviation of √N.BSignal-to-noise ratio improves with the square root of the number of detected photons.
True
The signal is proportional to N and the noise to √N, so SNR ∝ √N.CQuadrupling the dose halves the relative quantum noise.
True
Relative noise is 1/√N. Four times the photons gives 1/√4 = half the relative noise.DThe detective quantum efficiency (DQE) of a well-designed detector can exceed 1.
False
DQE compares output SNR² with input SNR². A detector cannot add information, so DQE is at most 1, the value for a perfect detector.EIncreasing the pixel size increases the quantum noise in each pixel.
False
A larger pixel collects more photons, so relative noise per pixel falls, at the cost of spatial resolution.
Question 35
Regarding scatter and anti-scatter grids:
AGrid ratio is the height of the lead strips divided by the width of the gaps between them.
True
Grid ratio = h / D. A higher ratio accepts a narrower range of angles, so it rejects more scatter.BA higher grid ratio removes more scatter.
True
Taller strips relative to the gap block more obliquely travelling photons, improving the scatter rejection and contrast.CUsing a grid increases patient dose.
True
The grid absorbs scatter and some primary radiation, so more mAs is needed for the same detector exposure. The ratio is the Bucky factor, typically 2–6.DGrids are routinely needed for adult extremity radiographs.
False
Thin body parts (under about 10–12 cm) generate little scatter, so a grid would raise dose for little benefit.EThe air-gap technique reduces scatter without increasing magnification.
False
An air gap works by moving the detector away from the patient, so obliquely scattered photons miss it. That increases magnification and geometric unsharpness.
Question 36
Regarding contrast in X-ray imaging:
ASubject contrast falls as kVp increases.
True
At higher energies the photoelectric effect contributes less and attenuation differences between tissues shrink, so subject contrast falls.BIodine contrast is most effective at photon energies just above 33 keV.
True
Iodine's K-edge is 33.2 keV. Just above it, photoelectric absorption in iodine rises sharply, maximising contrast with surrounding tissue.CScattered radiation reduces image contrast.
True
Scatter adds an almost uniform background signal that carries no anatomical information, reducing contrast.DCT has poorer contrast resolution than plain radiography.
False
CT removes overlying structures and uses tight collimation, so it has much better low-contrast resolution. Plain radiography has better spatial resolution.EThe K-edge of barium is lower than that of iodine.
False
Barium (Z = 56) has a K-edge of about 37.4 keV, higher than iodine's 33.2 keV (Z = 53).
Fluoroscopy questions
Question 37
Regarding patient dose in fluoroscopy:
APulsed fluoroscopy at a lower pulse rate reduces patient dose.
True
Fewer pulses per second means less radiation over the same screening time. Moving from 30 to 7.5 pulses per second can cut dose substantially.BLast-image hold helps reduce screening time.
True
Keeping the last image on the monitor lets the operator study it without further exposure, reducing fluoroscopy time.CElectronic magnification on an image intensifier increases the patient entrance dose rate.
True
A smaller field of view uses less of the input screen, so automatic brightness control raises the dose rate to keep the output brightness constant.DMoving the patient closer to the X-ray tube reduces skin dose.
False
By the inverse square law, the skin is then exposed to a more intense beam, so entrance dose increases. Keep the patient as far from the tube as practical.ERaising the image detector further from the patient reduces patient dose.
False
A larger patient-to-detector gap means automatic brightness control raises the output to compensate, increasing entrance dose. Keep the detector close to the patient.
Question 38
Regarding the image intensifier:
AThe input phosphor is usually caesium iodide.
True
CsI absorbs X-rays efficiently, and its needle-like crystals limit light spread, allowing a thick layer without much blur.BBrightness gain equals minification gain multiplied by flux gain.
True
Brightness gain is the product of minification gain (input diameter² / output diameter²) and flux gain from accelerating the electrons.CPincushion distortion arises from the curved input surface.
True
Projecting the image onto a curved input screen stretches the periphery more than the centre, giving pincushion distortion.DThe electron image is magnified onto the output screen.
False
The electron image is minified: the output phosphor (about 2.5 cm) is much smaller than the input screen (15–40 cm). Concentrating the electrons onto a small screen is the source of minification gain.EVeiling glare improves image contrast.
False
Veiling glare is light and electron scatter within the intensifier that spreads signal across the image, reducing contrast.
Question 39
Regarding the kerma-area product (KAP):
AKAP stays the same at any distance from the source, ignoring attenuation and scatter.
True
Air kerma falls as 1/d², but the beam area grows as d², so their product is independent of distance.BKAP is usually measured with a transmission ionisation chamber at the collimator.
True
A large-area ionisation chamber mounted on the tube collimator intercepts the whole beam, measuring KAP for every exposure.CLong interventional procedures can cause deterministic skin injury.
True
Prolonged screening on one skin area can exceed thresholds for erythema (about 2 Gy) and more serious injury at higher doses.DKAP is a good measure of peak skin dose.
False
KAP reflects the total energy entering the patient, not how concentrated it is. The same KAP from a small, fixed field gives a much higher peak skin dose than from a large, moving one.ECollimating the beam more tightly increases the KAP.
False
Tighter collimation reduces the beam area, so KAP falls, which is one reason collimation is such an effective dose reduction tool.
More on this topic: Fluoroscopy question bank
Mammography questions
Question 40
Regarding mammography:
ATypical tube voltages are in the range 25–35 kV.
True
Low energies maximise the small attenuation differences between fibroglandular tissue, fat and microcalcifications.BA molybdenum target produces characteristic X-rays at about 17.5 and 19.6 keV.
True
Molybdenum's K-lines at 17.5 and 19.6 keV sit in the ideal energy band for breast imaging, and a molybdenum filter removes much of the spectrum above its 20 keV K-edge.CThe cathode is positioned at the chest wall side.
True
The heel effect makes intensity higher on the cathode side, so the cathode is placed over the thicker chest-wall tissue.DStandard mammography uses a tungsten target with 2 mm of aluminium filtration.
False
2 mm aluminium would remove the useful low-energy photons. Mammography uses molybdenum, rhodium or tungsten targets with thin K-edge filters (molybdenum, rhodium or silver).EBreast compression increases the dose to the breast.
False
Compression reduces thickness, so less radiation is needed and scatter falls: dose goes down while contrast and sharpness improve.
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