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ABR Radiation Oncology study materials
ABR Radiation Oncology study resources

ABR Radiation Oncology practice questions

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Question 1Medical Physics for Radiation Oncology

The absorbed dose to a medium is defined as the energy deposited per unit mass. Which SI unit expresses absorbed dose?

  1. Coulomb per kilogram
  2. Gray Correct answer.
  3. Becquerel
  4. Sievert

Explanation

Absorbed dose is energy imparted per unit mass and its SI unit is the gray (1 Gy = 1 J/kg). The becquerel measures activity, the sievert measures equivalent/protection-weighted dose, and C/kg measures exposure.

Question 2Medical Physics for Radiation Oncology

A photon beam is calibrated in water. What does a dose of 1 Gy correspond to in terms of energy deposition?

  1. 1 electron volt per kilogram of medium
  2. 1 joule per kilogram of medium Correct answer.
  3. 1 erg per gram of medium
  4. 1 watt per kilogram of medium

Explanation

By definition, 1 Gy equals 1 J/kg of absorbed energy in the medium. The erg/g (the rad) is the older unit and equals 0.01 Gy.

Question 3Medical Physics for Radiation Oncology

In external beam radiotherapy, the inverse square law describes how photon fluence changes with distance from a point source. If the distance from the source is doubled, the fluence becomes:

  1. One half of the original value
  2. Unchanged
  3. Twice the original value
  4. One quarter of the original value Correct answer.

Explanation

For a point source, fluence falls off as the inverse of the square of the distance. Doubling distance increases the denominator by a factor of four, so fluence drops to one quarter.

Question 4Medical Physics for Radiation Oncology

The half-value layer (HVL) of a photon beam is best described as:

  1. The thickness of material that reduces the beam intensity to half its original value Correct answer.
  2. The distance over which beam intensity falls to half by inverse square
  3. The thickness of material that reduces beam energy to half its original value
  4. The thickness of material that doubles the beam intensity

Explanation

HVL is defined as the attenuating thickness that halves the photon beam intensity through exponential attenuation. It is related to the linear attenuation coefficient by HVL = ln2/μ.

Question 5Medical Physics for Radiation Oncology

Which quantity is used to specify the dose at a reference point in a phantom under reference conditions, forming the basis of clinical beam calibration?

  1. Reference air kerma rate
  2. Exposure in air
  3. Absorbed dose to water Correct answer.
  4. Equivalent dose to tissue

Explanation

Modern calibration protocols (e.g., TG-51, TRS-398) specify output as absorbed dose to water at a reference point under reference conditions. Air kerma and exposure are intermediate quantities used in older protocols.

Question 6Medical Physics for Radiation Oncology

The percent depth dose (PDD) of a megavoltage photon beam increases with field size primarily because:

  1. Larger fields produce more scatter contributing dose at depth Correct answer.
  2. Larger fields decrease the attenuation coefficient of water
  3. Larger fields reduce the source-to-surface distance
  4. Larger fields increase beam energy

Explanation

Increasing field size adds scattered photons that deposit dose at depth, raising PDD. The effect diminishes with increasing energy because scatter becomes less dominant.

Question 7Medical Physics for Radiation Oncology

For a fixed field size, as photon beam energy increases from 6 MV to 18 MV, the depth of maximum dose (dmax) in water:

  1. Decreases
  2. Increases Correct answer.
  3. Becomes zero
  4. Remains identical

Explanation

Higher-energy beams have greater penetrability and build-up extends deeper, so dmax shifts deeper (roughly 1.5 cm at 6 MV to about 3 cm at 18 MV). Surface dose also tends to decrease as energy rises.

Question 8Medical Physics for Radiation Oncology

The output factor of a linear accelerator is defined as the ratio of:

  1. Dose at depth to dose at dmax
  2. Dose at dmax for a given field size to dose at dmax for the reference field size Correct answer.
  3. Dose in phantom to dose in air
  4. Dose with wedge to dose without wedge

Explanation

Output factor (Sc,p or total scatter factor) compares machine output for a given field to the reference field under the same conditions. It captures both collimator scatter and phantom scatter.

Question 9Medical Physics for Radiation Oncology

In a photon beam, the region of electronic equilibrium occurs where:

  1. Scattered photons dominate the primary beam
  2. All photons have been fully absorbed
  3. The number of electrons entering a volume equals those leaving it Correct answer.
  4. Dose equals kerma exactly at all depths

Explanation

Electronic equilibrium exists when secondary electron fluence in equals fluence out of a small volume, so absorbed dose approximates collision kerma. It is approximately achieved near and beyond dmax.

Question 10Medical Physics for Radiation Oncology

Which detector is generally preferred for small-field photon dosimetry because of its small sensitive volume and near water equivalence?

  1. Large-volume Farmer chamber
  2. Unshielded diode or small-volume ionization chamber Correct answer.
  3. Plane-parallel chamber with 1 cm gap
  4. Thermoluminescent dosimeter chip of 3 mm

Explanation

Small fields require detectors with high spatial resolution and minimal volume averaging; small-volume chambers or unshielded diodes are used. Large chambers suffer volume averaging and perturbation in steep gradients.

Question 11Medical Physics for Radiation Oncology

In intensity-modulated radiation therapy, the primary purpose of inverse treatment planning is to:

  1. Manually adjust beam weights until the plan looks acceptable
  2. Optimize fluence maps to satisfy dose objectives specified for targets and organs at risk Correct answer.
  3. Reduce the number of CT slices required
  4. Eliminate the need for image guidance

Explanation

Inverse planning uses optimization algorithms to derive fluence distributions that best meet prescribed dose-volume objectives. Forward planning, in contrast, relies on iterative manual adjustment of beam parameters.

Question 12Medical Physics for Radiation Oncology

A treatment plan uses a 6 MV beam with a source-to-axis distance of 100 cm. If the field size at the isocenter is 10 x 10 cm, the collimator setting at the source is approximately:

  1. 5 x 5 cm
  2. 100 x 100 cm
  3. 20 x 20 cm
  4. 10 x 10 cm Correct answer.

Explanation

For SAD setups, field size is conventionally defined at the isocenter, and the collimator is set to produce that size at 100 cm. Thus the collimator setting corresponds to 10 x 10 cm at isocenter.

Question 13Medical Physics for Radiation Oncology

Which statement correctly describes the difference between kerma and absorbed dose?

  1. Kerma applies only to neutrons, absorbed dose only to photons
  2. Absorbed dose is measured in air, kerma only in tissue
  3. Kerma is energy transferred to charged particles per mass; absorbed dose is energy deposited per mass Correct answer.
  4. Kerma and absorbed dose are identical at all depths

Explanation

Kerma (kinetic energy released per mass) quantifies energy transferred from uncharged to charged particles, while absorbed dose quantifies energy actually deposited. They converge only under charged-particle equilibrium.

Question 14Medical Physics for Radiation Oncology

A wedged photon field is used to compensate for a sloping patient surface. The wedge factor is defined as the ratio of:

  1. Dose at the wedge heel to dose at the toe
  2. Dose at depth to dose at dmax
  3. Dose on central axis to dose off axis
  4. Dose with the wedge in place to dose without the wedge at the same point Correct answer.

Explanation

The wedge factor corrects output for the attenuating wedge and is measured as the ratio of dose with versus without the wedge at a reference point. It is applied to monitor unit calculations.

Question 15Medical Physics for Radiation Oncology

For electron beams, the depth of 90% dose (R90) is commonly used to select beam energy because:

  1. It equals the practical range Rp
  2. It estimates the depth of the target volume adequately covered by a therapeutic dose Correct answer.
  3. It marks the maximum surface dose
  4. It defines the X-ray contamination level

Explanation

Electron energy is chosen so that the 90% isodose covers the deepest extent of the target, providing therapeutic coverage while sparing distal tissue. R90 approximates this clinically useful depth.

Question 16Medical Physics for Radiation Oncology

Brachytherapy sources are often specified by air kerma strength. The primary reason for using air kerma strength rather than activity is that:

  1. It is identical for all radionuclides
  2. It characterizes the source output in a quantity measurable and traceable to standards independent of source construction Correct answer.
  3. It directly gives the absorbed dose at 1 cm in tissue
  4. It allows dose to be calculated without any source-specific data

Explanation

Air kerma strength (or reference air kerma rate) is a measurable, calibratable quantity that reflects photon output and is traceable to standards. It avoids reliance on nominal activity, which depends on encapsulation and manufacturing.

Question 17Medical Physics for Radiation Oncology

The tissue maximum ratio (TMR) differs from percent depth dose (PDD) primarily because TMR:

  1. Depends strongly on field size at the surface
  2. Is independent of source-to-surface distance Correct answer.
  3. Is defined only for electron beams
  4. Includes inverse square effects explicitly

Explanation

TMR is defined at a fixed source-to-axis distance and is largely independent of SSD, making it convenient for isocentric calculations. PDD varies with SSD because it incorporates inverse square and scatter changes.

Question 18Medical Physics for Radiation Oncology

In a CT simulator, the purpose of a scout (topogram) image is to:

  1. Provide a projection radiograph used to plan the extent of the axial scan Correct answer.
  2. Measure electron density directly
  3. Deliver the treatment planning dose calculation
  4. Calibrate the linear accelerator output

Explanation

A scout image is a projection radiograph used to define the scan range and positioning before axial acquisition. It is not used for dose calculation or machine calibration.

Question 19Medical Physics for Radiation Oncology

The Hounsfield unit of water and air on a CT scanner are, by definition, respectively:

  1. 0 and approximately -1000 Correct answer.
  2. -1000 and 0
  3. 1000 and 0
  4. 0 and 1000

Explanation

CT numbers are scaled so water is 0 HU and air is about -1000 HU, with dense bone positive. This scale is used to convert CT number to electron density for dose calculation.

Question 20Medical Physics for Radiation Oncology

A patient is treated with a 10 MV photon beam. Compared with a 6 MV beam, the 10 MV beam will generally produce:

  1. A smaller penumbra at all depths
  2. A higher surface dose and shallower penetration
  3. A lower surface dose and greater depth penetration Correct answer.
  4. Identical depth dose characteristics

Explanation

Higher photon energy reduces surface dose (deeper build-up) and increases penetrability, shifting dmax deeper. Penumbra is influenced by energy and lateral scatter but is not uniformly smaller at all depths.

Question 21Medical Physics for Radiation Oncology

In monitor unit calculations for a photon field, the tissue phantom ratio (TPR) or TMR is used together with the output factor and inverse square factor to:

  1. Convert CT numbers to electron density
  2. Determine the beam's flatness specification
  3. Account for depth-dependent attenuation and scatter along the central axis Correct answer.
  4. Correct for patient setup errors

Explanation

MU calculations combine machine output, field-size factor, depth factor (TPR/TMR or PDD), and inverse square to determine the dose at a point. These factors account for attenuation and scatter with depth.

Question 22Medical Physics for Radiation Oncology

Which of the following best describes the role of a flattening filter in a conventional linac photon beam?

  1. It converts electrons into photons
  2. It produces a relatively uniform lateral dose profile across the field Correct answer.
  3. It increases the beam energy to therapeutic levels
  4. It removes all scattered photons from the beam

Explanation

The flattening filter attenuates the forward-peaked bremsstrahlung to create a more uniform lateral profile. Its removal (flattening-filter-free beams) yields higher dose rates but a peaked profile.

Question 23Medical Physics for Radiation Oncology

A dose distribution is calculated on a CT data set with a 2 mm slice thickness. Which parameter most directly affects the accuracy of dose calculation at tissue interfaces such as lung?

  1. The linac's photon flatness specification
  2. The gantry rotation speed
  3. The number of available electron beams
  4. The accuracy of the CT-number-to-density conversion and heterogeneity correction algorithm Correct answer.

Explanation

Heterogeneity-corrected algorithms rely on accurate conversion of CT numbers to electron density, especially at low-density interfaces like lung. Errors in this conversion propagate into dose errors.

Question 24Medical Physics for Radiation Oncology

In quality assurance, the purpose of a daily output constancy check on a linac is to:

  1. Verify that machine output has not drifted beyond an acceptable tolerance from its baseline Correct answer.
  2. Measure the beam's penumbra precisely
  3. Determine the patient-specific MU
  4. Recalibrate the absolute dose to water each morning

Explanation

Daily constancy checks confirm output stability relative to a baseline established by absolute calibration. Absolute recalibration is performed less frequently (e.g., annually) with a reference chamber.

Question 25Medical Physics for Radiation Oncology

For a photon beam, the penumbra region at the field edge is primarily caused by:

  1. Finite source size and lateral scatter of photons and electrons Correct answer.
  2. The inverse square law alone
  3. The flattening filter's central attenuation
  4. Complete absence of scatter outside the field

Explanation

Penumbra arises from the finite focal spot size (geometric penumbra) and from lateral scatter of photons and secondary electrons. Reducing source size and using collimators narrows the penumbra.

Question 26Medical Physics for Radiation Oncology

The absorbed dose to water is defined as the energy imparted per unit mass, with the SI unit being:

  1. joule per kilogram (gray) Correct answer.
  2. joule per coulomb (sievert)
  3. coulomb per kilogram (roentgen)
  4. electron-volt per meter

Explanation

Absorbed dose is energy imparted per unit mass, and 1 J/kg equals 1 gray. The gray is the SI unit for absorbed dose.

Question 27Medical Physics for Radiation Oncology

In a linear accelerator, the function of the bending magnet is to:

  1. accelerate electrons to therapeutic energies
  2. deflect the electron beam so that it exits along the treatment axis Correct answer.
  3. flatten the photon beam profile
  4. monitor the dose rate during treatment

Explanation

The bending magnet changes the direction of the accelerated electron beam so it points toward the target and exits the gantry along the treatment axis. Acceleration occurs in the waveguide, not the bending magnet.

Question 28Medical Physics for Radiation Oncology

For a photon beam, the dose at the surface relative to the dose at depth is increased primarily by:

  1. placing a bolus material on the patient's skin Correct answer.
  2. using a higher photon energy
  3. increasing the field size
  4. increasing the source-to-surface distance

Explanation

Bolus material builds up dose at the surface by providing scattering material and reducing the depth of maximum dose. Higher photon energies reduce surface dose, and SSD and field size have smaller effects.

Question 29Medical Physics for Radiation Oncology

The inverse square law states that the dose rate from a point source is:

  1. independent of distance
  2. directly proportional to the square of the distance
  3. inversely proportional to the square of the distance Correct answer.
  4. inversely proportional to the distance

Explanation

The inverse square law describes the decrease in dose rate with distance from a point source as 1/d². This applies to photons and follows from geometric spreading.

Question 30Medical Physics for Radiation Oncology

In a photon beam, the build-up region is characterized by:

  1. a gradual increase in dose with depth until electronic equilibrium is reached Correct answer.
  2. a sharp decrease in dose at the surface due to backscatter
  3. a constant dose with depth
  4. a gradual decrease in dose with depth due to attenuation

Explanation

In the build-up region, dose increases with depth as secondary electrons are produced and electronic equilibrium is established. The depth of maximum dose (dmax) marks the end of build-up.

Question 31Medical Physics for Radiation Oncology

The main purpose of a multileaf collimator (MLC) in a linear accelerator is to:

  1. flatten the photon beam profile
  2. accelerate the electron beam
  3. monitor the delivered dose
  4. shape the radiation field to conform to the target volume Correct answer.

Explanation

The MLC consists of computer-controlled leaves that can be positioned to create irregular field shapes, allowing conformal treatment. It is used for field shaping, not for beam flattening or dose monitoring.

Question 32Medical Physics for Radiation Oncology

The tissue-air ratio (TAR) is defined as the ratio of:

  1. dose at a given depth in tissue to dose at dmax in tissue
  2. dose at dmax in tissue to dose at the same point in air
  3. dose at a given depth in tissue to dose at the same point in air Correct answer.
  4. dose in air to dose at the same point in tissue

Explanation

TAR is the ratio of absorbed dose at a given depth in a phantom to the absorbed dose at the same point in air for the same field size. It is used in isocentric calculations.

Question 33Medical Physics for Radiation Oncology

The primary function of the ion chamber in a linear accelerator is to:

  1. shape the radiation field
  2. flatten the photon beam
  3. monitor the delivered dose and terminate the beam when the preset monitor units are reached Correct answer.
  4. measure the energy of the electron beam

Explanation

Ion chambers in the monitor chamber measure the charge produced by radiation and are used to monitor dose and control beam termination. They are essential for accurate dose delivery.

Question 34Medical Physics for Radiation Oncology

A photon beam with a field size of 10x10 cm² at 100 cm SSD has a PDD of 65% at 10 cm depth. If the field size is increased to 20x20 cm², the PDD at 10 cm depth will most likely:

  1. decrease initially then increase
  2. remain unchanged
  3. decrease
  4. increase Correct answer.

Explanation

Larger field sizes produce more scattered radiation, which increases the dose at depth relative to dmax, thereby increasing PDD. This is a well-known effect in photon dosimetry.

Question 35Medical Physics for Radiation Oncology

A 6 MV photon beam has a TMR of 0.78 at 10 cm depth for a 10x10 cm² field. If the dose at dmax for this field is 200 cGy, what is the dose at 10 cm depth?

  1. 200 cGy
  2. 178 cGy
  3. 256 cGy
  4. 156 cGy Correct answer.

Explanation

TMR is the ratio of dose at depth to dose at dmax for the same field size. Thus, dose at 10 cm = 200 cGy × 0.78 = 156 cGy.

Question 36Medical Physics for Radiation Oncology

An electron beam with a nominal energy of 12 MeV has a practical range (R90) of approximately 4.8 cm in water. What is the approximate depth of the 90% isodose?

  1. 4.8 cm Correct answer.
  2. 6.4 cm
  3. 3.2 cm
  4. 1.6 cm

Explanation

R90 is defined as the depth at which the dose falls to 90% of the maximum dose. Therefore, the depth of the 90% isodose is approximately equal to R90, which is 4.8 cm.

Question 37Medical Physics for Radiation Oncology

In a 6 MV photon beam, the dose at 5 cm depth is 80% of the dose at dmax for a 10x10 cm² field at 100 cm SSD. What is the PDD at 5 cm depth?

  1. 80% Correct answer.
  2. 100%
  3. 20%
  4. 125%

Explanation

PDD is defined as the dose at depth expressed as a percentage of the dose at dmax. Given the dose at 5 cm is 80% of dmax, the PDD is 80%.

Question 38Medical Physics for Radiation Oncology

A treatment plan uses a 15 MV photon beam. Compared with a 6 MV beam, the 15 MV beam will have:

  1. a higher surface dose and a greater depth of maximum dose
  2. a higher surface dose and a shallower depth of maximum dose
  3. a lower surface dose and a greater depth of maximum dose Correct answer.
  4. a lower surface dose and a shallower depth of maximum dose

Explanation

Higher-energy photon beams have greater penetration, so dmax occurs deeper, and the surface dose is lower because of increased forward scatter and reduced backscatter. This is a standard characteristic of megavoltage beams.

Question 39Medical Physics for Radiation Oncology

In a CT image, the Hounsfield unit (HU) of a voxel is calculated using the linear attenuation coefficient of the voxel relative to that of water. If a tissue has a linear attenuation coefficient twice that of water, its HU is approximately:

  1. 1000 Correct answer.
  2. 500
  3. 0
  4. 2000

Explanation

HU is defined as 1000 × (μ_tissue - μ_water)/μ_water. If μ_tissue = 2 μ_water, then HU = 1000 × (2-1)/1 = 1000.

Question 40Medical Physics for Radiation Oncology

The main reason for using a flattening filter in a linear accelerator photon beam is to:

  1. produce a uniform dose distribution across the treatment field Correct answer.
  2. increase the dose rate at the isocenter
  3. harden the beam by removing low-energy photons
  4. reduce the penumbra at the field edges

Explanation

The flattening filter attenuates the central part of the beam more than the periphery, resulting in a flatter beam profile across the field. This improves dose uniformity.

Question 41Medical Physics for Radiation Oncology

A wedged photon field has a wedge factor of 0.70. If the prescribed dose at a point requires 100 cGy from the open beam, what is the dose at the same point with the wedge in place, assuming the same monitor units?

  1. 100 cGy
  2. 70 cGy Correct answer.
  3. 30 cGy
  4. 143 cGy

Explanation

The wedge factor is the ratio of dose with the wedge to dose without the wedge for the same monitor units. Thus, dose with wedge = 100 cGy × 0.70 = 70 cGy.

Question 42Medical Physics for Radiation Oncology

The main advantage of using a flattening filter free (FFF) photon beam is:

  1. lower penumbra
  2. better beam uniformity
  3. higher surface dose
  4. higher dose rate Correct answer.

Explanation

FFF beams remove the flattening filter, allowing higher dose rates because of reduced attenuation. This is beneficial for treatments like SBRT where high dose rates are desired.

Question 43Medical Physics for Radiation Oncology

In monitor unit calculations for a photon beam, the output factor is used to account for:

  1. the variation of output with energy
  2. the variation of output with SSD
  3. the variation of output with field size Correct answer.
  4. the variation of output with depth

Explanation

The output factor (or Scp) corrects for the change in dose rate at dmax as the field size changes. It is the ratio of dose at dmax for a given field size to that for the reference field size.

Question 44Medical Physics for Radiation Oncology

The penumbra of a photon beam is primarily caused by:

  1. the scattering of photons in the collimator
  2. the energy spectrum of the beam
  3. the finite size of the radiation source Correct answer.
  4. the attenuation of the beam in the patient

Explanation

Penumbra arises because the source has a finite size, so photons from different parts of the source project different edges at the field boundary. This creates a region of dose gradient.

Question 45Medical Physics for Radiation Oncology

A patient is treated with a 6 MV photon beam using a 10x10 cm² field at 100 cm SSD. The prescribed dose is 200 cGy at 5 cm depth. If the PDD at 5 cm is 85% and the output at dmax is 1 cGy/MU, what is the required monitor units? (Assume no other factors.)

  1. 170 MU
  2. 235 MU Correct answer.
  3. 250 MU
  4. 200 MU

Explanation

MU = prescribed dose / (output × PDD). Thus, MU = 200 / (1 × 0.85) = 235.3 MU, approximately 235 MU.

Question 46Medical Physics for Radiation Oncology

In a 3D conformal treatment plan, the dose distribution is calculated on a CT dataset. Which of the following most directly affects the accuracy of the dose calculation in the build-up region?

  1. The CT scanner's kVp setting
  2. The CT slice thickness Correct answer.
  3. The number of CT slices
  4. The CT reconstruction kernel

Explanation

Slice thickness determines the spatial resolution along the z-axis and affects the accurate modeling of surface curvature and tissue inhomogeneities in the build-up region. Thinner slices improve accuracy.

Question 47Medical Physics for Radiation Oncology

The main purpose of a daily quality assurance check of linac output is to:

  1. measure the beam profile
  2. verify that the output is within a specified tolerance of the baseline value Correct answer.
  3. calibrate the beam energy
  4. check the mechanical isocenter

Explanation

Daily output constancy checks ensure that the dose delivered per monitor unit remains consistent with the calibrated value. They are a quick check to detect significant deviations.

Question 48Medical Physics for Radiation Oncology

A 10 MV photon beam has a mass attenuation coefficient of 0.02 cm²/g in water. If the density of water is 1 g/cm³, what is the linear attenuation coefficient?

  1. 0.002 cm⁻¹
  2. 0.02 cm⁻¹ Correct answer.
  3. 2 cm⁻¹
  4. 0.2 cm⁻¹

Explanation

The linear attenuation coefficient (μ) is the product of the mass attenuation coefficient and the density: μ = (μ/ρ) × ρ = 0.02 cm²/g × 1 g/cm³ = 0.02 cm⁻¹.

Question 49Medical Physics for Radiation Oncology

In a photon beam, the dose at a depth of 10 cm in a patient is 50 cGy. If the tissue maximum ratio (TMR) at 10 cm is 0.80, what is the dose at dmax for the same field?

  1. 80 cGy
  2. 50 cGy
  3. 62.5 cGy Correct answer.
  4. 40 cGy

Explanation

TMR = dose at depth / dose at dmax. Rearranging gives dose at dmax = dose at depth / TMR = 50 / 0.80 = 62.5 cGy.

Question 50Medical Physics for Radiation Oncology

Which of the following detectors is most suitable for measuring the dose in a small photon field (e.g., 1x1 cm²) due to its small sensitive volume and minimal volume averaging?

  1. Radiochromic film
  2. Thermoluminescent dosimeter (TLD)
  3. Diamond detector Correct answer.
  4. Farmer-type ionization chamber

Explanation

Diamond detectors have a small sensitive volume and near-tissue equivalence, making them suitable for small-field dosimetry with minimal volume averaging. Farmer chambers have larger volumes and suffer from volume averaging.