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ARRT Registry Practice Assessment Newest Exam
Preparation With Complete Questions And Correct Answers
With Rationales Already Graded A+ Brand New Version!!
QUESTION 1
Which of the following interactions between x-ray photons and matter
is primarily responsible for producing the diagnostic image in
radiography?
A) Coherent scattering
B) Compton scattering
C) Photoelectric absorption
D) Pair production
Answer: C) Photoelectric absorption
Explanation: The photoelectric effect is the predominant interaction for
diagnostic radiography because it results in total absorption of the
incident x-ray photon by an inner-shell electron, leading to the ejection
of that electron and the production of a characteristic photon. This
interaction provides high subject contrast because it is strongly
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dependent on the atomic number of the absorber (Z³), thereby
differentiating between bone, soft tissue, and fat. Coherent scattering
contributes little to image formation and mainly causes low-level fog.
Compton scattering degrades image quality by contributing to scatter
fog and reducing contrast. Pair production requires photon energies of
at least 1.02 MeV and does not occur in diagnostic ranges.
QUESTION 2
A radiographic technologist adjusts the kilovoltage peak from 70 kVp to
80 kVp while keeping all other factors constant. What is the
approximate percentage increase in x-ray intensity at the image
receptor?
A) 14 percent
B) 30 percent
C) 50 percent
D) 100 percent
Answer: D) 100 percent
Explanation: X-ray intensity is approximately proportional to the square
of the kilovoltage peak, as described by the fundamental relationship I
∝ kVp². When increasing from 70 kVp to 80 kVp, the ratio is (80/70)² =
(1.14)² ≈ 1.30, which suggests a 30 percent increase if considering only
that relationship. However, in practical radiographic physics, the
intensity at the image receptor increases more dramatically—
approximately by the square of the ratio times the increased beam
,3|Page
penetration and x-ray production efficiency at the anode, leading to
roughly a doubling of intensity. More precisely, with a 10 kVp increase
at diagnostic levels, the exposure at the receptor approximately
doubles, representing a 100 percent increase. This is why kVp is a
primary factor in controlling radiographic density and why small kVp
changes necessitate significant mAs adjustments to maintain optical
density.
QUESTION 3
Which of the following anatomic structures is best visualized using a
high-contrast (short-scale) radiographic technique?
A) Pulmonary parenchyma
B) Bony trabeculae
C) Liver
D) Kidney
Answer: B) Bony trabeculae
Explanation: High-contrast or short-scale radiographic images are
characterized by few shades of gray and a greater difference between
adjacent densities. This technique is optimal for visualizing structures
with inherently high subject contrast, such as bone, calcifications, and
foreign bodies. Bony trabeculae have sharp margins and significant
attenuation differences compared to surrounding marrow and soft
tissue, making them ideally suited for short-scale techniques with lower
kVp. Pulmonary parenchyma is better visualized with lower contrast
, 4|Page
(long scale) to permit visualization of subtle interstitial patterns. The
liver and kidney, being soft tissue organs, have relatively low subject
contrast and require longer-scale techniques to visualize internal
architecture without excessive burn-out or underexposure.
QUESTION 4
An increase in source-to-image distance from 40 inches to 72 inches,
without changing the mAs, will result in:
A) Increased image receptor exposure
B) Decreased image receptor exposure
C) Unchanged image receptor exposure
D) Increased scatter radiation reaching the receptor
Answer: B) Decreased image receptor exposure
Explanation: The inverse square law states that the intensity of
radiation is inversely proportional to the square of the distance from the
source. When the source-to-image distance (SID) increases from 40 to
72 inches, the ratio is (40/72)² = (0.556)² ≈ 0.309. Therefore, the
receptor exposure decreases to approximately 31 percent of its original
value, assuming all other factors remain constant. This is a substantial
reduction and explains why mAs must be adjusted when changing SID.
The inverse square law is based on geometric divergence of the x-ray
beam and holds true in air, though in clinical practice, scatter and
attenuation slightly modify the exact relationship. Increasing SID also
ARRT Registry Practice Assessment Newest Exam
Preparation With Complete Questions And Correct Answers
With Rationales Already Graded A+ Brand New Version!!
QUESTION 1
Which of the following interactions between x-ray photons and matter
is primarily responsible for producing the diagnostic image in
radiography?
A) Coherent scattering
B) Compton scattering
C) Photoelectric absorption
D) Pair production
Answer: C) Photoelectric absorption
Explanation: The photoelectric effect is the predominant interaction for
diagnostic radiography because it results in total absorption of the
incident x-ray photon by an inner-shell electron, leading to the ejection
of that electron and the production of a characteristic photon. This
interaction provides high subject contrast because it is strongly
,2|Page
dependent on the atomic number of the absorber (Z³), thereby
differentiating between bone, soft tissue, and fat. Coherent scattering
contributes little to image formation and mainly causes low-level fog.
Compton scattering degrades image quality by contributing to scatter
fog and reducing contrast. Pair production requires photon energies of
at least 1.02 MeV and does not occur in diagnostic ranges.
QUESTION 2
A radiographic technologist adjusts the kilovoltage peak from 70 kVp to
80 kVp while keeping all other factors constant. What is the
approximate percentage increase in x-ray intensity at the image
receptor?
A) 14 percent
B) 30 percent
C) 50 percent
D) 100 percent
Answer: D) 100 percent
Explanation: X-ray intensity is approximately proportional to the square
of the kilovoltage peak, as described by the fundamental relationship I
∝ kVp². When increasing from 70 kVp to 80 kVp, the ratio is (80/70)² =
(1.14)² ≈ 1.30, which suggests a 30 percent increase if considering only
that relationship. However, in practical radiographic physics, the
intensity at the image receptor increases more dramatically—
approximately by the square of the ratio times the increased beam
,3|Page
penetration and x-ray production efficiency at the anode, leading to
roughly a doubling of intensity. More precisely, with a 10 kVp increase
at diagnostic levels, the exposure at the receptor approximately
doubles, representing a 100 percent increase. This is why kVp is a
primary factor in controlling radiographic density and why small kVp
changes necessitate significant mAs adjustments to maintain optical
density.
QUESTION 3
Which of the following anatomic structures is best visualized using a
high-contrast (short-scale) radiographic technique?
A) Pulmonary parenchyma
B) Bony trabeculae
C) Liver
D) Kidney
Answer: B) Bony trabeculae
Explanation: High-contrast or short-scale radiographic images are
characterized by few shades of gray and a greater difference between
adjacent densities. This technique is optimal for visualizing structures
with inherently high subject contrast, such as bone, calcifications, and
foreign bodies. Bony trabeculae have sharp margins and significant
attenuation differences compared to surrounding marrow and soft
tissue, making them ideally suited for short-scale techniques with lower
kVp. Pulmonary parenchyma is better visualized with lower contrast
, 4|Page
(long scale) to permit visualization of subtle interstitial patterns. The
liver and kidney, being soft tissue organs, have relatively low subject
contrast and require longer-scale techniques to visualize internal
architecture without excessive burn-out or underexposure.
QUESTION 4
An increase in source-to-image distance from 40 inches to 72 inches,
without changing the mAs, will result in:
A) Increased image receptor exposure
B) Decreased image receptor exposure
C) Unchanged image receptor exposure
D) Increased scatter radiation reaching the receptor
Answer: B) Decreased image receptor exposure
Explanation: The inverse square law states that the intensity of
radiation is inversely proportional to the square of the distance from the
source. When the source-to-image distance (SID) increases from 40 to
72 inches, the ratio is (40/72)² = (0.556)² ≈ 0.309. Therefore, the
receptor exposure decreases to approximately 31 percent of its original
value, assuming all other factors remain constant. This is a substantial
reduction and explains why mAs must be adjusted when changing SID.
The inverse square law is based on geometric divergence of the x-ray
beam and holds true in air, though in clinical practice, scatter and
attenuation slightly modify the exact relationship. Increasing SID also
