Question 1

In nuclear reactor engineering, an ideal coolant should enable efficient heat removal while remaining chemically compatible with materials: which property best captures this requirement?

Accepted Answer

Introduction / Context: An ideal nuclear reactor coolant must remove fission heat safely and efficiently without undermining the neutron economy or damaging plant materials. Besides thermal performance, chemical stability and radiation tolerance are crucial because coolants are exposed to intense neutron and gamma fields and contact structural alloys, fuel cladding, and seals for long periods. Given Data / Assumptions: We are considering generic reactor-coolant requirements, not a specific design. Key concerns include neutron absorption, thermal capacity, heat-transfer coefficient, viscosity, corrosion, and radiation stability. “Ideal” emphasises safety and compatibility as well as performance. Concept / Approach: A practical coolant should have low neutron absorption (to preserve reactivity), high specific heat and good thermal conductivity (to carry heat effectively), low viscosity (to limit pumping power), a high boiling point or high allowable temperature (to widen the operating window), and—critically—chemical compatibility with structural materials and resistance to radiolysis and radiation-induced degradation. Among competing statements, the property that most universally safeguards integrity and availability is freedom from radiation damage and non-corrosiveness. Step-by-Step Solution: Identify essential coolant traits: thermal efficiency, neutronic benignity, chemical compatibility, radiation stability.Evaluate each option against these traits.Select the option that is always beneficial and broadly necessary across reactor types.Conclude that “free from radiation damage and non-corrosive” is the best overarching requirement. Verification / Alternative check: Water (light or heavy) is favoured for many designs due to chemical manageability, while liquid metals (e.g., sodium) are selected for high-temperature performance; in all cases, corrosion control and radiation stability define maintenance cycles, component life, and safety margins. Thus, compatibility is a universal criterion. Why Other Options Are Wrong: Be a good absorber of neutrons: High absorption is undesirable; it poisons the neutron economy. Attain high temperature only when highly pressurised: Pressure dependence is a system design constraint, not an “ideal” coolant property. Have very high density but a low heat-transfer coefficient: Low heat-transfer coefficient is detrimental to heat removal. Exhibit high viscosity: Increases pumping power and reduces heat-transfer coefficients. Common Pitfalls: Overrating thermal metrics while ignoring long-term material compatibility; assuming any one property (e.g., density) dominates overall performance; neglecting radiolysis and corrosion product transport. Final Answer: Be free from radiation damage and be non-corrosive

Question 2

Atomic structure concept check: isotopes of the same chemical element share which identical property?

Accepted Answer

Introduction / Context: Isotopes are variants of a chemical element that differ in the number of neutrons but possess the same number of protons. Because chemical behaviour arises from electron arrangement determined by nuclear charge (atomic number), isotopes of an element show the same ground-state electronic configuration in neutral atoms and therefore very similar chemistry. Given Data / Assumptions: Isotopes: same atomic number Z, different neutron count N. Neutral atoms considered (no ionisation states) to discuss configuration cleanly. Chemical properties principally depend on electron configuration. Concept / Approach: The atomic number fixes the electron count of a neutral atom and thus its electronic configuration. Changing neutrons alters mass number (A = Z + N) and nuclear stability but not the electronic shell structure for the neutral atom. Consequently, isotopes have essentially identical valence structures and similar chemical reactions, with small kinetic isotope effects in some cases. Step-by-Step Solution: Define isotope: same Z, different N.Infer that A (mass number) changes with N.Recognise electronic configuration depends on Z for neutral atoms.Therefore, isotopes share electronic configuration. Verification / Alternative check: Consider chlorine-35 and chlorine-37: both have Z = 17, hence identical electron distribution (1s^2 2s^2 2p^6 3s^2 3p^5) and similar chemistry; only mass-dependent properties (diffusion rates, vibrational frequencies) vary slightly. Why Other Options Are Wrong: Number of neutrons: By definition, different. Mass number: Different because N differs. Atomic weight: A weighted average over isotopic distribution; not the same for individual isotopes and certainly not identical across isotopes. Nuclear mass exactly equal: Mass differs with neutron count. Common Pitfalls: Confusing atomic weight (periodic table average) with mass number of a single isotope; forgetting that ions can alter electron counts—here we are comparing neutral atoms for configuration. Final Answer: Electronic configuration

Question 3

Nuclear reactor classification: a heterogeneous reactor is one in which the moderator and the fuel are present in different phases or distinct regions. Which option best matches this definition?

Accepted Answer

Introduction / Context: Reactors are often categorised as homogeneous or heterogeneous based on how fuel and moderator are arranged. In a heterogeneous reactor, fuel is typically in discrete elements (rods, plates, pebbles) embedded in or surrounded by a moderator, yielding spatially separate regions and often different physical phases. Given Data / Assumptions: “Heterogeneous” refers to spatial/phase separation between fuel and moderator. Examples include solid fuel rods in graphite or heavy/light water moderators. Coolant may be the moderator or a separate fluid; this does not define heterogeneity by itself. Concept / Approach: The defining feature is that fuel and moderator are not uniformly mixed at the microscopic level. Homogeneous systems (e.g., solutions of fuel salts in molten salt reactors) contrast with solid-fuel rods in a fluid moderator. Thus, an option explicitly stating fuel and moderator are in different phases/regions captures the definition most directly. Step-by-Step Solution: Identify the key terms: “moderator” and “fuel.”Check which option states they are in different phases/regions.Select the option with explicit example: solid uranium with liquid heavy water. Verification / Alternative check: Graphite-moderated, gas-cooled reactors (solid fuel, solid moderator) are still heterogeneous because fuel elements and moderator blocks are distinct regions; phase difference is a common but not exclusive indicator of heterogeneity—distinct regions suffice. Why Other Options Are Wrong: Moderator and coolant different materials: Common but not definitional. Moderator and coolant different phases: Irrelevant to heterogeneity between fuel and moderator. None of these: Incorrect since (c) matches the definition. All same phase requirement: Opposite of heterogeneity. Common Pitfalls: Equating “heterogeneous” with “different materials anywhere”; the distinction is between fuel and moderator arrangements, not coolant identity alone. Final Answer: Moderator and fuel are present in different phases (e.g., solid uranium fuel with liquid heavy water moderator)

Question 4

Conceptual comparison: nuclear fission is most directly the opposite process to which of the following?

Accepted Answer

Introduction / Context: Nuclear energy can be released either by splitting heavy nuclei (fission) or by joining light nuclei (fusion). These processes occupy opposite ends of the binding-energy-per-nucleon curve: very heavy and very light nuclei can move toward the iron peak, releasing energy when transitioning to more tightly bound configurations. Given Data / Assumptions: Fission: a heavy nucleus (e.g., U-235) splits into lighter fragments plus neutrons. Fusion: light nuclei (e.g., deuterium and tritium) combine to form heavier nuclei. Energy release governed by changes in nuclear binding energy. Concept / Approach: On the mass–energy landscape, fission moves from heavy to intermediate masses; fusion moves from light to intermediate masses. Both release energy when products have higher binding energy per nucleon than the reactants. Therefore, the “opposite” of fission is fusion, not generic radioactive decay or non-nuclear phenomena. Step-by-Step Solution: Define fission and fusion in terms of nuclear mass numbers.Relate energy release to binding-energy differences.Identify the direct inverse transformation: heavy-to-lighter versus light-to-heavier.Select nuclear fusion as the opposite process. Verification / Alternative check: Binding-energy curves show a peak around iron–nickel; moving toward this peak from either side (heavy via fission, light via fusion) releases energy, reinforcing the conceptual opposition of the two processes. Why Other Options Are Wrong: Radioactive decay: Includes alpha, beta, gamma processes; not simply the inverse of fission. Thermionic emission: Electron emission from heated metals; not nuclear. Combustion: Chemical oxidation; unrelated to nuclear binding energies. Photoelectric emission: Quantum electron ejection; not nuclear. Common Pitfalls: Conflating “radioactivity” broadly with all nuclear processes; forgetting that decay can also occur in fission fragments but is not the inverse of fission itself. Final Answer: Nuclear fusion

Question 5

Thermal (slow-neutron) reactor versus fast breeder reactor: which statement correctly contrasts their neutron energy usage and typical characteristics?

Accepted Answer

Introduction / Context: Nuclear reactors are often categorised by the neutron spectrum in which most fissions occur. Thermal reactors rely on slow (thermalised) neutrons, typically using moderators such as light water, heavy water, or graphite. Fast breeder reactors operate with fast neutrons and are optimised to convert fertile isotopes (e.g., U-238) into fissile ones (e.g., Pu-239). Given Data / Assumptions: Thermal reactors: moderated spectrum; fuel commonly low-enriched uranium. Fast breeders: minimal or no moderation; liquid-metal coolants such as sodium are common. Comparisons are qualitative and typical rather than absolute for every design. Concept / Approach: Slow neutrons have higher fission cross-sections for isotopes like U-235, enabling stable chain reactions with moderation. Fast systems exploit higher-energy neutrons to achieve breeding ratios greater than one, often resulting in higher power density cores but stricter material and heat removal constraints. Step-by-Step Solution: Identify the defining neutron spectrum for thermal reactors: slow/thermalised.Recognise fast breeders use unmoderated or weakly moderated fast neutrons.Select the statement aligning with these fundamentals. Verification / Alternative check: Design documents and introductory reactor physics texts consistently define thermal reactors by their moderated neutron spectrum; this is independent of specific fuel cycles or coolants used. Why Other Options Are Wrong: Uses faster neutrons: Opposite of thermal definition. Higher power density for thermal vs fast breeder: Generally fast cores achieve higher power density. Requires less fuel at same power: Breeders are designed for fuel efficiency and conversion. Converts more fertile material: Breeders outperform thermal systems in conversion ratio. Common Pitfalls: Assuming coolant choice alone determines neutron spectrum; it is moderation and core design that dominate spectrum characteristics. Final Answer: A thermal reactor uses slower neutrons for fission

Question 6

Natural uranium composition check: what percentage of the fissile isotope U-235 is present in natural uranium?

Accepted Answer

Introduction / Context: Natural uranium is primarily U-238 with a small fraction of U-235, the fissile isotope that sustains chain reactions in many reactor types. Understanding this baseline composition is essential for topics such as enrichment, fuel fabrication, and heavy-water moderated reactor choices. Given Data / Assumptions: Refers to naturally occurring isotopic abundance prior to any enrichment. Values rounded to two decimal places for ease. Fissile here specifically means U-235 (ignoring minor U-234 content). Concept / Approach: Natural uranium contains about 0.71% U-235, about 99.28% U-238, and a trace of U-234. Light-water reactors typically require enriched fuel (e.g., 3–5% U-235), whereas heavy-water reactors can operate with natural uranium because the moderator has very low neutron absorption, preserving reactivity. Step-by-Step Solution: Recall standard isotopic abundances of natural uranium.Identify the U-235 percentage approximately equal to 0.71%.Select the matching option. Verification / Alternative check: The widespread use of enrichment facilities and the design of CANDU-type heavy-water reactors both hinge on this small U-235 fraction, corroborating the remembered value. Why Other Options Are Wrong: 6.31 and 12.73: Far above natural abundance; typical of enriched fuel ranges (though still low for some designs). 99.29: Closer to U-238 abundance, not U-235. 3.00: Represents low enrichment levels, not natural. Common Pitfalls: Confusing weight percent with atom percent; mixing up U-238 dominance with U-235 fraction. Final Answer: 0.71

Question 7

Reactor types and moderators: heavy water (D2O) is used as the moderator in which reactor type?

Accepted Answer

Introduction / Context: Moderators slow down neutrons to thermal energies to enhance fission probabilities in certain fuels. Heavy water (D2O) is an exceptionally effective moderator because of its very low neutron absorption while maintaining good moderating power. Given Data / Assumptions: We are matching reactor designs to their characteristic moderators. PWRs and BWRs employ ordinary light water as coolant and moderator. CANDU (Canada Deuterium Uranium) designs use heavy water as moderator—and often as coolant. Concept / Approach: Heavy water’s low absorption enables operation on natural uranium fuel. CANDU reactors exploit this to avoid enrichment while achieving adequate reactivity with pressure tubes and separate moderator calandria. Other listed designs rely on different moderator and coolant strategies. Step-by-Step Solution: Identify which standard design explicitly specifies D2O moderation.Recognise CANDU as the canonical heavy-water system.Select CANDU reactor. Verification / Alternative check: Public design descriptions consistently list D2O moderation for CANDU; PWR/BWR employ H2O; fast reactors avoid moderation and use liquid metals like sodium for cooling; graphite-moderated types use graphite, not D2O. Why Other Options Are Wrong: PWR/BWR: Light water moderators. Molten-sodium fast reactor: Minimal moderation; sodium is coolant. Graphite-moderated CO2-cooled: Graphite is the moderator. Common Pitfalls: Assuming all water-cooled reactors use the same moderator; overlooking the critical difference between light and heavy water in neutron absorption. Final Answer: CANDU reactor

Question 8

Properties of isotopes: isotopes of the same element differ in which fundamental quantity?

Accepted Answer

Introduction / Context: Isotopes are central to nuclear chemistry and applications such as radiometric dating, medical imaging, and reactor fuel cycles. They are nuclei of the same element with identical proton numbers but different neutron numbers, which changes their mass number and sometimes nuclear stability, while leaving electronic behaviour broadly similar in neutral atoms. Given Data / Assumptions: Same element implies same atomic number (same number of protons). Neutral atoms when discussing electron configurations. Chemical properties may show small isotope effects but remain qualitatively similar. Concept / Approach: Mass number A equals protons (Z) plus neutrons (N). Isotopes share Z and differ in N; therefore A differs. This alters nuclear mass and stability but not the fundamental charge that sets electron configuration for neutral atoms. Step-by-Step Solution: Note: isotopes ⇒ same Z, different N.Therefore A = Z + N differs.Electronic configuration depends on Z (same for neutral isotopes).Select “Mass number” as the differing quantity. Verification / Alternative check: Carbon-12 vs carbon-14 illustrate identical chemical valence with different nuclear mass and stability, confirming that mass number is the differentiator. Why Other Options Are Wrong: Electronic configuration: Same for neutral atoms of the same element. Nuclear charge/number of protons: By definition, equal in isotopes. All chemical properties: Not entirely different; differences are usually subtle kinetic isotope effects. Common Pitfalls: Confusing atomic weight averages with single-isotope properties; assuming large chemical differences where only small kinetic effects exist. Final Answer: Mass number

Question 9

Effect of beta (β) decay on nuclear composition: compared to the parent, the new nucleus formed after β-decay has which change?

Accepted Answer

Introduction / Context: Beta (β−) decay is a common nuclear transformation in which a neutron converts to a proton with the emission of an electron and an antineutrino. Understanding how β-decay affects atomic number and mass number is foundational in nuclear chemistry and decay-chain calculations. Given Data / Assumptions: We refer to β− decay (not β+ or electron capture). Products: daughter nucleus, electron (beta particle), and antineutrino. Neglect the tiny mass of the emitted electron relative to nucleon masses when discussing mass number. Concept / Approach: In β− decay, a neutron in the nucleus changes into a proton: n → p + e− + anti-ν. The number of protons (atomic number Z) increases by one, while the total number of nucleons (mass number A) remains the same because one neutron is replaced by one proton. Hence, atomic number increases, mass number is unchanged. Step-by-Step Solution: Start with parent nucleus (Z, A).After β− decay: (Z + 1, A) because neutron → proton.Compare to options: “More atomic number” matches exactly. Verification / Alternative check: Decay series diagrams consistently show β− steps moving one place to the right on the periodic table (higher Z) while staying in the same horizontal mass-number row (same A). Why Other Options Are Wrong: Less atomic number: Opposite direction; would match β+ or electron capture. Less or more atomic weight: Atomic/molar mass changes minimally; mass number unchanged. Lower mass number by one: Incorrect; A is constant in β− decay. Common Pitfalls: Confusing β− with β+; assuming mass number changes because a particle is emitted—mass number counts nucleons only. Final Answer: More atomic number

Question 10

Thermodynamic efficiency comparison at the same temperature and pressure limits: which reactor/coolant system is generally most efficient?

Accepted Answer

Introduction / Context: Thermal efficiency for power cycles is primarily governed by the temperature at which heat is added and rejected. Reactor systems that permit higher coolant and steam temperatures (for a given pressure constraint and material limits) can reach higher Rankine-cycle efficiencies. Given Data / Assumptions: Operating between comparable overall temperature and pressure limits for the core/steam cycle. Focus on thermodynamic potential, not licensing or economic factors. Heat-exchanger and materials performance are adequate for the stated limits. Concept / Approach: Liquid-metal coolants like sodium feature excellent thermal conductivity, low pumping power, and allow high core outlet temperatures at relatively low pressures, enabling higher steam temperatures and better thermal efficiencies. Water-cooled systems (PWR, BWR) operate at high pressures to prevent boiling (PWR) or at saturation conditions (BWR), which constrain peak steam temperatures. Gas-cooled systems can also achieve good temperatures, but pressure and heat-transfer limitations often reduce practical efficiencies compared with liquid-metal systems under the same constraints. Step-by-Step Solution: Recall efficiency trends: higher heat-addition temperature increases ideal Rankine efficiency.Identify coolant classes that safely support higher outlet temperatures.Recognise sodium-cooled reactors typically allow higher temperature operation.Select molten sodium cooled reactor as most efficient under the stated comparison. Verification / Alternative check: Fast sodium-cooled reactor concepts frequently target higher steam-generator outlet temperatures than light-water designs, improving cycle efficiency for similar pressure limits. This qualitative industry pattern supports the selection. Why Other Options Are Wrong: CO2 gas-cooled: Reasonable performance but generally less than liquid-metal systems for the same limits. PWR: High pressure but relatively lower steam temperature versus sodium-cooled potential. BWR: Saturation-temperature constraints limit main steam temperature. Heavy-water moderated, light-water cooled: Similar water constraints apply. Common Pitfalls: Equating pressure capability with efficiency; overlooking the role of achievable main steam temperature and heat-transfer characteristics. Final Answer: Molten sodium cooled reactor

Question 11

In nuclear engineering terminology, “heavy water” (D2O) is called “heavy” because it is the oxide of deuterium, the heavy isotope of hydrogen, which makes the molecule heavier than ordinary water (H2O). What is the most accurate reason for the term …

Accepted Answer

Introduction / Context: Heavy water, written as D2O, appears in many nuclear engineering applications, especially in reactors that use it as a moderator. Learners often mix up why it is called “heavy.” This explanation clarifies the nuclear-chemistry basis for the name and distinguishes it from merely being “denser.” Given Data / Assumptions: D symbolizes deuterium, the hydrogen isotope with one proton and one neutron. Heavy water formula: D2O; ordinary water formula: H2O. Deuterium roughly doubles the mass contribution of hydrogen in the molecule. Concept / Approach: The adjective “heavy” is rooted in isotopic composition, not thickness or viscosity. Because deuterium has an extra neutron relative to protium (normal hydrogen), D2O molecules are more massive than H2O molecules. This isotopic substitution yields a higher molecular weight and slightly higher density, but density is a consequence, not the defining reason for the term. Step-by-Step Solution: Identify isotopes: protium (H, mass ≈1) vs. deuterium (D, mass ≈2).Construct molecules: H2O vs. D2O.Compare molecular masses: D2O has greater mass because each hydrogen site is replaced by heavier deuterium.Conclude naming: “heavy” refers to the oxide of heavy hydrogen (deuterium), not to oxygen isotopes or viscosity. Verification / Alternative check: Heavy water’s physical properties (e.g., density about 10% higher than H2O at room temperature) are consistent with the isotopic mass increase. However, textbooks define “heavy water” specifically as deuterium oxide, confirming the naming basis is isotopic composition. Why Other Options Are Wrong: It is denser: true as a property, but not the naming reason.Heavy, viscous liquid: viscosity change is minor; not the naming origin.Heavier oxygen isotope: heavy water uses normal oxygen-16 unless specified; the “heavy” refers to hydrogen.Dissolved heavy salts: irrelevant to isotopic definition. Common Pitfalls: Confusing cause (isotopic substitution) with effect (density increase).Assuming “heavy” implies impurities or high viscosity. Final Answer: It is an oxide of deuterium (heavy hydrogen)

Question 12

At the Bhabha Atomic Research Centre (BARC) complex in Trombay (Mumbai), which of the following reactor types has historically been housed for experimentation and training purposes?

Accepted Answer

Introduction / Context: India’s nuclear ecosystem includes several specialized sites with different missions. The Bhabha Atomic Research Centre (BARC) at Trombay is the premier hub for nuclear research and development, materials, fuel cycles, and training—not a commercial power station. Understanding the role of Trombay helps aspirants correctly classify facilities in exam questions. Given Data / Assumptions: BARC Trombay’s mandate emphasizes research, development, and training. India’s commercial power reactors are spread across separate generating stations. Fast breeder development has been centered at Kalpakkam under a different unit. Concept / Approach: “Research reactor” refers to reactors designed for experiments, neutron activation, isotope production, and training rather than continuous grid electricity generation. BARC Trombay historically hosted research reactors (e.g., Apsara and subsequent units over time) aligned with this mission, distinct from large power reactors or heavy-water production plants. Step-by-Step Solution: Identify BARC’s mission: R&D and training.Map missions to reactor types: research reactors support experiments and isotope work.Exclude commercial power and breeder focus assigned to other sites.Therefore, the correct classification is “research reactor.” Verification / Alternative check: Standard references and Indian nuclear program overviews consistently list Trombay as a research complex, not a site for commercial power generation or heavy water manufacture (which are managed by other entities/sites). Why Other Options Are Wrong: Power reactor: those are located at designated nuclear power stations.Fast breeder reactor: development focused elsewhere (e.g., Kalpakkam).Heavy water plant: handled by production units at other industrial sites.Only fuel reprocessing: BARC does much more, including research reactors. Common Pitfalls: Assuming every nuclear campus houses power reactors.Confusing Trombay’s R&D mission with production operations. Final Answer: Research reactor

Question 13

Among the following nuclides, which is considered the best fissile nuclear fuel for thermal reactors under standard conditions?

Accepted Answer

Introduction / Context: Fissile fuels sustain a chain reaction with thermal (slow) neutrons in conventional reactor designs. Recognizing which nuclides are truly fissile and practical is foundational for nuclear engineering exams and industry practice. Given Data / Assumptions: Comparison is in the context of standard thermal reactor conditions. Focus on fissile character, fission cross section with thermal neutrons, and practical fuel-cycle use. Concept / Approach: Fissile nuclides include U-235, Pu-239, and U-233 (not in the list). Among the given options, Pu-239 is a well-known fissile isotope produced from neutron capture in U-238 followed by beta decays. It has a high thermal fission cross section and favorable neutron economy, making it a cornerstone of mixed-oxide (MOX) fuels and breeder strategies. Step-by-Step Solution: Evaluate each option’s fissility and practicality.Np-239: short-lived intermediate in Pu-239 production, not a practical fuel.U-234: not fissile under thermal neutrons; mainly a decay product and fertile in limited contexts.Pu-239: established fissile fuel with strong thermal fission characteristics.Th-236: not the key fissile; thorium breeds U-233 which is fissile.Therefore, Pu-239 is best among listed. Verification / Alternative check: Fuel-cycle literature confirms MOX fuels (UO2 + PuO2) leverage Pu-239’s fissile nature in LWRs and other designs, validating its selection here. Why Other Options Are Wrong: Np-239: transient nuclide, not used as reactor fuel.U-234: poor thermal fissility.Th-236: thorium’s valuable product is U-233, not Th-236.Am-241 (if considered): used in specialty sources; not principal thermal fuel. Common Pitfalls: Confusing fertile (U-238, Th-232) with fissile (U-235, Pu-239, U-233).Selecting intermediates that appear in breeding chains. Final Answer: Pu-239

Question 14

In a pressurized water reactor (PWR), pressurized light water in the primary circuit performs which role(s)? Choose the statement that best captures its function.

Accepted Answer

Introduction / Context: Pressurized water reactors (PWRs) dominate the global fleet of light-water reactors. Understanding which role the primary-circuit water plays clears up frequent confusion between “coolant,” “moderator,” and the secondary steam cycle that drives the turbine. Given Data / Assumptions: Standard PWR design with a closed, high-pressure primary loop. Separate secondary loop where steam drives the turbine. No graphite or heavy-water moderation in a conventional PWR. Concept / Approach: In a PWR, light water (H2O) at high pressure circulates through the reactor core, removing fission heat (coolant function) and slowing down fast neutrons to thermal energies (moderator function). It is kept under pressure to avoid boiling in the primary loop. Heat is transferred to the secondary side via steam generators to produce steam for the turbine, keeping radioactive primary water out of the turbine system. Step-by-Step Solution: Identify primary-loop functions: heat removal + moderation.Confirm turbine working fluid: secondary-side steam, not primary water.Exclude graphite/heavy water: those are design features of other reactor types.Hence, the correct role is “both coolant and moderator.” Verification / Alternative check: Reactor schematics universally show primary pressurized water through core and steam generators, with moderation achieved by the same water in-core. The turbine loop is separate and non-radioactive (under normal operation). Why Other Options Are Wrong: Primary coolant only with graphite: incorrect for PWRs; graphite is used in other designs.Working fluid in turbine: wrong; that is secondary steam.Neither coolant nor moderator: contradicts PWR fundamentals.Emergency-only use: primary water is continuously used, not only in emergencies. Common Pitfalls: Confusing PWR with BWR (where water boils in the core).Assuming the same loop drives the turbine in all LWRs. Final Answer: It serves as both the coolant and the neutron moderator in the primary circuit

Question 15

Uranium is obtained from its ores primarily by which broad method, considering real-world hydrometallurgical practice?

Accepted Answer

Introduction / Context: Nuclear fuel begins as ores from which uranium must be extracted economically and safely. Recognizing the dominant industrial pathway helps distinguish practical flowsheets from purely academic alternatives. Given Data / Assumptions: Common uranium minerals: uraninite, coffinite, carnotite, etc. Ore grades may be low, necessitating selective extraction. Goal: produce a concentrate (often called yellowcake) for conversion and enrichment steps. Concept / Approach: The primary industrial route is hydrometallurgy: chemical leaching (acidic or alkaline) to dissolve uranium selectively, followed by solid–liquid separation and chemical recovery (e.g., solvent extraction, ion exchange, or precipitation). Physical beneficiation alone rarely suffices for low-grade ores, while direct pyrometallurgy to metallic uranium from ore is not the initial, mainstream approach. Step-by-Step Solution: Crush and grind ore to liberate minerals.Leach uranium with sulfuric acid or carbonate systems (depending on gangue).Separate solids from pregnant solution.Recover uranium chemically (e.g., SX/IX) and precipitate as a concentrate.Dry and package yellowcake for conversion/enrichment. Verification / Alternative check: Industry case studies and standards describe hydrometallurgical leach plants as the backbone of uranium production, with in-situ recovery (ISR) being a variant that is likewise chemical in nature. Why Other Options Are Wrong: Pyrometallurgy directly to metal: not the first industrial step from ore.Physical beneficiation only: insufficient for most uranium ores.Electrometallurgy first: niche, not the mainstream starting point.Fermentation without chemicals: not an established uranium practice. Common Pitfalls: Equating “beneficiation” with “extraction” in low-grade ores.Ignoring the strong role of solution chemistry in uranium recovery. Final Answer: Chemical method (hydrometallurgical leaching and recovery)

Question 16

Radioactive decay problem: If 75% of a radioactive sample decays in 6 hours (i.e., only 25% remains), what is the half-life of the nuclide, expressed in hours?

Accepted Answer

Introduction / Context: Half-life problems are staple quantitative questions in nuclear science. They require translating a percentage of remaining material into an exponential-decay relationship to solve for the characteristic time constant—the half-life (t_1/2). Given Data / Assumptions: After 6 hours, 75% has decayed, so 25% remains. Decay is first order: N(t) = N0 * (1/2)^(t / t_1/2). No growth or additional sources; closed system for the nuclide. Concept / Approach: Set the fraction remaining equal to (1/2)^(t / t_1/2). Solve algebraically for t_1/2 by taking logarithms or matching powers of 1/2. Because 25% = 1/4 = (1/2)^2, the exponent reveals how many half-lives elapsed. Step-by-Step Solution: Write remaining fraction: N/N0 = 0.25.Express 0.25 as power of 1/2: 0.25 = (1/2)^2.Equate exponents: (1/2)^(t / t_1/2) = (1/2)^2 ⇒ t / t_1/2 = 2.Solve for half-life: t_1/2 = t / 2 = 6 / 2 = 3 hours. Verification / Alternative check: Two half-lives reduce a sample to 25%: After first half-life (3 h) → 50%; after second (another 3 h) → 25%. This matches the 6-hour condition, confirming t_1/2 = 3 h. Why Other Options Are Wrong: 3/4 or 1/6 hours: not consistent with two full half-lives in 6 hours.4 or 6 hours: would leave fractions of 0.3536 or 0.5 after 6 h, not 0.25. Common Pitfalls: Using linear instead of exponential decay.Confusing “decayed” fraction (75%) with “remaining” fraction (25%). Final Answer: 3

Question 17

Why is the atomic mass of an element usually a fractional (non-integer) value on the periodic table?

Accepted Answer

Introduction / Context: Students often wonder why the “atomic weight” shown in periodic tables is not an integer, even though mass numbers (A) are integers. This distinction hinges on isotopic composition and natural abundance, not on quantum uncertainty or the mere presence of neutrons. Given Data / Assumptions: Real elements exist as mixtures of isotopes in nature. Each isotope has an exact atomic mass (not necessarily an integer) due to binding energies. Periodic tables list average atomic weights reflecting natural isotopic abundances. Concept / Approach: The tabulated “atomic mass” (or atomic weight) is calculated as a weighted average: sum over (isotope mass * fractional natural abundance). Since different isotopes contribute different masses, the average is typically a non-integer, even when one isotope dominates. Step-by-Step Solution: Identify stable isotopes of the element and their natural abundances.Multiply each isotope’s mass by its abundance fraction.Add these products to obtain the atomic weight.Observe that the result is generally fractional, reflecting the mixture. Verification / Alternative check: Chlorine has two common isotopes (approximately Cl-35 and Cl-37) with abundances near 75% and 25%, yielding an average around 35.5 u—clearly fractional and matching textbook tables. Why Other Options Are Wrong: Uncertainty principle: governs measurement limits, not average atomic weights.Isobars: same mass number across elements, not the reason for fractional averages.Presence of neutrons: all nuclei except protium have neutrons; does not cause the averaging.Electron binding energies: relatively small compared to nuclear masses. Common Pitfalls: Confusing mass number (integer) with atomic weight (average mass).Assuming laboratory-prepared isotopically pure samples match periodic values. Final Answer: Because it is a weighted average over naturally occurring isotopes

Question 18

What is a positron in the context of nuclear and particle physics?

Accepted Answer

Introduction / Context: Positrons appear in beta-plus decay and in pair production. Knowing their identity is essential for balancing nuclear reactions and understanding annihilation phenomena in medical imaging (PET) and particle physics. Given Data / Assumptions: Positron symbol e⁺ or β⁺. Same rest mass as an electron; opposite charge (+e). Antiparticle of the electron. Concept / Approach: Classify common nuclear particles: alpha (He nucleus), deuteron (p + n), proton (p), neutron (n), electron (e⁻), and positron (e⁺). A positron is not a nucleus; rather, it is a lepton with positive charge whose interactions include annihilation with electrons to yield photons (typically two gamma rays of ~511 keV each in the rest frame case). Step-by-Step Solution: Identify mass-charge pairing: m(e⁺) = m(e⁻); q(e⁺) = +e.Relate to decay: β⁺ decay converts a proton to a neutron, emitting e⁺ and a neutrino.Relate to detection: annihilation photons enable PET imaging. Verification / Alternative check: Reaction balancing uses A and Z. Positron carries A = 0, Z = +1 (counted as +1 charge unit), keeping nucleon count unchanged while adjusting charge balance appropriately. Why Other Options Are Wrong: Two-proton nucleus: unbound and not a positron.Helium nucleus: that is an alpha particle.Deuteron: composite nucleon pair, not a lepton.Photon: massless boson; not a positron. Common Pitfalls: Confusing β⁺ (positron) with β⁻ (electron) in decay notation.Treating positrons as nuclear particles rather than leptons. Final Answer: An electron with positive charge (anti-electron)

Question 19

In reactor terminology, “light water” used as a coolant refers to which substance?

Accepted Answer

Introduction / Context: Light-water reactors (LWRs) include PWRs and BWRs that use ordinary water as coolant and, in many cases, as moderator. “Light” here distinguishes H2O from “heavy water,” which is D2O used in certain other reactor designs. Given Data / Assumptions: Terminology: light water = H2O with protium, not deuterium. Coolant quality control (chemistry) is important but does not change the “light vs heavy” classification. Concept / Approach: Light water is ordinary water composed of hydrogen-1 (protium) and oxygen-16 predominately. While plant chemistry maintains conductivity and pH within tight windows to protect components, those adjustments do not redefine it as something other than “light water.” Step-by-Step Solution: Define light vs heavy: H2O (light) vs D2O (heavy).Identify coolant in LWRs: ordinary water flows through core.Eliminate pH-based distractors: acidity/alkalinity control is separate from isotopic composition. Verification / Alternative check: Plant specifications and textbooks universally use “light water” to mean H2O. Heavy-water reactors (HWRs) explicitly specify D2O for moderation/cooling, underlining the distinction. Why Other Options Are Wrong: pH 6 or pH 8 descriptors: chemistry control does not redefine isotopic identity.D2O or “demineralized heavy water”: those are heavy water, not light water. Common Pitfalls: Equating water-treatment details with isotope-based terminology.Assuming “light” means low mineral content rather than isotopic composition. Final Answer: Ordinary water (H2O)

Question 20

Which of the following particles is generally not emitted as a characteristic product of radioactive decay processes?

Accepted Answer

Introduction / Context: Radioactive decay encompasses several common modes: alpha (He nucleus), beta-minus (electron), beta-plus (positron), and gamma emission. Understanding which emissions are typical helps quickly eliminate distractors in objective questions. Given Data / Assumptions: “Characteristic” or common decay modes are considered for naturally occurring and many artificial radionuclides. Exotic or very rare processes are not the focus unless explicitly stated. Concept / Approach: Most nuclides decay by alpha, beta-minus, beta-plus, electron capture, or emission of gamma photons from excited states. While proton emission exists as a nuclear process, it is comparatively rare and typically appears in highly proton-rich, short-lived nuclides near the drip line, not as a common feature of standard decay chains taught at an introductory level. Step-by-Step Solution: List typical emissions: α, β⁻, β⁺/electron capture, and γ are common.Assess proton emission: recognized in nuclear physics but rare and specialized.Therefore, among the listed options, “proton” is the least characteristic emission. Verification / Alternative check: Nuclear data tables show long series dominated by α and β decays, with γ de-excitation. Proton emission is observed but is not routine for the nuclides commonly discussed in general curricula. Why Other Options Are Wrong: Alpha: very common, especially in heavy nuclides.Positron (β⁺): common for proton-rich isotopes; paired with electron capture.Beta (β⁻): ubiquitous for neutron-rich isotopes.Gamma: frequent as an accompanying de-excitation. Common Pitfalls: Overgeneralizing rare proton emission as if it were common.Confusing β⁺ (positron) with a proton due to positive charge. Final Answer: Proton

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