NMTCB Radiation Physics & Detection 4 — Questions and Answers
Question 1: The photoelectric effect cross-section (τ) depends on atomic number Z approximately as:
- τ ∝ Z / E³
- τ ∝ Z⁴ to Z⁵ / E³·⁵ (Correct answer)
- τ ∝ Z² / E
- τ ∝ Z / E²
Correct answer: τ ∝ Z⁴ to Z⁵ / E³·⁵
Photoelectric cross-section scales approximately as Z⁴⁻⁵/E³·⁵, which is why high-Z materials like lead are effective gamma ray shields at low energies.
Question 2: In liquid scintillation counting, chemical quenching reduces efficiency by:
- Increasing the density of the scintillation cocktail
- Absorbing energy from excited scintillant molecules before photon emission occurs (Correct answer)
- Blocking the photomultiplier tube window with colored compounds
- Reducing the high voltage applied to the PMT dynodes
Correct answer: Absorbing energy from excited scintillant molecules before photon emission occurs
Chemical quenching agents absorb energy from excited fluorescent molecules via collisional deactivation, reducing the number of light photons produced per decay event.
Question 3: The effective half-life (T_eff) of a radionuclide in a biological system is calculated as:
- T_eff = T_physical + T_biological
- T_eff = (T_physical × T_biological) / (T_physical + T_biological) (Correct answer)
- T_eff = T_physical - T_biological
- T_eff = T_physical / T_biological
Correct answer: T_eff = (T_physical × T_biological) / (T_physical + T_biological)
The effective half-life combines physical decay and biological elimination as parallel first-order processes: 1/T_eff = 1/T_physical + 1/T_biological.
Question 4: Which statement correctly describes the Compton plateau in a pulse height spectrum?
- A sharp peak at the photopeak centroid created by full energy deposition
- A broad continuum of pulse heights below the photopeak from partial energy deposits in Compton interactions (Correct answer)
- A discrete peak at 0.511 MeV from positron annihilation
- A narrow peak representing characteristic X-ray fluorescence of the detector
Correct answer: A broad continuum of pulse heights below the photopeak from partial energy deposits in Compton interactions
Compton scattering results in scattered photons that may escape the detector, depositing only partial energy and creating a continuum of pulse heights below the photopeak.
Question 5: Fluorescence yield (ω) in X-ray emission describes the probability that a vacancy in an electron shell will result in:
- Characteristic X-ray emission rather than Auger electron emission (Correct answer)
- Internal conversion rather than gamma emission
- Beta particle emission following electron capture
- Positron emission versus electron capture branching ratio
Correct answer: Characteristic X-ray emission rather than Auger electron emission
Fluorescence yield ω = N_X-ray / (N_X-ray + N_Auger), representing the fraction of inner-shell vacancies that decay by emitting characteristic X-rays versus Auger electrons.
Question 6: A semiconductor detector (HPGe) offers superior energy resolution compared to NaI(Tl) because:
- Germanium has a higher density than iodine for better stopping power
- The average energy to create an electron-hole pair (~3 eV) is much lower than for a scintillation photon (~100 eV equivalent), giving more charge carriers per event (Correct answer)
- Semiconductor detectors operate at room temperature eliminating thermal noise
- Germanium emits more characteristic X-rays than iodine per interaction
Correct answer: The average energy to create an electron-hole pair (~3 eV) is much lower than for a scintillation photon (~100 eV equivalent), giving more charge carriers per event
More charge carriers per event means smaller statistical fluctuations (Fano-limited noise), resulting in energy resolutions of ~0.1% for HPGe versus ~6-8% for NaI(Tl) at 662 keV.
Question 7: The range of alpha particles in tissue is approximately:
- Several centimeters, similar to beta particles
- A few tens of micrometers to less than 100 micrometers (Correct answer)
- Several millimeters, similar to low-energy X-rays
- Less than 1 micrometer, depositing all dose at the nuclear membrane
Correct answer: A few tens of micrometers to less than 100 micrometers
Alpha particles are stopped within ~40-100 μm of tissue due to their high charge (+2) and mass, making them extremely high-LET radiation with very short range.
NMTCB Radiation Physics & Detection 4
The photoelectric effect cross-section (τ) depends on atomic number Z approximately as: