Radiation Physics and Protection Practice Questions

Q: Cobalt-60 is:

An artificial radioactive source. **Explanation:** **Cobalt-60 ($^{60}$Co)** is an **artificial radioactive source** produced by the neutron activation of stable Cobalt-59 in a nuclear reactor. It does not occur naturally in the earth's crust. It undergoes beta decay to stable Nickel-60, emitting two high-energy gamma rays (1.17 MeV and 1.33 MeV) in the process. * **Why Option C is correct:** Cobalt-60 is man-made (artificial) and unstable (radioactive), emitting ionizing radiation used extensively in external beam radiotherapy (Teletherapy). * **Why Options A & B are incorrect:** Natural radioactive materials include elements like Uranium, Radium, and Thorium, which exist inherently in nature. Cobalt-60 must be synthesized. * **Why Option D is incorrect:** A stable substance does not undergo radioactive decay. Cobalt-60 is inherently unstable with a specific half-life. **High-Yield Clinical Pearls for NEET-PG:** 1. **Half-life:** The half-life of Cobalt-60 is **5.27 years**. In clinical practice, the source is usually replaced every 5–7 years. 2. **Energy:** It emits two photons with an **average energy of 1.25 MeV**. 3. **Penumbra:** Cobalt-60 machines have a larger geometric penumbra compared to Linear Accelerators (LINAC) because the source size is larger (typically 1.5–2.0 cm in diameter). 4. **Dmax:** The depth of maximum dose (Dmax) for Cobalt-60 is **0.5 cm** below the skin surface (providing a modest skin-sparing effect). 5. **Application:** Primarily used in Teletherapy units and Gamma Knife radiosurgery.

Q: In radiography, regarding scattered radiation, which of the following statements is true?

Reduces image contrast. ### Explanation Scattered radiation is primarily produced by the **Compton effect**, where X-ray photons interact with outer-shell electrons of the patient's tissues, changing direction and losing energy. Unlike the primary beam, which travels in a straight line to represent anatomy, scattered photons hit the image receptor at random angles. **1. Why Option C is Correct:** Scattered radiation acts as a uniform "fog" or "noise" over the image. It adds a baseline density to both light and dark areas of the radiograph, thereby **reducing the difference in optical density** between adjacent structures. This decrease in the signal-to-noise ratio directly leads to **reduced image contrast**. **2. Analysis of Incorrect Options:** * **A & B:** Scatter degrades image quality. By adding non-diagnostic "noise," it masks anatomical details, thereby **decreasing** both contrast and spatial resolution. * **D:** Scattered radiation does not reduce patient dose; in fact, techniques used to *compensate* for scatter (like using a grid) often require increasing the mAs (exposure), which **increases** the patient's radiation dose. **3. NEET-PG High-Yield Pearls:** * **Factors increasing scatter:** Larger field size (collimation), thicker body parts, and higher kVp (kilovoltage). * **The Grid:** A device placed between the patient and the receptor to absorb scatter. While it improves contrast, it increases the **Bucky Factor**, necessitating a higher radiation dose. * **Collimation:** The most effective way to reduce scatter production is by limiting the field size. * **Air Gap Technique:** Increasing the distance between the patient and the film allows scattered photons to diverge away from the receptor, improving contrast without a grid.

Q: All of the following factors increase the density of an X-ray image except:

Decreasing the mA. ### Explanation In radiology, **Radiographic Density** refers to the degree of "blackness" on a processed film. It is primarily determined by the quantity of X-ray photons reaching the receptor. **1. Why "Decreasing the mA" is the correct answer:** Milliamperage (mA) controls the heating of the filament and, consequently, the number of electrons produced. **mA is directly proportional to the quantity of X-rays.** * **Increasing mA** increases the number of photons, leading to higher density (darker image). * **Decreasing mA** reduces the number of photons, resulting in a lighter image (lower density). Therefore, it does **not** increase density. **2. Analysis of Incorrect Options:** * **Increasing the kVp:** Peak kilovoltage (kVp) controls the quality (penetrating power) and quantity of the X-ray beam. Higher kVp increases the kinetic energy of electrons, producing more energetic photons that easily penetrate the patient and reach the film, thereby **increasing density**. * **Reducing the distance (Focal Spot to Film):** According to the **Inverse Square Law**, the intensity of the X-ray beam is inversely proportional to the square of the distance. If the distance is reduced, the intensity of the beam hitting the film increases significantly, thereby **increasing density**. **3. Clinical Pearls for NEET-PG:** * **mAs (mA × time):** This is the primary factor used to control density. If you want to double the density, you double the mAs. * **kVp:** This is the primary factor controlling **Radiographic Contrast**. While it affects density (15% rule), its main clinical role is managing the "gray scale." * **Inverse Square Law Formula:** $I_1/I_2 = (D_2/D_1)^2$. Even a small decrease in distance leads to a large increase in intensity/density. * **Grid use:** Using a grid (to reduce scatter) actually **decreases** density, requiring an increase in mAs to compensate.

Q: What should be the diameter of the tungsten filament used as a cathode?

2 mm in diameter. **Explanation:** In an X-ray tube, the **cathode** serves as the source of electrons. It consists of a spiral filament made of **Tungsten**, chosen for its high melting point ($3422^\circ\text{C}$) and high atomic number ($Z=74$), which facilitates efficient thermionic emission. 1. **Why 2 mm is correct:** The standard tungsten filament used in diagnostic X-ray tubes is approximately **2 mm in diameter** and about 1 cm to 2 cm in length. This specific diameter provides the necessary surface area to produce a sufficient cloud of electrons (space charge) while maintaining structural integrity under high heat. It is coiled into a small spiral to fit within the focusing cup, which directs the electron beam toward the focal spot on the anode. 2. **Why other options are incorrect:** * **1 mm (Option A):** Too thin; it would have insufficient surface area for thermionic emission and would be prone to "burn out" or breakage due to high thermal stress. * **1 cm and 2 cm (Options C & D):** These are far too large for a filament diameter. A filament this thick would require an impractical amount of current to heat and would result in a massive electron beam, making it impossible to focus the beam into a small "focal spot." This would lead to significant image blurring (penumbra). **High-Yield Clinical Pearls for NEET-PG:** * **Thermionic Emission:** The process of "boiling off" electrons from the filament by heating it. * **Focusing Cup:** Usually made of **Nickel**; it is negatively charged to repel electrons into a narrow stream. * **Dual Focus Tubes:** Most modern tubes have two filaments (small and large) to provide two different focal spot sizes, optimizing either detail or heat loading. * **Filament Evaporation:** The most common cause of X-ray tube failure is the thinning and eventual snapping of the tungsten filament.

Q: All of the following modalities use non-ionizing radiation, except:

Radiography. ### Explanation The core concept in this question is distinguishing between **ionizing** and **non-ionizing** radiation based on their energy levels and ability to displace electrons from atoms. **1. Why Radiography is the Correct Answer:** Radiography (X-rays) uses high-energy electromagnetic waves that possess enough energy to remove tightly bound electrons from the orbit of an atom, creating ions. This process is called **ionizing radiation**. Because it can alter atomic structures, it carries risks of deterministic effects (like skin erythema) and stochastic effects (like carcinogenesis). Other examples of ionizing modalities include CT scans, PET scans, and Mammography. **2. Why the Other Options are Incorrect:** * **Ultrasonography (A):** Uses high-frequency **sound waves** (mechanical energy), not electromagnetic radiation. It is completely non-ionizing and safe for fetal imaging. * **Thermography (B):** Detects **infrared radiation** (heat) emitted naturally by the body. Infrared falls below the ionizing threshold on the electromagnetic spectrum. * **MRI (C):** Utilizes strong **magnetic fields** and **Radiofrequency (RF) pulses**. RF waves are at the low-energy end of the spectrum and do not have enough energy to ionize atoms. **3. High-Yield Clinical Pearls for NEET-PG:** * **The Spectrum:** Remember the mnemonic for increasing energy: **R**adio waves < **M**icrowaves < **I**nfrared < **V**isible light < **U**ltraviolet < **X**-rays < **G**amma rays. Only UV (partially), X-rays, and Gamma rays are ionizing. * **ALARA Principle:** "As Low As Reasonably Achievable" is the fundamental rule for ionizing radiation protection. * **Dose Limits:** The annual effective dose limit for a radiation worker is **20 mSv** per year (averaged over 5 years). * **Radiosensitivity:** According to the Law of Bergonie and Tribondeau, cells that divide rapidly (e.g., bone marrow, gonads) are the most sensitive to ionizing radiation.

Q: All of the following are beta emitter particles except?

Technetium 99m. ### Explanation The core concept tested here is the difference between **particulate radiation** (Beta particles) used primarily for therapy and **electromagnetic radiation** (Gamma rays) used for diagnostic imaging. **Why Technetium 99m (Tc-99m) is the correct answer:** Technetium 99m is a **pure gamma emitter**. The "m" stands for metastable, meaning it exists in an excited nuclear state. When it decays to Technetium 99, it releases energy in the form of a **140 keV gamma photon** without emitting any particulate radiation (like alpha or beta particles). This property, combined with its short half-life (6 hours), makes it the "workhorse" of diagnostic nuclear medicine (e.g., Bone scans, MUGA scans) because it provides high-quality images with low radiation dose to the patient. **Analysis of Incorrect Options (Beta Emitters):** * **Strontium 89:** A pure beta emitter used for the **palliative treatment of painful bone metastases** (e.g., from prostate cancer). * **Phosphorus 32:** A pure beta emitter historically used for treating **Polycythemia Vera** and currently used for intracavitary therapy and bone pain palliation. * **Tin 117m:** A unique radioisotope that emits **conversion electrons** (which behave like beta particles) and is used for treating bone pain. **NEET-PG High-Yield Clinical Pearls:** 1. **Pure Beta Emitters:** Remember the mnemonic **"YPS"** — **Y**ttrium-90, **P**hosphorus-32, and **S**trontium-89. These are used for therapy, not imaging. 2. **Iodine-131:** Unlike Tc-99m, I-131 emits **both** beta particles (for treating thyroid cancer/thyrotoxicosis) and gamma rays (allowing for post-therapy imaging). 3. **Alpha Emitters:** Radium-223 is a notable alpha emitter used in metastatic prostate cancer. 4. **Positron Emitters:** Used in PET scans (e.g., Fluorine-18). A positron is essentially a "positive beta particle."

Q: What is the atomic number of tungsten?

74. **Explanation:** **Correct Option: C (74)** Tungsten (Wolfram) is the material of choice for the **anode target** in diagnostic X-ray tubes. Its atomic number (Z) is **74**. In X-ray production, the efficiency of Bremsstrahlung (braking radiation) is directly proportional to the atomic number of the target material ($Efficiency \approx Z \times V$). A high atomic number ensures efficient production of X-rays and provides high-energy characteristic radiation (K-shell binding energy is ~69.5 keV), which is essential for diagnostic imaging. **Analysis of Incorrect Options:** * **A. 42:** This is the atomic number of **Molybdenum (Mo)**. Molybdenum is used as the target material in **Mammography** because it produces lower-energy characteristic X-rays (approx. 17–20 keV) suitable for soft tissue imaging of the breast. * **B. 181:** This is the approximate atomic weight of Tantalum, not an atomic number relevant to standard X-ray anode materials. * **D. 82:** This is the atomic number of **Lead (Pb)**. While lead has a high Z-number, it is used for **radiation shielding** (aprons, walls) rather than as a target material because it has a very low melting point. **High-Yield Clinical Pearls for NEET-PG:** * **Why Tungsten?** It is chosen for its high atomic number (74) and its **high melting point (3422°C)**, which allows it to withstand the intense heat generated during X-ray production (where 99% of energy is converted to heat). * **Rhenium-Tungsten Alloy:** Modern rotating anodes often use an alloy of 90% Tungsten and 10% Rhenium to prevent surface thermal cracking (pitting). * **Mammography Targets:** Use Molybdenum (Z=42) or Rhodium (Z=45) for better contrast in fatty vs. glandular tissue.

Q: The magnetic field in an MRI machine is measured in what unit?

Tesla. **Explanation:** The correct answer is **Tesla (T)**. In MRI physics, the strength of the static magnetic field ($B_0$) is measured in Tesla. One Tesla is equal to 10,000 Gauss. Most clinical MRI scanners operate at field strengths of 1.5T or 3.0T. Higher magnetic field strengths result in a higher signal-to-noise ratio (SNR), allowing for better image resolution and faster scanning. **Analysis of Incorrect Options:** * **A. Hounsfield Units (HU):** This is the unit used in **Computed Tomography (CT)** to describe radiodensity. It represents the degree of X-ray attenuation (e.g., Water = 0 HU, Air = -1000 HU, Bone = +1000 HU). * **C. MHz (Megahertz):** This is a unit of **frequency**. In MRI, the Larmor frequency (the frequency at which protons precess) is measured in MHz. For example, at 1.5T, hydrogen protons precess at approximately 63.8 MHz. * **D. None of the above:** Incorrect, as Tesla is the standard SI unit for magnetic flux density. **High-Yield Clinical Pearls for NEET-PG:** * **Larmor Equation:** $f = \gamma B_0$ (Frequency is directly proportional to the magnetic field strength). * **Gyromagnetic Ratio ($\gamma$):** For Hydrogen, it is **42.58 MHz/Tesla**. * **Quenching:** The rapid, accidental, or intentional loss of superconductivity in an MRI magnet, leading to the release of liquid helium as gas. * **Safety:** MRI is non-ionizing, making it safer than CT for pregnant patients and children, though it is contraindicated in patients with non-MRI-compatible metallic implants or pacemakers.

Q: Digital radiography requires less radiation than conventional radiography because:

The sensor is more sensitive to x-rays.. **Explanation:** The fundamental reason digital radiography (DR) reduces patient radiation exposure is the **high detective quantum efficiency (DQE)** of digital sensors compared to traditional silver halide films. **1. Why Option B is Correct:** Digital sensors (such as CCD, CMOS, or Photostimulable Phosphor plates) are significantly **more sensitive to X-ray photons**. They can capture a higher percentage of the incident X-ray beam and convert it into an electronic signal with minimal loss. Because the sensor is more efficient at "counting" photons, a lower photon flux (lower mAs) is required to produce a diagnostic-quality image. This allows for a reduction in radiation dose by up to 50–80% in many clinical scenarios. **2. Why Other Options are Incorrect:** * **Option A:** The size of the sensor does not inherently reduce radiation. In fact, many digital sensors are smaller than conventional film (especially in intraoral radiography), but it is the *efficiency* of the surface, not the area, that dictates dose. * **Option C:** Increasing exposure time would **increase** the radiation dose and lead to motion blur. Digital radiography actually allows for *shorter* exposure times due to high sensitivity. * **Option D:** Since A and C are incorrect, this cannot be the answer. **Clinical Pearls for NEET-PG:** * **ALARA Principle:** Digital radiography is a key tool in practicing "As Low As Reasonably Achievable" radiation safety. * **Dynamic Range:** Digital sensors have a **wider dynamic range** (latitude) than film, meaning they can compensate for over- or under-exposure, reducing the need for "repeat" films. * **Image Post-processing:** Digital systems allow for contrast and brightness adjustments after the exposure, further preventing unnecessary re-exposures.

Question 1

Cobalt-60 is:

Accepted Answer

An artificial radioactive source

Answer

A natural radioactive source

Answer

A natural radioactive material

Answer

An artificial stable substance

Question 2

Radiological investigations in females of reproductive age group are generally restricted to which period of the menstrual cycle?

Accepted Answer

During the first 10 days of the menstrual cycle

Answer

During the menstrual period

Answer

Between the 10th and 20th days of the menstrual cycle

Answer

During the last 10 days of the menstrual cycle

Question 3

In radiography, regarding scattered radiation, which of the following statements is true?

Accepted Answer

Reduces image contrast

Answer

Improves image contrast

Answer

Improves image resolution

Answer

Reduces patient radiation dose

Question 4

All of the following factors increase the density of an X-ray image except:

Accepted Answer

Decreasing the mA

Answer

Increasing the kVp

Answer

Reducing the distance between the focal spot and film

Answer

None of the above

Question 5

What should be the diameter of the tungsten filament used as a cathode?

Accepted Answer

2 mm in diameter

Answer

1 mm in diameter

Answer

1 cm in diameter

Answer

2 cm in diameter

Question 6

All of the following modalities use non-ionizing radiation, except:

Accepted Answer

Radiography

Answer

Ultrasonography

Answer

Thermography

Answer

MRI

Question 7

All of the following are beta emitter particles except?

Accepted Answer

Technetium 99m

Answer

Strontium 89

Answer

Phosphorus 32

Answer

Tin 117m

Question 8

What is the atomic number of tungsten?

Accepted Answer

74

Answer

42

Answer

181

Answer

82

Question 9

The magnetic field in an MRI machine is measured in what unit?

Accepted Answer

Tesla

Answer

Housefield units

Answer

MHz

Answer

None of the above

Question 10

Digital radiography requires less radiation than conventional radiography because:

Accepted Answer

The sensor is more sensitive to x-rays.

Answer

The sensor is larger.

Answer

The exposure time is increased.

Answer

All of the above.

Radiation Physics and Protection — MCQs

Radiation Physics and Protection — MCQs

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