Q: Fissile vs fertile: which of the following is a non-fissile (fertile) material under thermal-neutron conditions?

Introduction / Context: Nuclear fuel classification distinguishes fissile isotopes (which readily fission with thermal neutrons) from fertile isotopes (which do not, but can be converted into fissile species by neutron capture and subsequent decay). Recognizing which is which is fundamental for reactor fuel-cycle questions and breeder concepts. Given Data / Assumptions: Thermal-neutron regime is implied unless otherwise noted. Common fissile isotopes include U-235, U-233, Pu-239, Pu-241. Common fertile isotopes include Th-232 and U-238. Concept / Approach: Thorium-232 is fertile: it captures a neutron to become Th-233 → Pa-233 → U-233 (fissile). In contrast, U-235, U-233, Pu-239, and Pu-241 are fissile and support thermal-neutron chain reactions. Therefore, within the given list, Th-232 is the non-fissile (fertile) choice. Step-by-Step Solution: List fissile isotopes: U-235, U-233, Pu-239, Pu-241.Identify Th-232 as fertile, not fissile, under thermal conditions.Select Thorium-232 accordingly.Note that fertile isotopes are essential in breeding strategies to produce new fissile fuel. Verification / Alternative check: Fuel-cycle diagrams universally show Th-232 breeding to U-233 and U-238 breeding to Pu-239 via neutron capture and beta decays; both are classic fertile pathways. Why Other Options Are Wrong: Pu-239 / Pu-241: well-known fissile isotopes.U-235: primary fissile constituent of natural uranium.U-233: fissile product bred from Th-232. Common Pitfalls: Confusing “fissile” with “fertile” due to similar wording; remember fertile needs conversion, while fissile sustains thermal-neutron fission directly. Final Answer: Thorium-232

Question 1

Radioactive decay of a substance follows which kinetic order in chemical-reaction terms (i.e., rate proportional to the amount remaining)?

Accepted Answer

Introduction / Context: Radioactive decay is a spontaneous nuclear process, but its time dependence can be described using the same mathematics as chemical reaction kinetics. Examinations often ask which “order” of reaction best models the decay law that leads to exponential decrease and the concept of half-life. Given Data / Assumptions: Nuclei decay independently and randomly. Decay probability per nucleus per unit time is constant (decay constant = lambda). No external forcing or replenishment of the parent nuclide. Concept / Approach: The defining kinetic equation is dN/dt = −lambda * N, where N is the number of undecayed nuclei. This is identical in form to a first-order rate law used in chemistry, where rate is directly proportional to the amount (or concentration) of the reactant present. Step-by-Step Solution: Start with: dN/dt = −lambda * N.Separate variables and integrate: ∫ dN/N = −∫ lambda dt.Obtain: ln(N) = −lambda t + constant ⇒ N = N0 * exp(−lambda * t).Define half-life t_1/2 from N/N0 = 1/2 ⇒ t_1/2 = 0.693 / lambda. Verification / Alternative check: First-order systems always produce a constant half-life regardless of starting amount, exactly as observed for radionuclides. Plotting ln(N) versus time gives a straight line with slope −lambda, a classic first-order test. Why Other Options Are Wrong: Zero order: would give linear decrease, not exponential.Second/Third order: rates would scale with higher powers of amount, not supported by decay statistics.Mixed order: unnecessary; a single first-order constant captures the behavior. Common Pitfalls: Confusing nuclear decay (first order) with surface or diffusion-limited processes.Thinking half-life depends on the initial quantity; it does not for first-order processes. Final Answer: First order

Question 2

Critical mass of a fissionable material: which measure most effectively reduces it for a given isotope and geometry?

Accepted Answer

Introduction / Context: Critical mass is the minimum amount of fissile material needed to sustain a chain reaction. Designers and safety engineers must understand how geometry, density, moderation, and surrounding materials change neutron economy and therefore the critical size or mass. Given Data / Assumptions: Same fissile isotope and core shape. No change to enrichment; focus on boundary conditions and environment. Standard fast or thermal spectrum as implied by the configuration. Concept / Approach: A chain reaction requires that each fission produces, on average, at least one neutron that causes another fission (k_eff ≥ 1). A reflector (e.g., beryllium, graphite, steel, or heavy water in thermal systems) returns escaping neutrons back into the core, increasing the effective neutron utilization and lowering the mass needed for criticality. Step-by-Step Solution: Identify loss channels: leakage, absorption in non-fuel, and non-fission captures.Place reflector: returning leaked neutrons reduces net leakage probability.Improved economy: more neutrons available for causing fission ⇒ lower critical mass.Other effects (e.g., cooling) modestly help via density increases but less than reflectors; absorbers do the opposite. Verification / Alternative check: Diffusion-theory and Monte Carlo models both show a substantial reduction in critical size with thick reflectors; published data sets list “bare” versus “reflected” critical masses, the latter being significantly lower. Why Other Options Are Wrong: Heating: thermal expansion lowers density, usually increasing the critical mass.Cooling: can reduce critical mass slightly by raising density, but effect is smaller than a reflector.Neutron absorber shell: increases critical mass by removing neutrons.Dilution with inert metal: lowers fissile density, raising critical mass. Common Pitfalls: Assuming temperature changes dominate; boundary neutron economy is far more impactful.Confusing moderators with reflectors; not all moderators are good reflectors in every spectrum. Final Answer: Surrounding it with a neutron-reflecting material (reflector)

Question 3

Uranium enrichment by gaseous diffusion: before diffusion through porous barriers, natural uranium is converted to which gaseous compound?

Accepted Answer

Introduction / Context: Isotopic enrichment increases the fraction of U-235 in uranium. Historical diffusion plants and many modern processes first convert uranium to a volatile compound so that mass transfer and separation can occur in the gas phase. Given Data / Assumptions: Gaseous diffusion or gas-centrifuge front-end chemistry. Need for a thermally stable, sufficiently volatile uranium compound at moderate temperatures. Compatibility with corrosion-resistant process materials (e.g., nickel, fluorinated alloys). Concept / Approach: UF6 (uranium hexafluoride) is uniquely suitable because it sublimes readily and remains molecular in the gas phase. The slight mass difference between 235UF6 and 238UF6 enables separation by diffusion or centrifugation. Other uranium compounds (oxides, carbides, sulfates) are nonvolatile under operating conditions and cannot be used in gaseous-diffusion trains. Step-by-Step Solution: Convert U3O8 (yellowcake) → UF6 via fluorination steps.Heat to obtain gaseous UF6 for separation.Pass through porous barriers (diffusion) or centrifuges (rotation) to exploit mass differences.Collect product and tails streams for further processing. Verification / Alternative check: Materials balance and phase-diagram data confirm UF6 is the only common gaseous uranium compound used industrially for enrichment, aligning with plant designs worldwide. Why Other Options Are Wrong: UO2, UC2, UO2SO4, UCl4: not suitably volatile under practical conditions for diffusion plants. Common Pitfalls: Confusing UF4 (green salt) used earlier in the chain with UF6 used for enrichment.Assuming oxide powders can be directly diffused; they cannot. Final Answer: Uranium hexafluoride (UF6)

Question 4

Monazite sand (rare-earth phosphate placer mineral): the typical uranium content by mass is approximately what percentage?

Accepted Answer

Introduction / Context: Monazite is a phosphate mineral rich in rare-earth elements and thorium, and it can contain trace uranium. Understanding typical assay values helps in resource evaluation and in distinguishing thorium-bearing sands from uranium ores. Given Data / Assumptions: Coastal placer monazite sands. Typical global averages; local deposits vary. Question seeks an order-of-magnitude estimate, not an exact assay. Concept / Approach: Monazite commonly contains significant thorium (often a few percent up to 7–10%), while uranium is much lower—generally a few tenths of a percent or less. Therefore, an estimate near 0.25% aligns with standard references and is widely used in exam keys to represent “trace-but-nonzero” uranium content. Step-by-Step Solution: Recall monazite composition: rare-earth phosphate with Th and minor U.Compare typical Th content (percent level) to U content (sub-percent).Select the closest sub-percent value among options.Hence, 0.25% is the accepted typical figure. Verification / Alternative check: Mining and mineral-processing texts list uranium in monazite on the order of several thousand ppm (i.e., a few tenths of a percent), corroborating the 0.25% choice, while thorium is notably higher. Why Other Options Are Wrong: 0.01%: too low for many monazite placers.1.2% and above (7%, 12%): characteristic of thorium, not uranium, in monazite. Common Pitfalls: Confusing uranium and thorium assay levels in monazite.Assuming a single deposit’s value is universal. Final Answer: 0.25%

Question 5

Isotopes and hydrogen: identify the incorrect statement among the following descriptions.

Accepted Answer

Introduction / Context: Isotopes are nuclides of the same element with identical atomic number but different mass numbers. Hydrogen is a classic teaching example because its isotopes—protium, deuterium, and tritium—are widely discussed in chemistry and nuclear engineering. Given Data / Assumptions: Neutral atoms considered (electrons equal protons). Natural abundance values are approximate. Periodic table placement is by atomic number Z. Concept / Approach: The key to spotting the incorrect statement is recalling that hydrogen has three well-known isotopes: 1H (protium), 2H or D (deuterium), and 3H or T (tritium). Therefore, the phrase “Hydrogen has two isotopes” is false. Deuterium abundance is roughly 0.015–0.016, giving an H:D ratio near 6400:1, which matches the figure given and is acceptable as an approximation. Step-by-Step Solution: Check isotope definition: same Z, different N.Hydrogen isotopes: protium, deuterium, tritium ⇒ not just two.Natural ratio H:D ≈ 6400:1 is reasonable.Conclusion: Option “Hydrogen has two isotopes” is incorrect. Verification / Alternative check: Standard references list three hydrogen isotopes; tritium is radioactive and less abundant in nature but well established, so counting only two is an error. Why Other Options Are Wrong: Option A: matches isotope definition.Option C: acceptable approximate ratio from natural abundance.Option E: isotopes share periodic position since Z is identical.Option D (“None of these”): cannot be correct because B is false. Common Pitfalls: Assuming tritium is excluded because it is radioactive; it is still an isotope.Confusing isotopes (same Z) with isobars (same A) or isotones (same N). Final Answer: Hydrogen has two isotopes.

Question 6

Isotopes of the same element differ in which fundamental quantity while sharing the same atomic number and (for neutral atoms) electronic configuration?

Accepted Answer

Introduction / Context: Isotopes are central to nuclear science and analytical chemistry. They are atoms of the same element (same atomic number Z) that vary in nucleon count, leading to different masses but essentially identical electronic structures for neutral atoms—hence similar chemistry. Given Data / Assumptions: Comparison at identical charge state (neutral). Chemical properties judged by electron configuration. Atomic number Z defines nuclear charge. Concept / Approach: Because isotopes have the same Z, they have the same number of electrons when neutral, so their electronic configuration and valence behavior are the same to first order. The distinguishing nuclear property is the number of neutrons, which changes the mass number A = Z + N. Small isotope effects in chemistry exist but do not alter the fundamental definition. Step-by-Step Solution: State isotope definition: same Z, different N.Identify differing quantity: mass number A changes with N.Check invariants: nuclear charge (Z) and neutral-atom electronic configuration remain the same.Conclude: “Mass number” is the correct distinguishing quantity. Verification / Alternative check: Periodic trends and chemical behavior are set by valence electrons; isotopes line up identically in the table because Z—hence valence configuration—does not change. Why Other Options Are Wrong: Electronic configuration: same for neutral isotopes of an element.Nuclear charge: by definition equal for isotopes.Chemical properties: essentially the same; only minor isotope effects occur.Valence electron count: identical for neutral atoms of the same element. Common Pitfalls: Confusing isotopes with ions (which do alter electron counts).Mixing up isotopes and isobars (same A, different Z). Final Answer: Mass number

Question 7

Control rods in nuclear reactors: which statements correctly describe their material choice and neutron-interaction requirement?

Accepted Answer

Introduction / Context: Control rods regulate reactor power by absorbing thermal (and sometimes epithermal) neutrons. Effective reactivity control requires materials with high absorption probabilities so that small rod motions produce measurable changes in neutron population. Given Data / Assumptions: Thermal reactor context (PWR/BWR/HWR). Standard rod materials and claddings. Goal: reliable shutdown and fine control during operation. Concept / Approach: Good control-rod materials combine a large microscopic absorption cross-section with acceptable mechanical, thermal, and radiation resistance. Boron (in B4C), silver–indium–cadmium (Ag–In–Cd), hafnium, and cadmium are classic choices because they remove neutrons efficiently, lowering reactivity (negative reactivity insertion). Step-by-Step Solution: Identify material requirement: high absorption cross-section desired.List common materials: B4C, Hf, Cd, Ag–In–Cd, dysprosium-titanate in some designs.Relate to function: inserting rods increases neutron captures, reducing k_eff.Thus, statements about boron/hafnium/cadmium and “large cross-section” are both correct. Verification / Alternative check: Shutdown margin and rod-worth calculations depend on these high capture cross-sections; licensing documents and vendor specs emphasize these materials for reactivity control. Why Other Options Are Wrong: Small cross-section (A): would make rods ineffective.Structural-only role (E): incorrect; nuclear absorption is their primary function. Common Pitfalls: Confusing control rods with burnable absorbers or soluble boron; all are absorbers but used differently.Ignoring temperature and radiation effects on rod materials over lifetime. Final Answer: Both (b) and (c)

Question 8

Among U-235, U-238, U-239, and Pu-239, which nucleus contains the largest number of neutrons?

Accepted Answer

Introduction / Context: Counting neutrons from nuclide notation A/Z X is a fundamental skill. The number of neutrons N equals A − Z, where A is mass number and Z is atomic number. Uranium has Z = 92 and plutonium has Z = 94, so comparing A values quickly reveals which has more neutrons. Given Data / Assumptions: Z(U) = 92 and Z(Pu) = 94. Mass numbers given: 235, 238, 239. We assume the common isotopes listed. Concept / Approach: Compute neutrons for each candidate with N = A − Z. Larger A at the same Z means more neutrons. Even when Z differs (U vs. Pu), a simple subtraction provides the neutron count directly. Step-by-Step Solution: U-235: N = 235 − 92 = 143.U-238: N = 238 − 92 = 146.U-239: N = 239 − 92 = 147.Pu-239: N = 239 − 94 = 145.Largest neutron count is 147 for U-239. Verification / Alternative check: Periodic knowledge confirms U has fewer protons than Pu; therefore, for the same A = 239, uranium must have more neutrons than plutonium (147 vs. 145), supporting the choice. Why Other Options Are Wrong: U-235: 143 neutrons—less than 147.U-238: 146 neutrons—still less than 147.Pu-239: 145 neutrons—less than U-239.U-233: 141 neutrons—much less than 147. Common Pitfalls: Mixing up A and Z or forgetting that Z differs between U and Pu.Assuming higher A always means more neutrons without checking Z. Final Answer: U-239

Question 9

During radioactive disintegration, beta (β) emission is characteristically accompanied by the emission of which particle or radiation?

Accepted Answer

Introduction / Context: Beta decay comes in two forms: β− (electron emission) and β+ (positron emission). Textbooks emphasize that lepton number and energy must be conserved, which introduced the neutrino concept historically. Many students mistakenly think γ rays always accompany β emission. Given Data / Assumptions: Typical nuclear decays as studied in introductory courses. Focus on what is “always” emitted with β decay. Gamma emission may occur from excited daughter nuclei but is not mandatory. Concept / Approach: In β− decay, a neutron converts to a proton with the emission of an electron and an antineutrino. In β+ decay, a proton converts to a neutron with the emission of a positron and a neutrino. Thus, a (anti)neutrino is the unavoidable partner in β processes; γ emission is optional and occurs only if the daughter nucleus remains excited. Step-by-Step Solution: β−: n → p + e− + anti-ν (antineutrino).β+: p → n + e+ + ν (neutrino).“Always” clause: neutrino/antineutrino is required; γ is not guaranteed.Given options do not include neutrino explicitly as the correct choice among A–D, hence “None of these.” Verification / Alternative check: Energy spectra of β particles are continuous because energy is shared with the (anti)neutrino, confirming its obligatory presence; discrete γ lines are not a universal feature of β decay. Why Other Options Are Wrong: Gamma rays: common but not guaranteed; many β decays go straight to ground state.Alpha particles and neutrons: define different decay modes altogether.Option E explains the truth but is not among A–D; the keyed answer among A–D is therefore “None of these.” Common Pitfalls: Memorizing “β always with γ” without the neutrino caveat.Confusing “often” with “always” in exam phrasing. Final Answer: None of these

Question 10

Structural difference between U-238 and U-235: by how many neutrons do these uranium isotopes differ?

Accepted Answer

Introduction / Context: Nuclide notation allows quick comparison of isotopes. Uranium’s atomic number is Z = 92 for all uranium atoms, so any difference between U-238 and U-235 must arise from the neutron count, not the proton count or electron configuration (for neutral atoms). Given Data / Assumptions: Uranium atomic number Z = 92. Mass numbers: 238 and 235. Neutral atoms considered for electron counts. Concept / Approach: The number of neutrons N is found by N = A − Z. Because Z is fixed for uranium, changes in A directly indicate changes in neutron number. Electrons adjust only with ionization, not isotope identity; protons define the element and therefore do not change among isotopes. Step-by-Step Solution: Compute for U-238: N = 238 − 92 = 146.Compute for U-235: N = 235 − 92 = 143.Difference: 146 − 143 = 3 neutrons.Thus, the isotopes differ by three neutrons, not by protons or electrons. Verification / Alternative check: All isotopes of uranium share Z = 92; any periodic table confirms this. Consequently, differing mass numbers must reflect neutron differences. Why Other Options Are Wrong: Protons: would change the element (not uranium anymore).Electrons: equal to protons in neutral atoms; isotope identity does not set ionization.Combined changes in electrons/protons: contradict isotope definition. Common Pitfalls: Confusing isotopes (same Z) with isobars (same A but different Z).Assuming electron count differs for isotopes; it does not for neutral atoms. Final Answer: Neutrons

Question 11

Fundamental particle masses: given that the mass of a neutron is about 1839 times the mass of an electron, what is the approximate ratio of the mass of a proton to the mass of an electron (use the standard rounded textbook value)?

Accepted Answer

Introduction / Context: The relative masses of the three basic sub-atomic particles (electron, proton, neutron) are staple facts in physics and chemistry. They underpin atomic structure, isotope behavior, spectroscopic shifts, and many order-of-magnitude estimates in nuclear and particle problems. This question asks you to recall (or quickly estimate) the proton-to-electron mass ratio when you are told the neutron-to-electron ratio for reference. Given Data / Assumptions: Neutron mass ≈ 1839 times electron mass (rounded classroom value). We need the proton mass ratio relative to the electron. Use widely taught rounded integers suitable for MCQ contexts. Concept / Approach: Standard constants provide these approximate ratios: m_p / m_e ≈ 1836 to 1837 and m_n / m_e ≈ 1839. The neutron is slightly heavier than the proton, so the proton ratio must be a little less than the neutron ratio. Textbook MCQs commonly offer 1836 or 1837; when one choice is 1837, it matches the expected rounded value and the inequality m_p < m_n remains consistent. Step-by-Step Solution: Identify known rounded ratios: m_p / m_e ≈ 1836–1837; m_n / m_e ≈ 1839.Compare: proton is slightly lighter than neutron, so ratio should be just below 1839.Scan the options and select 1837 as the nearest standard rounded value.Reject values that are implausibly small (159), excessively large (10000), or nonstandard (2537, 918.5). Verification / Alternative check: From physical constants, m_p ≈ 1.6726e−27 kg, m_e ≈ 9.109e−31 kg → m_p / m_e ≈ 1836.15. Rounded MCQ keys often use 1836 or 1837; with neutron given as 1839, picking 1837 maintains correct ordering m_e < m_p < m_n. Why Other Options Are Wrong: 159: Off by an order of magnitude; does not reflect known constant.2537: Arbitrary and far from accepted value.10000: Physically unrealistic for m_p / m_e.918.5: Roughly half the correct magnitude. Common Pitfalls: Memorizing only one ratio (e.g., neutron/electron) and assuming the proton has the same factor; forgetting the proton is slightly lighter than the neutron; or mixing up significant figures beyond what MCQs require. Final Answer: 1837

Question 12

Nuclear stability insight: an element that has a comparatively large number of which constituents is most readily susceptible to nuclear fission under neutron bombardment?

Accepted Answer

Introduction / Context: Nuclear fission behavior depends on nuclear composition and binding energy trends with mass and neutron-to-proton ratio. Heavier nuclei with relatively more neutrons tend to have lower binding energy per nucleon and are closer to instability thresholds, making them more susceptible to fission when perturbed by additional neutrons. Given Data / Assumptions: We compare which constituent count (protons, electrons, neutrons) most favors fission. Typical reactor-relevant fission involves heavy nuclides (U-235, Pu-239) absorbing neutrons. External electron configuration is largely irrelevant to nuclear fission thresholds. Concept / Approach: Fission is governed by nuclear forces, not electronic structure. Heavy elements with high neutron numbers often have larger neutron excess (n − z), which can make the nucleus less tightly bound and more likely to split after absorbing another neutron. A high neutron count also reduces Coulomb repulsion issues since neutrons add to mass without adding charge, allowing heavy nuclei to exist but at the cost of increased propensity to fission under excitation. Step-by-Step Solution: Recognize that electrons do not determine nuclear fission susceptibility.Protons add Coulomb repulsion; too many protons without compensating neutrons destabilize but do not alone favor controlled fissionability.A relatively larger number of neutrons correlates with heavy, fissile or fertile nuclei that can undergo fission after neutron capture.Choose “Neutrons” as the controlling constituent in this qualitative comparison. Verification / Alternative check: Isotopes like U-235 and Pu-239 have substantial neutron excess compared with lighter elements. The capture of an additional neutron creates a compound nucleus at higher excitation that can overcome the fission barrier. Why Other Options Are Wrong: Protons: Excess charge increases Coulomb repulsion but does not alone favor induced fission readiness.Electrons: Atomic electrons are essentially irrelevant to the nuclear fission mechanism.Nucleons “in equal proportion”: Stable light nuclei are near 1:1; easy fission is a heavy-nuclei property with neutron excess.Positrons: Not nuclear constituents of stable matter. Common Pitfalls: Conflating chemical properties with nuclear stability; forgetting that fission is driven by nuclear composition and excitation energy, not electron shells. Final Answer: Neutrons

Question 13

Control rods in nuclear reactors: identify the primary purpose—what do they directly control to regulate reactor power safely during operation?

Accepted Answer

Introduction / Context: Control rods are the “throttle” of a nuclear reactor. By absorbing neutrons, they determine how many neutrons remain available to trigger further fissions. This directly governs the multiplication factor and thus reactor power, ensuring safe start-up, load follow, and shutdown. Given Data / Assumptions: Control rods are made of strong neutron absorbers (e.g., boron, cadmium, hafnium). Reactivity k_eff depends on neutron population and losses. Goal is to manage the self-sustaining chain reaction. Concept / Approach: Reactor power scales with the rate of fission events, which depends on the number of neutrons that survive capture and leakage to induce new fissions. Inserting rods increases absorption, reducing available neutrons; withdrawing rods does the opposite. Rods do not primarily control radiation emissions directly, nor do they convert materials; they act on neutron economy. Step-by-Step Solution: Relate power to neutron population and effective multiplication factor.Recognize control rods change reactivity by absorbing neutrons.Conclude their main role is regulating the chain reaction and, therefore, power output.Select the option that explicitly states this neutron-control mechanism. Verification / Alternative check: Start-up procedures, scram (rapid shutdown), and power maneuvers all rely on rod insertion/withdrawal to alter neutron absorption and keep k_eff at or below 1 as required. Why Other Options Are Wrong: Radiation emission: a consequence of core physics, not directly “controlled” by rods at the boundary.Conversion to fertile/fissile: a blanket/fuel-cycle process, not a rod function.Changing only velocity: rods change number via absorption; spectrum shaping is secondary.Temperature-only control: temperature feedback exists but rods act on neutrons, not temperature in isolation. Common Pitfalls: Assuming rods are like thermal regulators; they change neutron balance first, with temperature responding as a result of power changes. Final Answer: The chain reaction and hence power by regulating the number of secondary neutrons causing fission

Question 14

In heavy-water reactor technology, heavy water (D2O) primarily serves which roles within the nuclear reactor system?

Accepted Answer

Introduction / Context: Heavy water reactors, notably Pressurised Heavy Water Reactors (PHWRs), rely on deuterium oxide (D2O) to achieve efficient thermal-neutron spectra while allowing the use of natural or low-enriched uranium fuel. Understanding the dual role of D2O clarifies why these designs differ from light-water reactors. Given Data / Assumptions: Heavy water has low neutron absorption and effective moderation properties. In PHWRs, D2O appears both in the moderator system and (often) in the primary coolant circuit. Question asks for the combined functional roles. Concept / Approach: Moderation slows fast fission neutrons to thermal energies to enhance fission probability in U-235. Deuterium has a small absorption cross-section, enabling high neutron economy. In PHWR layouts, D2O moderates in a separate calandria vessel and also circulates as coolant in pressure tubes, carrying heat from the fuel to the steam generators. It is not intended as a neutron “poison”; rather, its low absorption is a key advantage. Step-by-Step Solution: Identify moderation function: D2O slows neutrons with minimal losses.Identify cooling function: D2O removes fission heat in pressure tubes.Confirm dual role in canonical PHWR designs (e.g., CANDU lineage).Select “Both coolant and moderator.” Verification / Alternative check: Standard PHWR schematics show separate moderator calandria filled with D2O and a D2O primary heat transport system, confirming the dual functionality in many implementations. Why Other Options Are Wrong: Coolant only / Moderator only: incomplete—PHWRs commonly use D2O for both roles.Neutron absorber: contradicts low absorption property of deuterium.Gamma shield: shielding is mainly via dense materials and concrete, not D2O as a prime function. Common Pitfalls: Confusing PHWR (heavy water) with PWR/BWR (light water) where H2O acts as both coolant and moderator; in PHWRs, the moderator and coolant are both D2O but often in separated circuits. Final Answer: Both coolant and moderator

Question 15

Fissile vs fertile: which of the following is a non-fissile (fertile) material under thermal-neutron conditions?

Accepted Answer

Introduction / Context: Nuclear fuel classification distinguishes fissile isotopes (which readily fission with thermal neutrons) from fertile isotopes (which do not, but can be converted into fissile species by neutron capture and subsequent decay). Recognizing which is which is fundamental for reactor fuel-cycle questions and breeder concepts. Given Data / Assumptions: Thermal-neutron regime is implied unless otherwise noted. Common fissile isotopes include U-235, U-233, Pu-239, Pu-241. Common fertile isotopes include Th-232 and U-238. Concept / Approach: Thorium-232 is fertile: it captures a neutron to become Th-233 → Pa-233 → U-233 (fissile). In contrast, U-235, U-233, Pu-239, and Pu-241 are fissile and support thermal-neutron chain reactions. Therefore, within the given list, Th-232 is the non-fissile (fertile) choice. Step-by-Step Solution: List fissile isotopes: U-235, U-233, Pu-239, Pu-241.Identify Th-232 as fertile, not fissile, under thermal conditions.Select Thorium-232 accordingly.Note that fertile isotopes are essential in breeding strategies to produce new fissile fuel. Verification / Alternative check: Fuel-cycle diagrams universally show Th-232 breeding to U-233 and U-238 breeding to Pu-239 via neutron capture and beta decays; both are classic fertile pathways. Why Other Options Are Wrong: Pu-239 / Pu-241: well-known fissile isotopes.U-235: primary fissile constituent of natural uranium.U-233: fissile product bred from Th-232. Common Pitfalls: Confusing “fissile” with “fertile” due to similar wording; remember fertile needs conversion, while fissile sustains thermal-neutron fission directly. Final Answer: Thorium-232

Question 16

Isotopes refresher: for neutral atoms of isotopes of the same element, which property is different while the chemical identity remains the same?

Accepted Answer

Introduction / Context: Isotopes are variants of the same chemical element that differ in mass but not in elemental identity. This concept explains phenomena from radioisotopic labeling to nuclear stability trends and is basic to both chemistry and nuclear engineering. Given Data / Assumptions: “Isotopes” share the same atomic number Z. Neutral atoms of isotopes have the same number of electrons as protons. The mass (A) differs because the neutron count differs. Concept / Approach: Atomic number Z equals the number of protons and defines the element (e.g., Z = 92 → uranium). Isotopes of an element share Z but have different neutron counts, altering the mass number A, hence the atomic weight. In neutral atoms, electrons equal protons, so electron number is unchanged. Therefore, the property that differs is atomic weight (mass number). Step-by-Step Solution: Fix identity: same Z ⇒ same element and same proton count.Neutral atoms: electrons equal protons ⇒ same electron count.Only neutron count varies ⇒ different A and atomic weight.Select “Atomic weights (mass numbers).” Verification / Alternative check: Examples: C-12 and C-13; U-235 and U-238. Element is unchanged, chemistry is similar, but nuclear properties and masses differ. Why Other Options Are Wrong: Atomic number / protons: same for isotopes.Electrons (neutral): same as protons; therefore unchanged.Element identity: determined by Z, which is unchanged. Common Pitfalls: Confusing isotopes (same Z, different A) with isobars (same A, different Z) or isotones (same N, different Z). Final Answer: Atomic weights (mass numbers)

Question 17

History of discovery: who first reported the experimental observation of nuclear fission in 1938, opening the path to chain-reaction physics?

Accepted Answer

Introduction / Context: Nuclear fission’s experimental discovery transformed physics and engineering, enabling both civilian nuclear power and nuclear weapons. Knowing the key contributors and timeline is common in GK and engineering entrance exams. Given Data / Assumptions: Year of discovery: 1938. Need the experimental observers who identified fission products. Other scientists listed contributed to related fields or theoretical insights. Concept / Approach: Otto Hahn and Fritz Strassmann bombarded uranium with neutrons and chemically identified barium among the products, implying the uranium nucleus had split. Lise Meitner and Otto Frisch provided the correct physical interpretation and coined “fission,” while Niels Bohr and John Wheeler developed theoretical frameworks. Rutherford pioneered nuclear physics earlier; Fermi led neutron physics and chain-reaction experiments subsequently. The first experimental observation belongs to Hahn and Strassmann. Step-by-Step Solution: Recall timeline: late 1938 chemical identification of barium from irradiated uranium.Associate the lab work with Hahn and Strassmann.Recognize interpretive contributions by Meitner and Frisch soon after.Select “Hahn and Strassmann.” Verification / Alternative check: Historical accounts consistently credit Hahn–Strassmann with the experimental proof; Meitner–Frisch gave the theoretical explanation and name; Fermi later achieved sustained chain reaction (1942). Why Other Options Are Wrong: Planck: Quantum theory pioneer, not fission discoverer.Rutherford: Earlier nuclear scattering & nucleus discovery, not fission observation.Fermi: Chain reaction and neutron physics, not the first fission observation.Bohr and Wheeler: Theoretical analysis, not the 1938 experiment. Common Pitfalls: Conflating first observation with later theoretical explanations or with the first controlled chain reaction (Chicago Pile-1, 1942). Final Answer: Hahn and Strassmann

Question 18

Fast breeder reactor essentials: which option best summarizes core features of a typical fast breeder reactor (FBR) used in nuclear power R&D?

Accepted Answer

Introduction / Context: Fast breeder reactors (FBRs) are designed to operate with a fast-neutron spectrum and breed more fissile fuel than they consume. They are central to long-term fuel-cycle strategies because they can convert abundant fertile isotopes into useful fissile material while generating electricity. Given Data / Assumptions: Fast spectrum implies no moderator. Fertile blanket commonly uses U-238 → Pu-239 breeding. Liquid sodium is the canonical coolant due to high thermal conductivity and low pressure operation. Concept / Approach: FBR physics depends on maintaining high-energy neutrons; moderators would thermalise neutrons and defeat this purpose. The blanket captures excess neutrons to transmute U-238 to Pu-239, providing breeding. Sodium coolant supports excellent heat transfer and does not significantly slow neutrons, preserving the fast spectrum. Step-by-Step Solution: Confirm (a): fast neutrons are the defining feature.Confirm (b): breeding from U-238 to Pu-239 is standard.Confirm (c): molten sodium is a common FBR coolant.Therefore, choose “All of the above.” Verification / Alternative check: Design summaries of experimental and prototype FBRs (e.g., FBTR, BN-series, Phénix) consistently show these three features, with minor variations across programs. Why Other Options Are Wrong: Each single statement (a), (b), (c) is true but incomplete alone; the best summary is their combination.Heavy water moderation: inconsistent with a fast spectrum; moderators are absent in FBRs. Common Pitfalls: Assuming “breeder” implies any moderated reactor with a fertile blanket; true fast breeders must avoid moderators to maintain fast-neutron physics. Final Answer: All of the above: (a), (b) and (c)

Question 19

Hydrogen isotopes: how many neutrons are present in the nucleus of tritium (written as 1H3)?

Accepted Answer

Introduction / Context: Isotope notation quickly conveys the composition of a nucleus. Hydrogen’s three common isotopes—protium, deuterium, and tritium—are classic examples for counting protons and neutrons and practicing the relation between mass number and atomic number. Given Data / Assumptions: Tritium symbol: 1H3 (sometimes written as H-3). Atomic number Z for hydrogen = 1 (protons). Mass number A = 3 (protons + neutrons). Concept / Approach: For any nuclide, neutrons N = A − Z. For tritium, A = 3 and Z = 1, so N = 3 − 1 = 2. Protium (1H1) has 0 neutrons; deuterium (1H2) has 1 neutron; tritium (1H3) has 2 neutrons. This counting is fundamental to nuclear reaction bookkeeping and decay chain analysis. Step-by-Step Solution: Write N = A − Z.Substitute A = 3 and Z = 1.Compute N = 3 − 1 = 2.Select “2.” Verification / Alternative check: Textbook tables for hydrogen isotopes confirm the neutron counts: 0, 1, and 2 for protium, deuterium, tritium respectively. Why Other Options Are Wrong: 1 / 3 / 5: Incorrect arithmetic from N = A − Z for 1H3.0: This would correspond to protium, not tritium. Common Pitfalls: Mixing up mass number with atomic number or forgetting that the hydrogen symbol “H” always has Z = 1 regardless of isotope. Final Answer: 2

Question 20

Calder Hall–type gas-cooled reactor (Magnox): the circulating carbon-dioxide coolant typically operates at approximately what pressure (expressed in kgf/cm^2) under standard reference descriptions?

Accepted Answer

Introduction / Context: Early British gas-cooled reactors (Magnox type, exemplified by Calder Hall) used carbon dioxide as the primary coolant and graphite as moderator. Recognizing typical operating pressure ranges is valuable for quick design-type identification and safety or materials questions in general knowledge exams. Given Data / Assumptions: Coolant: CO2 gas in a graphite-moderated, gas-cooled reactor. Question requests a representative pressure in kgf/cm^2 for exam purposes. We select the commonly cited textbook value used in legacy MCQs. Concept / Approach: Magnox reactors operated CO2 at pressures substantially above atmospheric but far below the very high system pressures of pressurised water reactors. Typical GK sources quote a single-digit to low-tens value in kgf/cm^2 for such early designs. Among the choices, 7 kgf/cm^2 aligns with widely taught figures in classic exam prep material, indicating moderately pressurised gas flow adequate for heat removal without the high structural demands of water at dozens of bar. Step-by-Step Solution: Identify reactor type: gas-cooled, graphite moderated (Magnox/Calder Hall lineage).Recall representative CO2 system pressure from standard GK references.Map to options: 7 kgf/cm^2 is the standard single-digit pressure often quoted.Select “7.” Verification / Alternative check: Comparative reasoning: water-cooled reactors operate at many tens to over a hundred bar; early CO2-cooled reactors used markedly lower pressures. Among the MCQ options, 7 is the consistent teaching value. Why Other Options Are Wrong: 0.2 / 0.02: Near-atmospheric; inadequate for efficient CO2 heat transport.35 / 50: Much higher than typical GK values taught for early Magnox systems. Common Pitfalls: Assuming gas-cooled implies extremely low pressure like air-cooled systems; in practice, CO2 pressure is elevated to improve density and heat capacity while maintaining manageable mechanical loads. Final Answer: 7

Nuclear Power Engineering Questions