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Test Bank for Khan’s The Physics of Radiation Therapy 6th Edition Gibbons | All Chapters (1–28) | 2025 | A+ Verified

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Access the Test Bank for Khan’s The Physics of Radiation Therapy, 6th Edition by Gibbons (2025 version). Includes all 28 chapters with A+ verified questions and answers to help students and professionals master radiation physics, strengthen clinical knowledge, and prepare confidently for exams.

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Institución
Physics of Radiation Therapy
Grado
Physics of Radiation Therapy

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Subido en
25 de agosto de 2025
Número de páginas
235
Escrito en
2025/2026
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Examen
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,Chapter 1: Structure of Matter



Q1. Atomic Structure Scenario

A radiation therapy physicist explains to residents that electron binding energy plays a crucial
role in X-ray production.
Which factor has the greatest influence on the magnitude of electron binding energy in an
atom?

A. The number of neutrons in the nucleus
B. The distance of the electron from the nucleus
C. The mass of the orbiting electron
D. The number of isotopes the element has

✅ Correct Answer: B. The distance of the electron from the nucleus

Rationale: Electron binding energy increases as electrons are closer to the nucleus due to the
strong Coulombic attraction. While nuclear charge (protons) is the ultimate source, the effective
factor determining binding strength for a particular shell is the distance of the electron from the
nucleus.

Keywords: electron binding energy, Coulomb force, atomic shells



Q2. Numerical Problem – Binding Energy

The binding energy of a K-shell electron in tungsten (Z=74) is approximately 69.5 keV. If a 120
keV photon ejects this electron, what will be the kinetic energy of the ejected photoelectron?

A. 50.5 keV
B. 69.5 keV
C. 120 keV
D. 189.5 keV

✅ Correct Answer: A. 50.5 keV

Rationale: Photoelectric effect equation:
Kinetic Energy = Incident photon energy – Binding energy
= 120 keV – 69.5 keV = 50.5 keV.

Keywords: photoelectric effect, binding energy, kinetic energy calculation

,Q3. Professional Context

During an orientation session, a resident asks why neutrons are important in nuclear stability.
What is the primary role of neutrons within the nucleus?

A. Providing electrostatic repulsion
B. Reducing proton-proton repulsion by nuclear forces
C. Contributing negative charge balance
D. Increasing atomic number

✅ Correct Answer: B. Reducing proton-proton repulsion by nuclear forces

Rationale: Neutrons stabilize the nucleus by providing strong nuclear force without adding
charge, offsetting the repulsion between positively charged protons.

Keywords: neutrons, nuclear stability, strong force



Q4. Concept Check – Atomic Mass

A radiation oncology dosimetrist is calculating dose based on electron density. Which atomic
component contributes most significantly to the mass of an atom?

A. Protons and neutrons
B. Electrons only
C. Protons and electrons
D. Shell binding energies

✅ Correct Answer: A. Protons and neutrons

Rationale: Nearly all the atomic mass is concentrated in the nucleus, contributed by protons and
neutrons. Electrons are extremely light and contribute negligibly.

Keywords: atomic mass, nucleus, protons, neutrons



Q5. Clinical Physics Scenario

In a lecture on radiation interactions, a physicist notes that the nucleus remains unchanged during
electron transitions between shells.
Why is this statement correct?

,A. Electron transitions occur outside the nucleus
B. Proton numbers adjust simultaneously
C. Neutrons absorb the transition energy
D. Binding energy modifies the nucleus directly

✅ Correct Answer: A. Electron transitions occur outside the nucleus

Rationale: Electron transitions involve movement between atomic shells, affecting electron
configuration but not altering the nucleus (which determines the element identity).

Keywords: electron transitions, nucleus, atomic shells



Q6. Applied Problem

If carbon-12 has 6 protons, 6 neutrons, and 6 electrons, what is the atomic mass number of
carbon-12?

A. 6
B. 12
C. 18
D. 24

✅ Correct Answer: B. 12

Rationale: Mass number = protons + neutrons = 6 + 6 = 12. Electrons are excluded as they have
negligible mass.

Keywords: mass number, protons, neutrons



Q7. Clinical Analogy

When explaining energy levels to medical residents, a physicist compares electron shells to
"stairs in a building."
Why is it correct to say that an electron in an inner shell has greater binding energy?

A. It is further away from other electrons
B. It experiences stronger attraction to the nucleus
C. It has higher kinetic energy than outer electrons
D. It balances the charge neutrality

✅ Correct Answer: B. It experiences stronger attraction to the nucleus

,Rationale: Inner-shell electrons are closer to the positively charged nucleus, resulting in stronger
Coulombic attraction and thus greater binding energy.

Keywords: binding energy, electron shells, Coulomb force



Q8. Nuclear Medicine Context

A patient receives an isotope for PET imaging. The decay involves emission of positrons. Which
particle transformation occurs inside the nucleus to produce a positron?

A. Neutron → Proton + Beta– + antineutrino
B. Proton → Neutron + Beta+ + neutrino
C. Proton + Electron → Neutron + neutrino
D. Proton → Alpha particle + Beta–

✅ Correct Answer: B. Proton → Neutron + Beta+ + neutrino

Rationale: Positron emission (β+) occurs when a proton is converted into a neutron, releasing a
positron and a neutrino.

Keywords: positron emission, nuclear transformation, beta decay



Q9. Applied Calculation

If an electron has a rest mass energy of 0.511 MeV, what is the equivalent mass of the electron
in kg?
(Use E=mc2E = mc^2E=mc2, c=3.0×108m/sc = 3.0 × 10^8 m/sc=3.0×108m/s, 1 eV =
1.6×10−19J1.6 × 10^{-19} J1.6×10−19J).

A. 9.11 × 10^-31 kg
B. 1.67 × 10^-27 kg
C. 5.11 × 10^-4 kg
D. 1.60 × 10^-19 kg

✅ Correct Answer: A. 9.11 × 10^-31 kg

Rationale: 0.511 MeV = 0.511×106×1.6×10−19J0.511 × 10^6 × 1.6 × 10^{-19}
J0.511×106×1.6×10−19J.
m=E/c2≈9.11×10−31kgm = E/c^2 ≈ 9.11 × 10^-31 kgm=E/c2≈9.11×10−31kg.

Keywords: Einstein equation, electron rest mass, MeV to kg

, Q10. Radiation Safety Application

Why does high atomic number (Z) material such as lead serve as an effective radiation shield?

A. High neutron absorption
B. High electron density increases photon interactions
C. Low binding energies absorb energy efficiently
D. Strong nuclear binding reduces secondary emission

✅ Correct Answer: B. High electron density increases photon interactions

Rationale: Lead has a high Z and many tightly bound electrons, increasing the probability of
photoelectric effect interactions, making it excellent for photon shielding.

Keywords: radiation shielding, photoelectric effect, atomic number



Q11. Concept Check

A resident calculates binding energy per nucleon for carbon-12. Why is this value important in
nuclear physics?

A. It predicts how many isotopes exist
B. It measures the stability of the nucleus
C. It defines the number of valence electrons
D. It controls X-ray spectrum output

✅ Correct Answer: B. It measures the stability of the nucleus

Rationale: Binding energy per nucleon indicates how tightly nucleons are bound together.
Higher values mean greater nuclear stability.

Keywords: binding energy per nucleon, nuclear stability



Q12. Clinical Application

In radiation oncology, energy deposition is related to atomic structure. Which interaction most
strongly depends on the electron binding energy of inner shells?

A. Compton scattering
B. Photoelectric effect
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