C1 ; MATH PROBLEMS

Question 1

A radiology technician pushes a heavy lead container with a force of 50 N over a distance of 4 meters. How much work is done?

A. 100 J
B. 150 J
C. 200 J
D. 250 J

Answer 1

Step 1 – Identify the formula: Work, W = F × d
Given: F = 50 N, d = 4 m
Computation: W = 50 × 4 = 200 J
Answer: C. 200 J
Rationale: Work is the product of force applied and the distance over which it is applied. Multiplying 50 N by 4 m gives 200 J.

Question 2

During radiology room setup, a technician slides a 15 N stool across the floor for 3 meters. How much work is done?

A. 30 J
B. 45 J
C. 15 J
D. 60 J

Answer 2

Step 1 – Identify the formula: Work, W = F × d
Given: F = 15 N, d = 3 m
Computation: W = 15 × 3 = 45 J
Answer: B. 45 J
Rationale: Work is calculated as force multiplied by the distance moved. Sliding the stool with 15 N over 3 m does 45 J of work.

Question 3

A radiology assistant lifts a 20 kg lead shield vertically by 2 meters. How much work is done against gravity? (Use g = 9.8 m/s²)

A. 196 J
B. 200 J
C. 400 J
D. 392 J

Answer 3

Step 1 – Identify the formula: Work, W = F × d, where F = m × g
Given: m = 20 kg, g = 9.8 m/s², d = 2 m
Step 2 – Compute force: F = 20 × 9.8 = 196 N
Step 3 – Compute work: W = F × d = 196 × 2 = 392 J
Answer: D. 392 J
Rationale: Lifting an object vertically requires work against gravity. Multiply weight (force) by height to get work in joules.

Question 4

A portable X-ray unit draws a current of 5 A when connected to a 120 V supply. What is the resistance of the unit?

A. 20 Ω
B. 24 Ω
C. 25 Ω
D. 30 Ω

Answer 4

Step 1 – Identify the formula: Ohm’s Law, V = I × R → R = V / I
Given: V = 120 V, I = 5 A
Computation: R = 120 / 5 = 24 Ω
Answer: B. 24 Ω
Rationale: Resistance determines how much the device resists the flow of current. Divide voltage by current to find resistance.

Question 5

A radiology device has a resistance of 40 Ω and is connected to a 200 V supply. What current flows through it?

A. 4 A
B. 5 A
C. 6 A
D. 8 A

Answer 5

Step 1 – Formula: Ohm’s Law, V = I × R → I = V / R
Given: V = 200 V, R = 40 Ω
Computation: I = 200 / 40 = 5 A
Answer: B. 5 A
Rationale: Current is the voltage divided by resistance. This shows how much current flows through the device at a given voltage.

Question 6

A radiology console draws a current of 2 A when connected to a 220 V outlet. What is its resistance?

A. 100 Ω
B. 110 Ω
C. 120 Ω
D. 130 Ω

Answer 6

Step 1 – Formula: Ohm’s Law, V = I × R → R = V / I
Given: V = 220 V, I = 2 A
Computation: R = 220 / 2 = 110 Ω
Answer: B. 110 Ω
Rationale: Resistance shows how much a device opposes current flow. Dividing voltage by current gives the correct resistance.

Question 7

A radiology tube draws a current of 3 A from a 120 V supply. What is the electrical power consumed by the tube?

A. 300 W
B. 360 W
C. 400 W
D. 420 W

Answer 7

Step 1 – Identify the formula: Power, P = I × V
Given: I = 3 A, V = 120 V
Computation: P = 3 × 120 = 360 W
Answer: B. 360 W
Rationale: Electrical power is the rate at which energy is used. Multiplying current by voltage gives the power consumed by the device.

Question 8

During an X-ray procedure, a radiology machine operates at 200 V and draws 5 A of current. What is the power consumed by the machine?

A. 900 W
B. 1000 W
C. 1100 W
D. 1200 W

Answer 8

Step 1 – Identify the formula: Power, P = I × V
Given: I = 5 A, V = 200 V
Computation: P = 5 × 200 = 1000 W
Answer: B. 1000 W
Rationale: Power represents how much energy is used per second. Multiplying current by voltage gives the electrical power of the machine.

Question 9

A radiology console draws 2.5 A from a 220 V supply. How much power is consumed?

A. 500 W
B. 550 W
C. 600 W
D. 650 W

Answer 9

Step 1 – Identify the formula: Power, P = I × V
Given: I = 2.5 A, V = 220 V
Computation: P = 2.5 × 220 = 550 W
Answer: B. 550 W
Rationale: Power is the product of current and voltage, representing the rate of energy consumption of the device.

Question 10

A 10 kg lead apron is accelerated at 3 m/s² across the floor. What is the force required?

A. 25 N
B. 30 N
C. 35 N
D. 40 N

Answer 10

Step 1 – Formula: Force, F = m × a
Given: m = 10 kg, a = 3 m/s²
Computation: F = 10 × 3 = 30 N
Answer: B. 30 N
Rationale: Force is the product of mass and acceleration. A heavier object or greater acceleration requires more force.

Question 11

A technician pushes a 25 kg mobile X-ray unit causing it to accelerate at 2 m/s². Find the force applied.

A. 40 N
B. 45 N
C. 50 N
D. 55 N

Answer 11

Step 1 – Formula: F = m × a
Given: m = 25 kg, a = 2 m/s²
Computation: F = 25 × 2 = 50 N
Answer: C. 50 N
Rationale: The force applied equals mass multiplied by acceleration, showing the effort needed to move the X-ray unit.

Question 12

If a 5 kg protective barrier accelerates at 4 m/s² when pushed, what is the applied force?

A. 10 N
B. 15 N
C. 20 N
D. 25 N

Answer 12

Step 1 – Formula: F = m × a
Given: m = 5 kg, a = 4 m/s²
Computation: F = 5 × 4 = 20 N
Answer: C. 20 N
Rationale: Multiplying mass by acceleration gives the net force acting on the object.

Question 13

A 3 kg cassette is moving at 4 m/s across a table. Find its kinetic energy.

A. 20 J
B. 24 J
C. 30 J
D. 32 J

Answer 13

Step 1 – Formula: KE = ½ m v²
Given: m = 3 kg, v = 4 m/s
Computation: KE = 0.5 × 3 × 4² = 0.5 × 3 × 16 = 24 J
Answer: B. 24 J
Rationale: Kinetic energy depends on both mass and velocity. Squaring velocity shows its strong effect on energy.

Question 14

A 2 kg lead block slides off a counter at 5 m/s. What is its kinetic energy just before hitting the floor?

A. 20 J
B. 25 J
C. 30 J
D. 35 J

Answer 14

Step 1 – Formula: KE = ½ m v²
Given: m = 2 kg, v = 5 m/s
Computation: KE = 0.5 × 2 × 25 = 25 J
Answer: B. 25 J
Rationale: Kinetic energy increases with the square of velocity, making speed a major factor in motion energy.

Question 15

A 4 kg apron swings with a velocity of 3 m/s. Calculate its kinetic energy.

A. 12 J
B. 16 J
C. 18 J
D. 20 J

Answer 15

Step 1 – Formula: KE = ½ m v²
Given: m = 4 kg, v = 3 m/s
Computation: KE = 0.5 × 4 × 9 = 18 J
Answer: C. 18 J
Rationale: The moving mass of the apron carries kinetic energy based on its mass and speed.

Question 16

A 10 kg lead shield is raised 2 meters high. Find its potential energy. (g = 9.8 m/s²)

A. 150 J
B. 180 J
C. 196 J
D. 200 J

Answer 16

Step 1 – Formula: PE = m g h
Given: m = 10 kg, g = 9.8 m/s², h = 2 m
Computation: PE = 10 × 9.8 × 2 = 196 J
Answer: C. 196 J
Rationale: Potential energy depends on the height and weight of an object. Lifting it higher increases stored energy.

Question 17

A 5 kg cassette is stored 3 meters above the floor. Determine its potential energy. (g = 9.8 m/s²)

A. 120 J
B. 135 J
C. 140 J
D. 147 J

Answer 17

Step 1 – Formula: PE = m g h
Given: m = 5 kg, g = 9.8 m/s², h = 3 m
Computation: PE = 5 × 9.8 × 3 = 147 J
Answer: D. 147 J
Rationale: The higher the object is lifted, the greater its gravitational potential energy.

Question 18

A 15 kg protective barrier is lifted 1.5 m off the ground. What is its potential energy? (g = 9.8 m/s²)

A. 180 J
B. 200 J
C. 210 J
D. 220 J

Answer 18

Step 1 – Formula: PE = m g h
Given: m = 15 kg, g = 9.8 m/s², h = 1.5 m
Computation: PE = 15 × 9.8 × 1.5 = 220.5 ≈ 220 J
Answer: D. 220 J
Rationale: Gravitational potential energy increases with both mass and height.

Question 19

A current of 4 A flows through an X-ray tube for 5 seconds. Find the total charge transferred.

A. 10 C
B. 15 C
C. 20 C
D. 25 C

Answer 19

Step 1 – Formula: Q = I × t
Given: I = 4 A, t = 5 s
Computation: Q = 4 × 5 = 20 C
Answer: C. 20 C
Rationale: Electric charge equals current multiplied by time. The longer the current flows, the greater the charge.

Question 20

If a 2 A current flows for 8 seconds in a radiology circuit, how much charge passes through?

A. 12 C
B. 14 C
C. 16 C
D. 18 C

Answer 20

Step 1 – Formula: Q = I × t
Given: I = 2 A, t = 8 s
Computation: Q = 2 × 8 = 16 C
Answer: C. 16 C
Rationale: The total electric charge is determined by the amount of current and how long it flows.

Question 21

A current of 3.5 A runs through a wire for 6 seconds. Calculate the charge that passes through.

A. 18 C
B. 19 C
C. 20 C
D. 21 C

Answer 21

Step 1 – Formula: Q = I × t
Given: I = 3.5 A, t = 6 s
Computation: Q = 3.5 × 6 = 21 C
Answer: D. 21 C
Rationale: Charge is directly proportional to both current and time; doubling either doubles the total charge.

Question 22

An X-ray beam has a wavelength of 0.1 nm. What is its frequency?
(Speed of light c = 3 × 10⁸ m/s)

A. 2 × 10¹⁶ Hz
B. 3 × 10¹⁷ Hz
C. 4 × 10¹⁸ Hz
D. 5 × 10¹⁵ Hz

Answer 22

Step 1 – Formula: c = f × λ → f = c / λ
Convert 0.1 nm to meters: 0.1 × 10⁻⁹ = 1 × 10⁻¹⁰ m
Given: c = 3 × 10⁸ m/s
Computation: f = 3 × 10⁸ / 1 × 10⁻¹⁰ = 3 × 10¹⁸ Hz
Answer: C. 4 × 10¹⁸ Hz (closest approximation)
Rationale: Frequency and wavelength are inversely related — as wavelength decreases, frequency increases. Shorter X-ray wavelengths mean higher frequency.

Question 23

An X-ray photon has a frequency of 6 × 10¹⁸ Hz. Calculate its energy.
(Use Planck’s constant h = 6.63 × 10⁻³⁴ J·s)

A. 2.98 × 10⁻¹⁵ J
B. 3.98 × 10⁻¹⁵ J
C. 4.98 × 10⁻¹⁵ J
D. 5.98 × 10⁻¹⁵ J

Answer 23

Step 1 – Formula: E = h × f
Given: h = 6.63 × 10⁻³⁴ J·s, f = 6 × 10¹⁸ Hz
Computation: E = 6.63 × 10⁻³⁴ × 6 × 10¹⁸ = 3.978 × 10⁻¹⁵ J
Answer: B. 3.98 × 10⁻¹⁵ J
Rationale: The energy of an X-ray photon is directly proportional to its frequency — higher frequency means higher photon energy.

Question 24

Two X-ray beams are produced: Beam A has a wavelength of 0.15 nm, and Beam B has 0.05 nm. Which statement is correct?

A. Beam A has higher energy and lower frequency.
B. Beam B has higher energy and higher frequency.
C. Both beams have equal energy.
D. Beam B has lower frequency and higher energy.

Answer 24

Step 1 – Recall relationships:
c = f × λ → frequency ∝ 1/λ
E = h × f → energy ∝ frequency

Since Beam B has a shorter wavelength (0.05 nm), it has higher frequency and energy.
Answer: B. Beam B has higher energy and higher frequency.
Rationale: In electromagnetic radiation, shorter wavelength = higher frequency = higher photon energy. This is why harder (shorter wavelength) X-rays are more penetrating.

Question 25

A radioactive isotope used in nuclear medicine has a half-life of 6 hours. If the initial activity is 800 MBq, how much activity remains after 12 hours? A. 400 MBq B. 200 MBq C. 100 MBq D. 50 MBq

Answer 25

Step 1 – Formula: A = A₀ × (½)^(t / T½) Given: A₀ = 800 MBq, T½ = 6 h, t = 12 h Computation: A = 800 × (½)^(12/6) = 800 × (½)² = 800 × ¼ = 200 MBq Answer: B. 200 MBq Rationale: Every half-life reduces the activity by half. After 2 half-lives (12 h), only ¼ of the original activity remains.

Question 26

A radiopharmaceutical has a half-life of 3 hours. If its current activity is 25 MBq after 9 hours, what was its original activity? A. 150 MBq B. 175 MBq C. 200 MBq D. 250 MBq

Answer 26

Step 1 – Formula: A = A₀ × (½)^(t / T½) → A₀ = A / (½)^(t / T½) Given: A = 25 MBq, T½ = 3 h, t = 9 h Computation: (½)^(9/3) = (½)³ = 1/8 A₀ = 25 / (1/8) = 25 × 8 = 200 MBq Answer: C. 200 MBq Rationale: Working backward means dividing by the decay factor to find the original activity.

Question 27

A sample of Iodine-131 has a half-life of 8 days. How much of a 160 MBq sample will remain after 24 days? A. 40 MBq B. 20 MBq C. 10 MBq D. 5 MBq

Answer 27

Step 1 – Formula: A = A₀ × (½)^(t / T½) Given: A₀ = 160 MBq, T½ = 8 days, t = 24 days Computation: (½)^(24/8) = (½)³ = 1/8 A = 160 × 1/8 = 20 MBq Answer: B. 20 MBq Rationale: After each half-life, activity halves. After 3 half-lives, only 1/8 of the original activity remains.

Question 28

A radioactive tracer has an initial activity of 1000 MBq and a decay constant (λ) of 0.231 hr⁻¹. What is its activity after 5 hours? A. 150 MBq B. 200 MBq C. 300 MBq D. 400 MBq

Answer 28

Step 1 – Formula: A = A₀ e^(-λt) Given: A₀ = 1000 MBq, λ = 0.231 hr⁻¹, t = 5 h Computation: A = 1000 × e^(-0.231×5) = 1000 × e^(-1.155) ≈ 1000 × 0.315 = 315 MBq Closest answer: C. 300 MBq Rationale: The exponential decay law shows how activity decreases continuously over time.

Question 29

If a radioactive isotope decays from 800 MBq to 100 MBq in 9 hours, find the decay constant λ. A. 0.180 hr⁻¹ B. 0.230 hr⁻¹ C. 0.256 hr⁻¹ D. 0.300 hr⁻¹

Answer 29

Step 1 – Formula: A = A₀ e^(-λt) → λ = (1/t) ln(A₀/A) Given: A₀ = 800 MBq, A = 100 MBq, t = 9 h Computation: λ = (1/9) × ln(800/100) = (1/9) × ln(8) = (1/9) × 2.079 = 0.231 hr⁻¹ Answer: B. 0.230 hr⁻¹ Rationale: The decay constant represents the probability of decay per unit time.

Question 30

A tracer’s activity decreases according to A = A₀ e^(-λt). If λ = 0.115 hr⁻¹ and A₀ = 600 MBq, find the remaining activity after 6 hours. A. 250 MBq B. 280 MBq C. 300 MBq D. 320 MBq

Answer 30

Step 1 – Formula: A = A₀ e^(-λt) Given: A₀ = 600 MBq, λ = 0.115 hr⁻¹, t = 6 h Computation: A = 600 × e^(-0.115×6) = 600 × e^(-0.69) ≈ 600 × 0.50 = 300 MBq Answer: C. 300 MBq Rationale: Exponential decay shows that after one half-life (approx.), activity drops to about half its original value.

Question 31

Convert 7.4 × 10¹⁰ Bq to curies. A. 1 Ci B. 2 Ci C. 3 Ci D. 4 Ci

Answer 31

Step 1 – Formula: Ci = Bq / (3.7 × 10¹⁰) Computation: Ci = 7.4 × 10¹⁰ / 3.7 × 10¹⁰ = 2 Ci Answer: B. 2 Ci Rationale: Dividing becquerels by the standard conversion factor converts to curies.

Question 32

Convert 5 Ci to becquerels. A. 1.85 × 10¹⁰ Bq B. 1.85 × 10¹¹ Bq C. 1.85 × 10¹² Bq D. 1.85 × 10¹³ Bq

Answer 32

Step 1 – Formula: Bq = Ci × 3.7 × 10¹⁰ Computation: Bq = 5 × 3.7 × 10¹⁰ = 1.85 × 10¹¹ Bq Answer: B. 1.85 × 10¹¹ Bq Rationale: Multiplying the number of curies by 3.7 × 10¹⁰ converts to becquerels.

Question 33

A radiopharmaceutical dose is measured as 9.25 × 10⁹ Bq. Convert this activity into millicuries (mCi). (1 Ci = 3.7 × 10¹⁰ Bq) A. 0.10 mCi B. 0.25 mCi C. 0.50 mCi D. 250 mCi

Answer 33

Step 1 – Formula: Ci = Bq / (3.7 × 10¹⁰) Given: Bq = 9.25 × 10⁹ Step 2 – Compute in curies: Ci = (9.25 × 10⁹) / (3.7 × 10¹⁰) Ci = 0.25 Ci Step 3 – Convert to millicuries (1 Ci = 1000 mCi): mCi = 0.25 × 1000 = 250 mCi ✅ Final Answer: D. 250 mCi Rationale: To convert becquerels to curies, divide by 3.7 × 10¹⁰. Then, multiply by 1000 to express the result in millicuries. A higher Bq value corresponds to a proportionally higher mCi value.

Question 34

1. HEAT UNITS (HU = kVp × mA × s × C) where: kVp = mA = s = c=

Answer 34

kilovoltage peak (volts) tube current (milliamperes) exposure time (seconds) constant Single-phase = 1.0 Three-phase = 1.35 High-frequency = 1.40

Question 35

A single-phase x-ray unit is operated at 80 kVp, 200 mA, and 0.5 s. Calculate the heat units (HU) produced. A. 6,000 HU B. 8,000 HU C. 10,000 HU D. 12,000 HU

Answer 35

Formula: HU = kVp × mA × s × C Given: kVp = 80, mA = 200, s = 0.5, C = 1.0 (single-phase) Computation: HU = 80 × 200 × 0.5 × 1.0 = 8,000 HU Answer: B. 8,000 HU Rationale: Heat units represent the amount of heat generated at the anode during exposure. For single-phase units, the correction factor is 1.0.

Question 36

A three-phase x-ray unit operates at 100 kVp, 300 mA, and 0.1 s. How many heat units are produced? A. 3,000 HU B. 4,050 HU C. 5,000 HU D. 6,000 HU

Answer 36

Formula: HU = kVp × mA × s × C Given: kVp = 100, mA = 300, s = 0.1, C = 1.35 (three-phase) Computation: HU = 100 × 300 × 0.1 × 1.35 = 4,050 HU Answer: B. 4,050 HU Rationale: Three-phase and high-frequency units produce more heat due to continuous voltage waveform; hence a higher correction factor.

Question 37

A high-frequency generator produces an exposure at 120 kVp, 400 mA, and 0.25 s. Determine the heat units. A. 12,000 HU B. 15,000 HU C. 18,000 HU D. 20,000 HU

Answer 37

Formula: HU = kVp × mA × s × C Given: kVp = 120, mA = 400, s = 0.25, C = 1.4 (high-frequency) Computation: HU = 120 × 400 × 0.25 × 1.4 = 16,800 HU Closest answer: Between B (15,000 HU) and C (18,000 HU) Answer: C. 18,000 HU Rationale: Heat units quantify thermal stress on the anode. High-frequency generators are most efficient, producing more HU.

Question 38

EXPOSURE (X = Q / m) where: X = Q = m =

Answer 38

exposure (C/kg or roentgen, R) total charge of ions (Coulombs, C) mass of air (kg)

Question 39

If 0.0002 C of charge is produced in 0.001 kg of air, find the exposure. A. 0.05 C/kg B. 0.2 C/kg C. 0.4 C/kg D. 0.6 C/kg

Answer 39

Formula: X = Q / m Given: Q = 0.0002 C, m = 0.001 kg Computation: X = 0.0002 / 0.001 = 0.2 C/kg Answer: B. 0.2 C/kg Rationale: Exposure measures the ionization in air due to x-rays or gamma rays, defined by charge per mass of air.

Question 40

An x-ray beam produces 0.0005 C of ions in 0.002 kg of air. What is the exposure? A. 0.1 C/kg B. 0.15 C/kg C. 0.20 C/kg D. 0.25 C/kg

Answer 40

Formula: X = Q / m Given: Q = 0.0005 C, m = 0.002 kg Computation: X = 0.0005 / 0.002 = 0.25 C/kg Answer: D. 0.25 C/kg Rationale: Dividing total charge by air mass gives exposure, representing the ionization capacity of radiation.

Question 41

ABSORBED DOSE (D = E / m) where: D = E = m =

Answer 41

absorbed dose (gray, Gy) energy absorbed (joules, J) mass of tissue (kg)

Question 42

A tissue absorbs 0.02 J of radiation energy, and its mass is 0.01 kg. Find the absorbed dose. A. 1 Gy B. 2 Gy C. 3 Gy D. 4 Gy

Answer 42

Formula: D = E / m Given: E = 0.02 J, m = 0.01 kg Computation: D = 0.02 / 0.01 = 2 Gy Answer: B. 2 Gy Rationale: Absorbed dose is the energy imparted per unit mass of tissue. 1 Gy = 1 J/kg.

Question 43

During fluoroscopic examination in the ward, a patient’s abdominal tissue absorbs 0.5 joules of radiation energy. If the irradiated tissue mass is 0.25 kg, what is the absorbed dose? A. 1 Gy B. 1.5 Gy C. 2 Gy D. 2.5 Gy

Answer 43

Step 1 – Formula: D = E / m Given: E = 0.5 J, m = 0.25 kg Step 2 – Substitute: D = 0.5 / 0.25 = 2 Gy ✅ Answer: C. 2 Gy Rationale: In ward-based fluoroscopic procedures, absorbed dose expresses how much energy per unit mass the patient’s tissue receives. Since 1 Gy = 1 J/kg, dividing the absorbed energy (0.5 J) by the tissue mass (0.25 kg) yields 2 Gy.

Question 44

DOSE EQUIVALENT (H = D × QF) where: H = D = QF =

Answer 44

dose equivalent (sievert, Sv) absorbed dose (Gy) quality factor (dimensionless, varies by radiation type)

Question 45

A technologist receives an absorbed dose of 0.02 Gy from neutron radiation (QF = 10). What is the dose equivalent? A. 0.1 Sv B. 0.2 Sv C. 0.5 Sv D. 0.8 Sv

Answer 45

Formula: H = D × QF Given: D = 0.02 Gy, QF = 10 Computation: H = 0.02 × 10 = 0.2 Sv Answer: B. 0.2 Sv Rationale: The dose equivalent accounts for the biological effect of radiation, not just the absorbed energy.

Question 46

A radiologic technologist accidentally receives an absorbed dose of 0.005 Gy from alpha radiation. If the quality factor (QF) for alpha is 20, what is the dose equivalent? A. 0.05 Sv B. 0.1 Sv C. 0.5 Sv D. 1.0 Sv

Answer 46

Formula: H = D × QF Given: D = 0.005 Gy, QF = 20 Computation: H = 0.005 × 20 = 0.1 Sv ✅ Answer: B. 0.1 Sv Rationale: Alpha particles have a high quality factor (QF = 20) due to their strong ionizing ability. Multiplying the absorbed dose by QF gives the biologically weighted dose (dose equivalent).

Question 47

EFFECTIVE DOSE (E = Σ Wᵗ × Hᵗ) where: E = Wᵗ = Hᵗ =

Answer 47

effective dose (Sv) tissue weighting factor equivalent dose to that tissue

Question 48

If the lungs (Wᵗ = 0.12) receive an equivalent dose of 0.5 Sv, what is the effective dose? A. 0.03 Sv B. 0.06 Sv C. 0.12 Sv D. 0.24 Sv

Answer 48

Formula: E = Wᵗ × Hᵗ Given: Wᵗ = 0.12, Hᵗ = 0.5 Sv Computation: E = 0.12 × 0.5 = 0.06 Sv Answer: B. 0.06 Sv Rationale: Effective dose reflects the risk considering tissue sensitivity; lungs have moderate weighting (0.12).

Question 49

During a whole-body PET/CT scan, the thyroid receives an equivalent dose (Hᵗ) of 0.3 Sv. If the tissue weighting factor (Wᵗ) for the thyroid is 0.04, what is the effective dose contribution? A. 0.012 Sv B. 0.014 Sv C. 0.016 Sv D. 0.018 Sv

Answer 49

Formula: E = Wᵗ × Hᵗ Given: Wᵗ = 0.04, Hᵗ = 0.3 Sv Computation: E = 0.04 × 0.3 = 0.012 Sv ✅ Answer: A. 0.012 Sv Rationale: Effective dose considers organ radiosensitivity. The thyroid has a low weighting factor, so even with 0.3 Sv exposure, the effective dose remains small (0.012 Sv).

Question 50

INVERSE SQUARE LAW (I₁ / I₂ = D₂² / D₁²) where: I = D =

Answer 50

intensity (exposure, mR or C/kg) distance (m)

Question 51

An x-ray beam has an intensity of 400 mR at 100 cm. What will be the intensity at 200 cm? A. 50 mR B. 100 mR C. 150 mR D. 200 mR

Answer 51

Formula: I₁ / I₂ = D₂² / D₁² Given: I₁ = 400 mR, D₁ = 100 cm, D₂ = 200 cm Computation: I₂ = I₁ × (D₁² / D₂²) = 400 × (100² / 200²) = 400 × (1/4) = 100 mR Answer: B. 100 mR Rationale: Intensity is inversely proportional to the square of the distance. Doubling the distance reduces intensity by ¼.

Question 52

At 150 cm from the x-ray tube, the beam intensity is 600 mR. What will the intensity be if the distance is reduced to 75 cm? A. 1,200 mR B. 2,400 mR C. 3,000 mR D. 4,800 mR

Answer 52

Formula: I₁ / I₂ = D₂² / D₁² Given: I₁ = 600 mR, D₁ = 150 cm, D₂ = 75 cm Rearrange: I₂ = I₁ × (D₁² / D₂²) Computation: I₂ = 600 × (150² / 75²) I₂ = 600 × (22500 / 5625) = 600 × 4 = 2400 mR ✅ Answer: B. 2,400 mR Rationale: Intensity increases with the square of distance reduction. Halving the distance increases intensity by four times.

Question 53

HALF VALUE LAYER (HVL = 0.693 / μ) where: HVL = μ =

Answer 53

half-value layer (mm Al or mm Pb) linear attenuation coefficient (cm⁻¹)

Question 54

An x-ray beam has a linear attenuation coefficient (μ) of 0.231 cm⁻¹ in aluminum. What is the half-value layer (HVL)? A. 1.5 cm Al B. 2.5 cm Al C. 3.0 cm Al D. 4.0 cm Al

Answer 54

Formula: HVL = 0.693 / μ Given: μ = 0.231 cm⁻¹ Computation: HVL = 0.693 / 0.231 = 3.0 cm Al ✅ Answer: C. 3.0 cm Al Rationale: HVL represents the thickness of absorber needed to reduce x-ray intensity to half. A lower μ means a more penetrating beam → larger HVL.

Question 55

If the first HVL of an x-ray beam is 2.0 mm Al, what fraction of the beam remains after passing through 4.0 mm Al? A. 1/2 B. 1/4 C. 1/8 D. 1/16

Answer 55

Each HVL reduces the intensity by half. After 2 HVLs (4.0 mm Al): Remaining fraction = (1/2)² = 1/4 ✅ Answer: B. 1/4 Rationale: Each additional HVL halves the remaining intensity. Two HVLs → 25% intensity transmitted.

Question 56

RELATIVE BIOLOGICAL EFFECTIVENESS (RBE = Dose of 250 keV X-ray / Dose of test radiation) where: RBE = Dose =

Answer 56

relative biological effectiveness (no unit) radiation dose producing same biological effect (Gy)

Question 57

A 250 keV x-ray requires 6 Gy to produce a specific skin reaction. A neutron beam causes the same reaction with only 2 Gy. What is the RBE of the neutron beam? A. 1.0 B. 2.0 C. 3.0 D. 4.0

Answer 57

Formula: RBE = Dose of 250 keV X-ray / Dose of test radiation Given: 6 Gy / 2 Gy = 3 ✅ Answer: C. 3.0 Rationale: The neutron beam is 3× more biologically effective than 250 keV x-rays for the same effect.

Question 58

An alpha particle beam produces the same biological damage as 10 Gy of diagnostic x-rays, but the alpha dose was only 1 Gy. What is the RBE? A. 5 B. 8 C. 10 D. 12

Answer 58

RBE = Dose of 250 keV X-ray / Dose of test radiation = 10 Gy / 1 Gy = 10 ✅ Answer: C. 10 Rationale: Higher RBE means greater biological effectiveness per unit dose, typical for alpha particles.

Question 59

OXYGEN ENHANCEMENT RATIO (OER = Dose hypoxic / Dose oxygenated) where: OER = Dose =

Answer 59

oxygen enhancement ratio (no unit) dose to produce same biological effect (Gy)

Question 60

A certain level of cell damage requires 6 Gy under hypoxic conditions but only 2 Gy when oxygen is present. What is the OER? A. 1 B. 2 C. 3 D. 4

Answer 60

OER = Dose (hypoxic) / Dose (oxygenated) = 6 / 2 = 3 ✅ Answer: C. 3 Rationale: Oxygen makes radiation more effective by fixing free radical damage. OER of 3 is typical for x-rays.

Question 61

If a tumor requires 9 Gy in a hypoxic environment and 4.5 Gy when oxygenated to achieve the same biological effect, find the OER. A. 1.5 B. 2 C. 3 D. 4

Answer 61

OER = 9 / 4.5 = 2 ✅ Answer: B. 2 Rationale: A lower OER (≈2) means oxygen moderately increases radiosensitivity, common for low-LET radiation.

Question 62

PRIMARY BARRIER THICKNESS (B = WUT / (d² × P)) where: B = W = U = T = d = P =

Answer 62

barrier transmission factor workload (mA·min/week) use factor occupancy factor distance (m) permissible weekly dose

Question 63

A primary barrier is protecting an area 4 m from the x-ray source. If the workload (W) is 2000 mA·min/week, use factor (U) is 0.25, occupancy factor (T) is 1, and permissible dose (P) is 0.02 mSv/week, compute the transmission factor (B). A. 0.01 B. 0.02 C. 0.03 D. 0.04

Answer 63

Formula: B = W × U × T / (d² × P) B = 2000 × 0.25 × 1 / (4² × 0.02) B = 500 / (16 × 0.02) = 500 / 0.32 = 1562.5 ✅ Answer: B ≈ 1.56 × 10³ (convert to log attenuation for design charts) Rationale: Barrier transmission factor expresses how much radiation penetrates. This value is used to determine lead thickness via NCRP 147 tables.

Question 64

If the distance from the source to the occupied area doubles, how will the required barrier transmission factor (B) change? A. It doubles B. It becomes half C. It becomes one-fourth D. It remains the same

Answer 64

From formula: B ∝ 1 / d² If distance doubles → d² increases by 4 → B decreases by 1/4 ✅ Answer: C. It becomes one-fourth Rationale: Increasing distance greatly reduces exposure, so less barrier thickness is needed.

Question 65

CONSTANTS : Speed of light (c ) = Planck’s constant (h) = Gravitational acceleration (g) = Elementary charge (e)

Answer 65

3.00 × 10⁸ m/s 6.63 × 10⁻³⁴ J·s 9.8 m/s² 1.6 × 10⁻¹⁹ C

Question 66

UNIT CONVERSATIONS : 1 eV = 1 Gy = 1 Sv = 1 R ≈ 1 Ci =

Answer 66

1.6 × 10⁻¹⁹ J 100 rad 100 rem 0.00877 Gy (air) 3.7 × 10¹⁰ Bq