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PCN NDT Radiographic Testing Level 2 Exam study materials
PCN NDT Radiographic Testing Level 2 Exam study resources

PCN NDT Radiographic Testing Level 2 Exam practice questions

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Question 1Principles of Radiography and Radiation Physics

In the SI system, the quantity "radiation exposure" (ionization produced in air) is expressed in which unit?

  1. Coulomb per kilogram (C/kg) Correct answer.
  2. Sievert (Sv)
  3. Becquerel (Bq)
  4. Gray (Gy)

Explanation

Exposure is defined as the charge of one sign produced by ionization per unit mass of air, so its SI unit is the coulomb per kilogram. The gray measures absorbed dose, the sievert measures equivalent dose, and the becquerel measures activity.

Question 2Principles of Radiography and Radiation Physics

What is the principal function of the tungsten target inside an X-ray tube?

  1. To produce X-rays when energetic electrons are decelerated within it Correct answer.
  2. To filter low-energy photons out of the emerging beam
  3. To circulate cooling oil around the tube envelope
  4. To electrostatically focus the electron beam onto a small area

Explanation

When high-speed electrons strike the target they lose kinetic energy as bremsstrahlung and by ejecting inner-shell electrons, producing the X-ray beam. Tungsten is chosen because its high atomic number and high melting point favour efficient production and heat tolerance.

Question 3Principles of Radiography and Radiation Physics

A radiation source is moved so that its distance from the film is tripled. According to the inverse square law, the intensity at the film becomes:

  1. One sixth of the original value
  2. One ninth of the original value Correct answer.
  3. One third of the original value
  4. One twelfth of the original value

Explanation

For a point source in a non-absorbing medium, intensity falls with the square of distance: I is proportional to 1/d². Tripling d therefore divides intensity by 3² = 9.

Question 4Principles of Radiography and Radiation Physics

The half-value layer of an absorbing material is best described as the thickness that:

  1. Reduces the beam intensity to half of its original value Correct answer.
  2. Brings the transmitted intensity to zero
  3. Doubles the intensity of the transmitted beam
  4. Reduces the beam intensity to one tenth of its original value

Explanation

By definition the half-value layer is the thickness of a specified material that attenuates a narrow beam to 50% of its incident intensity. It is a measure of beam quality and is related to the linear attenuation coefficient by HVL = 0.693/mu.

Question 5Principles of Radiography and Radiation Physics

Photoelectric absorption becomes most significant under which combination of conditions?

  1. High photon energy and high atomic number of the absorber
  2. Low photon energy and low atomic number of the absorber
  3. Low photon energy and high atomic number of the absorber Correct answer.
  4. High photon energy and low atomic number of the absorber

Explanation

The photoelectric probability rises steeply as photon energy falls and as the atomic number of the absorber rises, approximately as Z⁴/E³ between absorption edges. This is why low-energy photons are removed rapidly by dense, high-Z materials such as lead.

Question 6Principles of Radiography and Radiation Physics

Gamma radiation differs from X-radiation mainly because gamma rays:

  1. Are produced by decelerating electrons in a target
  2. Travel at a lower speed than visible light
  3. Originate from the nucleus of a radioactive atom Correct answer.
  4. Carry no energy and therefore cannot expose film

Explanation

Gamma rays are electromagnetic photons emitted by excited nuclei during radioactive decay, whereas X-rays arise from electron transitions or electron deceleration outside the nucleus. Both are photons with no rest mass and travel at the speed of light.

Question 7Principles of Radiography and Radiation Physics

The absorbed dose delivered to tissue by ionizing radiation is expressed in which SI unit?

  1. Becquerel (Bq)
  2. Coulomb per kilogram (C/kg)
  3. Sievert (Sv)
  4. Gray (Gy) Correct answer.

Explanation

Absorbed dose is energy imparted per unit mass and is measured in joules per kilogram, which is given the name gray. The sievert results from applying a radiation weighting factor to absorbed dose.

Question 8Principles of Radiography and Radiation Physics

What happens to a photon that undergoes Compton scattering in the specimen?

  1. It is absorbed completely and the atom is ionized
  2. It transfers part of its energy to an outer-shell electron and is deflected Correct answer.
  3. It is converted into an electron-positron pair
  4. It transfers all its energy to a tightly bound inner-shell electron

Explanation

In Compton scattering the photon interacts with a loosely bound outer electron, ejecting it and continuing in a new direction with reduced energy. The energy lost by the photon equals the kinetic energy given to the recoil electron.

Question 9Principles of Radiography and Radiation Physics

Bremsstrahlung production in an X-ray tube is most efficient when the tube uses:

  1. A high-atomic-number target and low electron energy
  2. A high-atomic-number target and high electron energy Correct answer.
  3. A low-atomic-number target and low electron energy
  4. A low-atomic-number target and high electron energy

Explanation

The fraction of electron energy converted to bremsstrahlung rises with the atomic number of the target and with the energy of the incident electrons. This is why tungsten targets and higher accelerating potentials yield a greater continuous output.

Question 10Principles of Radiography and Radiation Physics

Characteristic X-rays are generated when:

  1. An outer-shell electron drops into a vacancy in an inner shell Correct answer.
  2. A photon is deflected by a free electron with no energy loss
  3. An electron and a positron annihilate each other
  4. A fast electron is slowly decelerated by the nuclear field

Explanation

An incident electron ejects an inner-shell electron, and when another electron falls into that vacancy a photon is emitted whose energy equals the difference between the two shell binding energies. The resulting lines are discrete and characteristic of the target element.

Question 11Principles of Radiography and Radiation Physics

Pair production in the field of a nucleus can occur only when the incident photon energy is at least about:

  1. 1.022 MeV Correct answer.
  2. 2.50 MeV
  3. 0.662 MeV
  4. 0.511 MeV

Explanation

Pair production converts a photon into an electron and a positron, whose combined rest mass corresponds to 1.022 MeV. Any energy above this threshold appears as kinetic energy shared by the two particles.

Question 12Principles of Radiography and Radiation Physics

A sealed source has an activity of 2 curies. Expressed in SI units, this corresponds to approximately:

  1. 7.4 GBq
  2. 37 GBq
  3. 3.7 GBq
  4. 74 GBq Correct answer.

Explanation

One curie equals 3.7 × 10¹⁰ becquerels, which is 37 GBq. Multiplying by two curies gives 74 GBq.

Question 13Principles of Radiography and Radiation Physics

A worker's whole body receives an absorbed dose of 5 mGy from gamma radiation. Using a radiation weighting factor of 1 for gamma rays, the equivalent dose is:

  1. 5 mSv Correct answer.
  2. 100 mSv
  3. 50 mSv
  4. 0.5 mSv

Explanation

Equivalent dose in sieverts is obtained by multiplying absorbed dose in grays by the radiation weighting factor. With a weighting factor of 1 for gamma radiation, 5 mGy gives 5 mSv.

Question 14Principles of Radiography and Radiation Physics

Geometric unsharpness of a radiograph is increased by which change in the exposure geometry?

  1. Using a smaller focal spot
  2. Moving the source closer to the specimen while keeping the film position fixed Correct answer.
  3. Placing the film in closer contact with the specimen
  4. Increasing the source-to-film distance with the specimen in contact with the film

Explanation

Geometric unsharpness follows Ug = F·t/d, where F is the focal spot size, t the object-to-film distance and d the source-to-object distance. Decreasing d while keeping t fixed increases Ug.

Question 15Principles of Radiography and Radiation Physics

A specimen is radiographed with a source-to-film distance of 100 cm and the source is 80 cm from the specimen. The geometric magnification is:

  1. 1.80 Correct answer.
  2. 1.25
  3. 2.25
  4. 0.80

Explanation

Magnification is the ratio of source-to-film distance to source-to-object distance, M = SID/SOD. Here M = 100/80 = 1.25.

Question 16Principles of Radiography and Radiation Physics

With kVp held constant, increasing the tube current (mA) in an X-ray set mainly:

  1. Increases the penetrative power of the beam
  2. Raises the average energy of the photons produced
  3. Increases the number of X-ray photons produced Correct answer.
  4. Increases the half-value layer of the beam

Explanation

Tube current determines how many electrons strike the target per second, and therefore the quantity of photons emitted per unit time. Photon energy, and hence penetrating power, is governed by the accelerating potential (kVp).

Question 17Principles of Radiography and Radiation Physics

Raising the tube potential (kVp) affects the emergent X-ray beam by:

  1. Producing more photons with unchanged energy
  2. Producing photons of higher energy and shorter wavelength Correct answer.
  3. Reducing photon energy while increasing the beam intensity
  4. Producing photons of longer wavelength and lower frequency

Explanation

Accelerating electrons through a larger potential difference gives them more kinetic energy, so the bremsstrahlung spectrum shifts toward higher photon energies. Higher energy corresponds to higher frequency and shorter wavelength, which improves penetration.

Question 18Principles of Radiography and Radiation Physics

Lead intensifying screens are used in high-energy radiography because they:

  1. Absorb scattered radiation and so improve image contrast
  2. Protect the film from mechanical damage during handling
  3. Emit electrons that expose the film, permitting a shorter exposure Correct answer.
  4. Fluoresce visible light that exposes the film more efficiently

Explanation

When high-energy photons strike a lead screen, photoelectrons and Compton electrons are ejected and these expose the emulsion in addition to the direct photon action. The resulting intensification allows a shorter exposure time or a lower activity source.

Question 19Principles of Radiography and Radiation Physics

Restricting the primary beam with a collimator improves radiographic image quality chiefly because it:

  1. Reduces the exposure time needed for a given density
  2. Reduces the volume of material irradiated and therefore the scattered radiation reaching the film Correct answer.
  3. Increases the energy of the primary radiation reaching the film
  4. Removes low-energy photons from the primary beam before it reaches the specimen

Explanation

Scattered radiation is generated throughout the irradiated volume, so limiting the beam to the area of interest reduces the amount of scatter produced and reaching the detector. The result is improved contrast, together with a lower dose to the surrounding material.

Question 20Principles of Radiography and Radiation Physics

A monoenergetic beam with a linear attenuation coefficient of 0.2 cm⁻¹ passes through 10 cm of material. The transmitted fraction I/I₀ is approximately:

  1. 0.135 Correct answer.
  2. 0.20
  3. 0.82
  4. 0.50

Explanation

Narrow-beam attenuation follows I = I₀e^(−mu x). With mu x = 0.2 × 10 = 2, the transmitted fraction is e⁻² ≈ 0.135.

Question 21Principles of Radiography and Radiation Physics

A geometric unsharpness of 0.2 mm is acceptable, the specimen-to-film distance is 20 mm, and the source-to-object distance is 600 mm. The largest focal spot that may be used is about:

  1. 2.4 mm
  2. 0.6 mm
  3. 6.0 mm Correct answer.
  4. 12 mm

Explanation

Rearranging Ug = F·t/d to solve for focal spot size gives F = Ug·d/t. Substituting gives F = 0.2 × 600 / 20 = 6.0 mm.

Question 22Principles of Radiography and Radiation Physics

For 10 MeV photons interacting with a lead absorber, the dominant interaction process is:

  1. Coherent (Rayleigh) scattering
  2. Compton scattering
  3. Pair production Correct answer.
  4. Photoelectric absorption

Explanation

In a high-atomic-number absorber such as lead, pair production becomes the principal mode of interaction once photon energies well exceed the 1.022 MeV threshold. At 10 MeV the photoelectric effect is negligible and Compton scattering is secondary.

Question 23Principles of Radiography and Radiation Physics

A 75 mm thick steel section must be radiographed with adequate penetration and reasonable exposure time. Which gamma source is the most appropriate choice?

  1. Cobalt-60, because its higher energy penetrates thick steel Correct answer.
  2. Iridium-192, because its lower energy gives better contrast in thick steel
  3. Ytterbium-169, because its very low energy suits thick sections
  4. Caesium-137, because it provides the highest available gamma energy

Explanation

Cobalt-60 emits gamma rays at about 1.17 and 1.33 MeV, which give the penetration needed for thick steel sections. Lower-energy sources such as Ir-192 or Yb-169 are strongly attenuated in thick steel and would require impractically long exposures.

Question 24Principles of Radiography and Radiation Physics

A cobalt-60 source with a half-life of about 5.3 years has an activity of 20 Ci. After 10.6 years, its activity will be approximately:

  1. 1.25 Ci
  2. 5 Ci Correct answer.
  3. 2.5 Ci
  4. 10 Ci

Explanation

The elapsed time of 10.6 years equals two half-lives. Activity therefore falls by a factor of two twice: 20 Ci becomes 10 Ci, then 5 Ci.

Question 25Radiation Safety, Protection and Dose Control

What is the SI unit used to express equivalent dose to a person?

  1. Becquerel (Bq)
  2. Coulomb per kilogram (C/kg)
  3. Gray (Gy)
  4. Sievert (Sv) Correct answer.

Explanation

The sievert is the SI unit of equivalent dose; it is obtained by multiplying the absorbed dose in gray by the radiation weighting factor. For gamma and X-radiation the weighting factor is 1, so the numerical values in Gy and Sv coincide.

Question 26Radiation Safety, Protection and Dose Control

Which statement best describes the ALARA principle as applied to industrial radiography?

  1. ALARA applies only to members of the public, not to classified workers.
  2. Only the annual dose limit needs to be considered, and the actual dose received by the worker is unimportant.
  3. Doses should be kept as low as reasonably achievable, with economic and social factors considered. Correct answer.
  4. Radiation exposure must be reduced to zero whatever the cost.

Explanation

ALARA requires that all exposures be kept as low as reasonably achievable, with economic and social factors taken into account. It is a optimisation principle applied on top of, not instead of, the legal dose limits. It applies to everyone, including classified radiation workers.

Question 27Radiation Safety, Protection and Dose Control

Which description correctly identifies the international warning symbol used to mark a radiography controlled area?

  1. A green circle containing a black hand symbol
  2. A magenta trefoil on a yellow background Correct answer.
  3. A blue triangle containing a white cross
  4. A white square containing a large red letter R

Explanation

The international ionising radiation warning sign is a magenta or purple trefoil on a yellow background. It is used on source containers, barriers, warning notices and doors to indicate the presence of ionising radiation.

Question 28Radiation Safety, Protection and Dose Control

What is the main purpose of the lead or depleted-uranium shielding built into a gamma source container?

  1. To focus the beam onto the film
  2. To increase the energy of the emitted gamma radiation and improve penetration
  3. To attenuate radiation so that dose rates around the container are reduced Correct answer.
  4. To keep the source capsule cool during transport

Explanation

The shielding absorbs photons so that the dose rate at the surface of, and around, the container is low enough for safe handling and storage. It does not change the energy of the useful beam or focus it. The radiation that escapes the container is simply attenuated in intensity.

Question 29Radiation Safety, Protection and Dose Control

When does an X-ray tube produce ionising radiation?

  1. Continuously from the moment the set is connected to the mains
  2. Only when a film cassette is positioned in the beam
  3. Only while the tube is energised and an exposure is in progress Correct answer.
  4. Only after the anode has reached its operating temperature

Explanation

X-radiation is generated only when electrons are accelerated across the tube and strike the target, that is, while the set is energised for an exposure. Once the high voltage is removed, the output stops immediately, which is a key safety advantage over gamma sources.

Question 30Radiation Safety, Protection and Dose Control

Within the context of site radiography, a controlled area is best described as:

  1. Any part of the site within 100 metres of the site office
  2. The area in which the radiographer stores protective clothing, barriers and equipment
  3. An area where access is restricted and exposure is supervised and monitored Correct answer.
  4. Any area in which a film badge is worn

Explanation

A controlled area is a defined region where access is restricted and where radiation exposure is kept under supervision and monitored in accordance with local rules. Boundaries are set from measured or calculated dose rates, not from arbitrary distances or from where equipment happens to be kept.

Question 31Radiation Safety, Protection and Dose Control

If the distance from a small gamma source is doubled, the dose rate in air at the new position will be:

  1. half the original rate
  2. the same as the original rate
  3. twice the original rate
  4. a quarter of the original rate Correct answer.

Explanation

For a point source the dose rate follows the inverse square law, so doubling the distance divides the dose rate by four. This is why distance is one of the most effective controls available to a radiographer.

Question 32Radiation Safety, Protection and Dose Control

A point gamma source produces a dose rate of 400 µSv/h at 2 m. Assuming no shielding, what is the dose rate at 8 m?

  1. 50 µSv/h
  2. 200 µSv/h
  3. 25 µSv/h Correct answer.
  4. 100 µSv/h

Explanation

The distance increases from 2 m to 8 m, a factor of four, so the inverse square law divides the dose rate by 4² = 16. Dividing 400 µSv/h by 16 gives 25 µSv/h.

Question 33Radiation Safety, Protection and Dose Control

A barrier currently transmits 800 µSv/h. If the half-value layer of the barrier material for the source in use is 10 mm, what dose rate results from adding 30 mm of the same material?

  1. 50 µSv/h
  2. 100 µSv/h Correct answer.
  3. 200 µSv/h
  4. 267 µSv/h

Explanation

Each half-value layer halves the dose rate, and 30 mm represents three half-value layers. Three successive halvings of 800 µSv/h give 800 ÷ 8 = 100 µSv/h.

Question 34Radiation Safety, Protection and Dose Control

What is the annual whole-body dose limit recommended by ICRP for a classified radiation worker?

  1. 50 mSv in any single year without averaging
  2. 1 mSv per year
  3. 500 mSv per year
  4. 20 mSv per year averaged over five years Correct answer.

Explanation

The ICRP recommended limit for occupational exposure is an effective dose of 20 mSv per year averaged over five consecutive years, with no single year exceeding 50 mSv. The 1 mSv figure applies to members of the public.

Question 35Radiation Safety, Protection and Dose Control

Which combination of controls will reduce the dose received by a radiographer during a series of exposures?

  1. Keep the same distance, use a smaller source and dispense with barriers
  2. Increase the distance from the source, shorten the time spent near it and add shielding Correct answer.
  3. Increase the exposure time and use a larger source
  4. Reduce the distance from the source, lengthen the exposure time and remove all shielding

Explanation

Occupational dose is controlled by the three basic factors: maximising distance, minimising time near the source and interposing adequate shielding. Each of these independently reduces the accumulated dose.

Question 36Radiation Safety, Protection and Dose Control

What is the primary radiation-protection benefit of fitting a collimator to an X-ray tube or gamma source?

  1. It measures the dose rate at the boundary of the controlled area
  2. It removes low-energy photons from the beam and therefore increases film contrast only
  3. It restricts the beam to the area of interest, reducing scatter and unnecessary dose Correct answer.
  4. It raises the energy of the beam so that shorter exposures are possible

Explanation

A collimator confines the beam to the region being examined. This reduces the amount of material or tissue irradiated and therefore reduces scattered radiation, which is a major contributor to occupational dose.

Question 37Radiation Safety, Protection and Dose Control

Immediately before making an exposure, why does the radiographer use a survey meter around the barrier?

  1. To check the dose rates and confirm the controlled area boundary is correct Correct answer.
  2. To check the temperature of the developer in the darkroom before processing the film
  3. To measure the activity of the sealed source in becquerels
  4. To confirm the film density that will be produced in the cassette

Explanation

A survey meter is used to verify that measured dose rates agree with the planned boundary and that the barrier position, signs and guards are adequate. If the reading is higher than expected, the boundary must be moved or additional shielding provided before the exposure starts.

Question 38Radiation Safety, Protection and Dose Control

How is the integrity of a sealed gamma source and its container normally checked for leakage?

  1. By carrying out a careful visual inspection of the outside of the container only
  2. By pressurising the container and looking for leaks
  3. By wiping the source and container and checking for removable contamination Correct answer.
  4. By radiographing the source capsule with the source itself

Explanation

Leak testing is done by taking wipe samples from the source, guide tube and container and checking them for removable radioactive contamination at prescribed intervals. A positive result above the permitted level means the source must be withdrawn from use.

Question 39Radiation Safety, Protection and Dose Control

During a gamma exposure the source fails to wind back into its container. What is the correct immediate action?

  1. Increase the drive-cable tension repeatedly until the source frees itself from the guide tube
  2. Switch off the lights and wait for the next shift to take over
  3. Evacuate and secure the area, post warnings and inform the radiation protection supervisor Correct answer.
  4. Grasp the guide tube and pull the source back by hand

Explanation

A stuck source is an emergency: the area must be evacuated and secured, warning signs and barriers posted, and dose rates monitored from a safe distance while the radiation protection supervisor is informed. No attempt should be made to handle the source or guide tube directly.

Question 40Radiation Safety, Protection and Dose Control

Which type of personal dosimeter gives an immediate, direct reading of dose and dose rate to the wearer?

  1. Optically stimulated luminescence badge
  2. Electronic personal dosimeter Correct answer.
  3. Film badge
  4. Thermoluminescent dosemeter

Explanation

An electronic personal dosimeter displays accumulated dose and dose rate as they occur, allowing the wearer to react immediately. Film, thermoluminescent and optically stimulated luminescence badges are passive devices that are processed later and provide a retrospective record.

Question 41Radiation Safety, Protection and Dose Control

After cranking the source back, which check gives the best confirmation that it is fully shielded?

  1. The crank handle turning freely once the wind-in has been completed
  2. A survey meter reading at the guide tube showing a return to background Correct answer.
  3. The film badge reading recorded at the end of the working shift
  4. The elapsed exposure time matching the exposure time calculated from the source activity

Explanation

Only a radiation measurement confirms that the source is back in the shielded position. A freely turning crank or an elapsed timer can be misleading if the cable has become detached or jammed.

Question 42Radiation Safety, Protection and Dose Control

For a radiographer positioned behind the source during a pipeline exposure, the largest contribution to personal dose usually comes from:

  1. Leakage from the generator cooling circuit
  2. The primary beam transmitted through the pipe wall
  3. Scattered radiation from the workpiece, floor and surrounding structures Correct answer.
  4. Natural background radiation

Explanation

Most of the radiation reaching a radiographer standing behind the source has been scattered by the workpiece, floor, walls and nearby structures. Scattered radiation is therefore the main target of barriers, collimation and distance control.

Question 43Radiation Safety, Protection and Dose Control

The equivalent dose limit for the hands and feet of a classified radiation worker is:

  1. 500 mSv per year Correct answer.
  2. 20 mSv per year
  3. 50 mSv per year
  4. 150 mSv per year

Explanation

The ICRP limit for the extremities (hands and feet) is 500 mSv per year in occupational exposure. This is higher than the whole-body limit because the hands and feet are less sensitive to stochastic effects.

Question 44Radiation Safety, Protection and Dose Control

A dose rate of 640 µSv/h must be reduced to 20 µSv/h by adding lead shielding whose half-value layer for the radiation is 12 mm. What thickness is required?

  1. 48 mm
  2. 36 mm
  3. 60 mm Correct answer.
  4. 24 mm

Explanation

The required attenuation factor is 640 ÷ 20 = 32, and 32 equals 2⁵, so five half-value layers are needed. Five half-value layers of 12 mm give a total of 60 mm of lead.

Question 45Radiation Safety, Protection and Dose Control

Which statement correctly relates absorbed dose and equivalent dose?

  1. Absorbed dose is expressed in sieverts while equivalent dose is expressed in grays
  2. Equivalent dose (Sv) equals absorbed dose (Gy) times the radiation weighting factor, which is 1 for gamma rays Correct answer.
  3. For every type of radiation the numerical values of absorbed dose and equivalent dose are identical
  4. Equivalent dose applies only to alpha particles and neutrons, which have high radiation weighting factors

Explanation

Absorbed dose is the physical energy deposited per unit mass, measured in gray. Equivalent dose adjusts this for biological effectiveness by multiplying by the radiation weighting factor, which is 1 for gamma and X-rays and therefore makes the numerical values equal for those radiations only.

Question 46Radiation Safety, Protection and Dose Control

A personal dosimeter worn by a classified radiographer returns a dose of 150 mSv for a monitoring period. Which response is required?

  1. Reassign the worker to non-radiation duties and take no other action
  2. Take no action because the reading is below the annual whole-body limit for a classified worker
  3. Investigate and record the dose, inform the worker and notify the enforcing authority Correct answer.
  4. Re-issue a new dosimeter without further action and continue the same work programme

Explanation

A dose of this magnitude greatly exceeds the annual limit and must be treated as a suspected overexposure. The employer must investigate the circumstances, formally assess and record the dose, inform the individual, arrange medical advice as appropriate and report the event to the regulator.

Question 47Radiation Safety, Protection and Dose Control

When broad-beam gamma radiation is attenuated by successive layers of the same material, the second half-value layer is usually thicker than the first. Why?

  1. Because the first layer becomes radioactive and adds to the dose rate
  2. Because the radiation energy increases as it passes through the shielding material
  3. Because scatter builds up, so attenuation is less than a simple exponential Correct answer.
  4. Because the source activity falls during the exposure

Explanation

In broad-beam geometry, photons scattered within the shield and surroundings add to the radiation reaching the detector or person. This build-up makes the measured attenuation less than a simple exponential, so each successive half-value layer is slightly thicker than the one before.

Question 48Radiation Safety, Protection and Dose Control

An iridium-192 source has a half-life of about 74 days. A source currently rated at 3.7 TBq will have what approximate activity after 148 days?

  1. 1.2 TBq
  2. 0.93 TBq Correct answer.
  3. 1.85 TBq
  4. 2.8 TBq

Explanation

148 days is two half-lives, so the activity falls by a factor of four. Dividing 3.7 TBq by four gives approximately 0.93 TBq, which means exposure times must be recalculated as the source decays.

Question 49Radiographic Equipment, X-ray Units and Gamma Sources

In a conventional X-ray tube, the filament is normally made of which material?

  1. Tungsten Correct answer.
  2. Graphite
  3. Molybdenum
  4. Copper

Explanation

Tungsten is used because it combines a very high melting point with efficient thermionic emission, so it can be heated to release electrons without rapidly vaporising. The filament is heated by a low-voltage current, and the emitted electrons form the tube current.

Question 50Radiographic Equipment, X-ray Units and Gamma Sources

The SI unit of radioactivity is the:

  1. becquerel Correct answer.
  2. roentgen
  3. curie
  4. gray

Explanation

One becquerel is one nuclear transformation per second and is the SI unit of activity. The curie is the older non-SI unit, equal to 3.7 × 10^10 decays per second.