ASNT IRRSP CERTIFICATION PRACTICE EXAM 2026/2027 |
100 Questions & Correct Answers | Industrial Radiography
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Section 1: Radiation Fundamentals & Physics (Questions 1-15)
Q1. A radiographer is working with an Ir-192 source. If the initial activity is 100 Ci,
what is the approximate activity after 148 days?
A. 75 Ci
B. 50 Ci
C. 25 Ci
D. 12.5 Ci
Correct Answer: C. 25 Ci [CORRECT]
Rationale: Ir-192 has a half-life of approximately 74 days. After one half-life (74 days),
100 Ci decays to 50 Ci. After two half-lives (148 days), 50 Ci decays to 25 Ci. The
number of half-lives method is the simplest approach: N = N₀ × (1/2)ⁿ where n =
148/74 = 2, so N = 100 × (1/2)² = 25 Ci. Distractor A (75 Ci) incorrectly assumes linear
decay. Distractor B (50 Ci) represents activity after only one half-life (74 days).
Distractor D (12.5 Ci) represents three half-lives (222 days). The IRRSP exam frequently
tests half-life calculations using this exact method per ANSI/HPS N43.6 and ASNT
training materials.
Q2. Which photon interaction with matter is most dominant for Ir-192 gamma rays
(average 380 keV) in lead shielding?
A. Photoelectric effect
B. Compton scattering
,C. Pair production
D. Coherent (Rayleigh) scattering
Correct Answer: B. Compton scattering [CORRECT]
Rationale: For photon energies between approximately 100 keV and 10 MeV, Compton
scattering is the dominant interaction mechanism. Ir-192 emits gamma rays with an
average energy of 380 keV, which falls squarely within this range. The photoelectric
effect (A) dominates at lower energies (<100 keV) and is strongly dependent on atomic
number (Z³). Pair production (C) requires photon energies above 1.022 MeV, which
exceeds Ir-192's maximum emission energy. Coherent scattering (D) is negligible at
these energies. For IRRSP exam purposes, remember: low energy = photoelectric
(diagnostic X-rays), medium energy = Compton (industrial radiography), high energy
(>1.022 MeV) = pair production.
Q3. Using the inverse square law, if the exposure rate at 2 meters from a Co-60 source is
100 mR/hr, what is the exposure rate at 4 meters?
A. 50 mR/hr
B. 25 mR/hr
C. 12.5 mR/hr
D. 200 mR/hr
Correct Answer: B. 25 mR/hr [CORRECT]
Rationale: The inverse square law states I₁/I₂ = D₂²/D₁² or I₂ = I₁ × (D₁²/D₂²). Here, I₂
= 100 mR/hr × (2²/4²) = 100 × (4/16) = 100 × 0.25 = 25 mR/hr. Doubling the distance
from 2m to 4m reduces the exposure rate to one-fourth. Distractor A (50 mR/hr)
incorrectly uses a linear inverse relationship (I₁/I₂ = D₂/D₁) instead of the square
relationship. Distractor C (12.5 mR/hr) incorrectly applies the square to the wrong side
of the equation. Distractor D (200 mR/hr) inverts the relationship entirely. The inverse
square law is fundamental to ALARA distance calculations and appears on virtually
every IRRSP exam.
,Q4. What is the approximate gamma constant (Γ) for Co-60?
A. 0.48 R·m²/hr·Ci
B. 1.3 R·m²/hr·Ci
C. 0.59 R·m²/hr·Ci
D. 2.2 R·m²/hr·Ci
Correct Answer: B. 1.3 R·m²/hr·Ci [CORRECT]
Rationale: The gamma constant for Co-60 is 1.3 R·m²/hr·Ci, meaning 1 Curie of Co-60
produces 1.3 Roentgen per hour at 1 meter. This is significantly higher than Ir-192's
gamma constant of 0.48 R·m²/hr·Ci (distractor A) due to Co-60's higher energy gamma
emissions (1.17 and 1.33 MeV). Distractor C (0.59) is the gamma constant for Cs-137.
Distractor D (2.2) is not a standard gamma constant for common industrial radiography
sources. For the IRRSP exam, memorize: Ir-192 = 0.48, Co-60 = 1.3, Cs-137 = 0.59
R·m²/hr·Ci.
Q5. A radiographer needs to reduce the exposure rate from 80 mR/hr to 10 mR/hr
using lead shielding. If the HVL of lead for this energy is 4 mm, how many HVLs are
required?
A. 2 HVLs
B. 3 HVLs
C. 4 HVLs
D. 5 HVLs
Correct Answer: B. 3 HVLs [CORRECT]
Rationale: Each HVL reduces exposure by 50%. After 1 HVL: 80 → 40 mR/hr. After 2
HVLs: 40 → 20 mR/hr. After 3 HVLs: 20 → 10 mR/hr. Alternatively, using the formula: I
= I₀ × (1/2)ⁿ, solving 10 = 80 × (1/2)ⁿ gives n = 3. Distractor A (2 HVLs) would reduce to
20 mR/hr. Distractor C (4 HVLs) would reduce to 5 mR/hr. Distractor D (5 HVLs) would
reduce to 2.5 mR/hr. Remember: 1 HVL = 50% reduction, 1 TVL = 90% reduction
(approximately 3.32 HVLs, NOT 2 HVLs—a common IRRSP trap).
, Q6. What is the relationship between the half-value layer (HVL) and tenth-value layer
(TVL)?
A. TVL = 2 × HVL
B. TVL = 3.32 × HVL
C. TVL = 5 × HVL
D. TVL = 10 × HVL
Correct Answer: B. TVL = 3.32 × HVL [CORRECT]
Rationale: The TVL reduces radiation exposure by 90% (factor of 10), while the HVL
reduces it by 50% (factor of 2). Mathematically: (1/2)ⁿ = 1/10, solving for n gives n =
log(10)/log(2) ≈ 3.32. Therefore, TVL ≈ 3.32 × HVL. Distractor A (2 × HVL) is a common
misconception—2 HVLs only reduce exposure to 25%, not 10%. Distractor C (5 × HVL)
would reduce exposure to about 3.1%. Distractor D (10 × HVL) has no physical basis.
This relationship is critical for shielding calculations in industrial radiography.
Q7. Which type of radioactive decay emits a helium nucleus (2 protons + 2 neutrons)?
A. Beta decay
B. Gamma emission
C. Alpha decay
D. Neutron emission
Correct Answer: C. Alpha decay [CORRECT]
Rationale: Alpha decay emits an alpha particle, which consists of 2 protons and 2
neutrons—identical to a helium-4 nucleus. Alpha particles have low penetration
(stopped by paper or skin) but high ionization potential. Beta decay (A) emits an
electron or positron. Gamma emission (B) emits high-energy photons with no mass or
charge. Neutron emission (D) releases free neutrons, which are uncharged and highly
penetrating. While alpha emitters are not used in industrial radiography, understanding
decay modes is required per 10 CFR 34 training requirements.
100 Questions & Correct Answers | Industrial Radiography
Radiation Safety Personnel | Latest Brand New | Pass
Guaranteed - A+ Graded
Section 1: Radiation Fundamentals & Physics (Questions 1-15)
Q1. A radiographer is working with an Ir-192 source. If the initial activity is 100 Ci,
what is the approximate activity after 148 days?
A. 75 Ci
B. 50 Ci
C. 25 Ci
D. 12.5 Ci
Correct Answer: C. 25 Ci [CORRECT]
Rationale: Ir-192 has a half-life of approximately 74 days. After one half-life (74 days),
100 Ci decays to 50 Ci. After two half-lives (148 days), 50 Ci decays to 25 Ci. The
number of half-lives method is the simplest approach: N = N₀ × (1/2)ⁿ where n =
148/74 = 2, so N = 100 × (1/2)² = 25 Ci. Distractor A (75 Ci) incorrectly assumes linear
decay. Distractor B (50 Ci) represents activity after only one half-life (74 days).
Distractor D (12.5 Ci) represents three half-lives (222 days). The IRRSP exam frequently
tests half-life calculations using this exact method per ANSI/HPS N43.6 and ASNT
training materials.
Q2. Which photon interaction with matter is most dominant for Ir-192 gamma rays
(average 380 keV) in lead shielding?
A. Photoelectric effect
B. Compton scattering
,C. Pair production
D. Coherent (Rayleigh) scattering
Correct Answer: B. Compton scattering [CORRECT]
Rationale: For photon energies between approximately 100 keV and 10 MeV, Compton
scattering is the dominant interaction mechanism. Ir-192 emits gamma rays with an
average energy of 380 keV, which falls squarely within this range. The photoelectric
effect (A) dominates at lower energies (<100 keV) and is strongly dependent on atomic
number (Z³). Pair production (C) requires photon energies above 1.022 MeV, which
exceeds Ir-192's maximum emission energy. Coherent scattering (D) is negligible at
these energies. For IRRSP exam purposes, remember: low energy = photoelectric
(diagnostic X-rays), medium energy = Compton (industrial radiography), high energy
(>1.022 MeV) = pair production.
Q3. Using the inverse square law, if the exposure rate at 2 meters from a Co-60 source is
100 mR/hr, what is the exposure rate at 4 meters?
A. 50 mR/hr
B. 25 mR/hr
C. 12.5 mR/hr
D. 200 mR/hr
Correct Answer: B. 25 mR/hr [CORRECT]
Rationale: The inverse square law states I₁/I₂ = D₂²/D₁² or I₂ = I₁ × (D₁²/D₂²). Here, I₂
= 100 mR/hr × (2²/4²) = 100 × (4/16) = 100 × 0.25 = 25 mR/hr. Doubling the distance
from 2m to 4m reduces the exposure rate to one-fourth. Distractor A (50 mR/hr)
incorrectly uses a linear inverse relationship (I₁/I₂ = D₂/D₁) instead of the square
relationship. Distractor C (12.5 mR/hr) incorrectly applies the square to the wrong side
of the equation. Distractor D (200 mR/hr) inverts the relationship entirely. The inverse
square law is fundamental to ALARA distance calculations and appears on virtually
every IRRSP exam.
,Q4. What is the approximate gamma constant (Γ) for Co-60?
A. 0.48 R·m²/hr·Ci
B. 1.3 R·m²/hr·Ci
C. 0.59 R·m²/hr·Ci
D. 2.2 R·m²/hr·Ci
Correct Answer: B. 1.3 R·m²/hr·Ci [CORRECT]
Rationale: The gamma constant for Co-60 is 1.3 R·m²/hr·Ci, meaning 1 Curie of Co-60
produces 1.3 Roentgen per hour at 1 meter. This is significantly higher than Ir-192's
gamma constant of 0.48 R·m²/hr·Ci (distractor A) due to Co-60's higher energy gamma
emissions (1.17 and 1.33 MeV). Distractor C (0.59) is the gamma constant for Cs-137.
Distractor D (2.2) is not a standard gamma constant for common industrial radiography
sources. For the IRRSP exam, memorize: Ir-192 = 0.48, Co-60 = 1.3, Cs-137 = 0.59
R·m²/hr·Ci.
Q5. A radiographer needs to reduce the exposure rate from 80 mR/hr to 10 mR/hr
using lead shielding. If the HVL of lead for this energy is 4 mm, how many HVLs are
required?
A. 2 HVLs
B. 3 HVLs
C. 4 HVLs
D. 5 HVLs
Correct Answer: B. 3 HVLs [CORRECT]
Rationale: Each HVL reduces exposure by 50%. After 1 HVL: 80 → 40 mR/hr. After 2
HVLs: 40 → 20 mR/hr. After 3 HVLs: 20 → 10 mR/hr. Alternatively, using the formula: I
= I₀ × (1/2)ⁿ, solving 10 = 80 × (1/2)ⁿ gives n = 3. Distractor A (2 HVLs) would reduce to
20 mR/hr. Distractor C (4 HVLs) would reduce to 5 mR/hr. Distractor D (5 HVLs) would
reduce to 2.5 mR/hr. Remember: 1 HVL = 50% reduction, 1 TVL = 90% reduction
(approximately 3.32 HVLs, NOT 2 HVLs—a common IRRSP trap).
, Q6. What is the relationship between the half-value layer (HVL) and tenth-value layer
(TVL)?
A. TVL = 2 × HVL
B. TVL = 3.32 × HVL
C. TVL = 5 × HVL
D. TVL = 10 × HVL
Correct Answer: B. TVL = 3.32 × HVL [CORRECT]
Rationale: The TVL reduces radiation exposure by 90% (factor of 10), while the HVL
reduces it by 50% (factor of 2). Mathematically: (1/2)ⁿ = 1/10, solving for n gives n =
log(10)/log(2) ≈ 3.32. Therefore, TVL ≈ 3.32 × HVL. Distractor A (2 × HVL) is a common
misconception—2 HVLs only reduce exposure to 25%, not 10%. Distractor C (5 × HVL)
would reduce exposure to about 3.1%. Distractor D (10 × HVL) has no physical basis.
This relationship is critical for shielding calculations in industrial radiography.
Q7. Which type of radioactive decay emits a helium nucleus (2 protons + 2 neutrons)?
A. Beta decay
B. Gamma emission
C. Alpha decay
D. Neutron emission
Correct Answer: C. Alpha decay [CORRECT]
Rationale: Alpha decay emits an alpha particle, which consists of 2 protons and 2
neutrons—identical to a helium-4 nucleus. Alpha particles have low penetration
(stopped by paper or skin) but high ionization potential. Beta decay (A) emits an
electron or positron. Gamma emission (B) emits high-energy photons with no mass or
charge. Neutron emission (D) releases free neutrons, which are uncharged and highly
penetrating. While alpha emitters are not used in industrial radiography, understanding
decay modes is required per 10 CFR 34 training requirements.
