Question 1

In nuclear power systems, why is liquid water generally a more effective reactor coolant than gases such as CO2, He, or N2? Consider heat-transfer performance, equipment sizing, operating pressure, and its concurrent role in common reactor designs.

Accepted Answer

Introduction / Context: Choosing a reactor coolant is a multidisciplinary engineering decision involving thermohydraulics, materials, neutronics, and plant economics. This question compares liquid water with common gases (CO2, He, N2) from the standpoint of heat removal and practical plant design, as encountered in light-water reactors and other thermal systems. Given Data / Assumptions: Comparable core thermal power must be removed in steady state. Coolant properties are evaluated at typical reactor conditions. Focus is on generic traits: density, heat capacity, thermal conductivity, and pressurisation. Concept / Approach: Water’s very high density and specific heat capacity mean each unit volume carries far more energy than a gas. Its thermal conductivity is higher than most gases at comparable temperatures, giving larger heat-transfer coefficients for similar flow geometries and velocities. Because water can be readily pressurised, its saturation temperature can be raised substantially, enabling high core outlet temperatures without phase change (PWR) or controlled boiling (BWR). These traits cascade into smaller pumps and heat-exchange surfaces for a given duty. Additionally, in thermal reactors, ordinary or heavy water also moderates neutrons, providing a dual role that simplifies the overall plant layout. Step-by-Step Solution: 1) Compare bulk properties: water has higher rho and Cp, so m_dot * Cp * ΔT per channel is large.2) Higher thermal conductivity and favorable Prandtl numbers increase h (convective coefficient) at similar Reynolds numbers.3) With higher heat capacity per volume, required volumetric flow and surface area drop → smaller pumps and exchangers.4) Pressurisation raises boiling point (PWR) or enables controlled boiling (BWR), keeping temperature margins manageable.5) In LWRs, water also moderates, reducing separate moderator needs. Verification / Alternative check: Helium has excellent thermal conductivity among gases, but its very low density demands high volumetric flow and large compressors. CO2 avoids activation issues and was used historically, yet its lower heat capacity per volume increases equipment size. Benchmarked plant designs consistently show more compact primary circuits with water for the same thermal rating. Why Other Options Are Wrong: (a) Correct: water is a strong moderator in thermal reactors.(b) Correct: higher volumetric heat capacity reduces pump and exchanger size.(c) Correct: high h and ease of pressurisation support high-temperature operation. Common Pitfalls: Ignoring corrosion/radiolysis control; assuming moderation is always desirable (it is not for fast reactors); overlooking activation/chemistry differences between coolants. Final Answer: All of the above.

Question 2

Order of magnitude check: the energy released per fission event in a typical heavy nucleus (e.g., U-235 or Pu-239) is approximately how many MeV?

Accepted Answer

Introduction / Context: When a heavy nucleus such as U-235 undergoes fission after absorbing a neutron, it splits into two fission fragments plus neutrons and gamma rays. The mass defect between reactants and products is converted to energy according to E = m * c^2. Engineers frequently use a rounded order-of-magnitude value for rapid calculations. Given Data / Assumptions: Thermal-neutron fission of U-235 or Pu-239. Energy includes kinetic energy of fission fragments, prompt/delayed neutrons, and gamma radiation. Interest is per fission event, not per mole or per kilogram. Concept / Approach: Experimental tally of product energies gives a figure near 200 MeV per fission. Of this, roughly 165–170 MeV appears as kinetic energy of the two fragments; a few MeV appear as prompt neutrons; and the remainder is carried as gamma rays and beta decays of fission products (delayed energy). Step-by-Step Solution: 1) Recognise that fission fragments carry most kinetic energy (largest term).2) Add contributions from prompt neutrons and gamma rays to reach ~200 MeV total.3) Convert to macroscopic units if desired: 1 MeV ≈ 1.602e-13 J; therefore 200 MeV ≈ 3.2e-11 J per fission.4) Multiply by Avogadro-scale fission counts to estimate reactor thermal power from fission rate. Verification / Alternative check: Back-calculating from 1 GW(th) requires ~3.1e19 fissions per second using 200 MeV per fission, which matches standard reactor physics back-of-the-envelope checks. Why Other Options Are Wrong: 20 MeV underestimates by an order of magnitude.1000 or 2000 MeV exceed typical fission energy and are closer to GeV-scale particle processes. Common Pitfalls: Confusing per-fission energy with binding energy per nucleon; mixing MeV per fission with MeV per neutron capture in other reactions. Final Answer: 200

Question 3

Typical thermal cycle performance: the net efficiency (electrical output divided by reactor thermal power) of an ordinary light-water-cooled, steam-cycle nuclear power plant is about what percent?

Accepted Answer

Introduction / Context: Light-water reactors (LWRs) such as BWRs and PWRs produce thermal energy in the core and convert a fraction of that heat into electricity through a Rankine steam cycle. Net efficiency accounts for all auxiliary loads and realistic turbine-condenser temperatures. Given Data / Assumptions: Condenser operates near ambient cooling-water temperatures. Turbine inlet steam conditions typical of LWRs, not supercritical fossil units. Auxiliary power consumption and heat losses are included in “net.” Concept / Approach: Because LWR core outlet temperatures are modest versus modern fossil supercritical plants, the achievable Carnot fraction is lower, leading to net plant efficiencies around the low-30% range. Typical values cluster around 30–35% depending on site cooling temperature, turbine design, and plant age. Step-by-Step Solution: 1) Recognise Rankine limitations with saturated or slightly superheated steam at moderate pressure/temperature.2) Account for auxiliary loads (pumps, fans, control/cleanup systems) to move from gross to net efficiency.3) The representative textbook value is about 32%. Verification / Alternative check: Thermodynamic models with typical PWR primary outlet ~315–325 C and BWR steam conditions yield net electrical efficiencies ≈ 0.30–0.33 under standard cooling. Why Other Options Are Wrong: 52% and 72% are characteristic of advanced combined cycles or idealised models, not LWR Rankine plants.88% is physically impossible for a heat engine with realistic temperature limits. Common Pitfalls: Confusing thermal efficiency (boiler/turbine) with capacity factor; ignoring seasonal condenser temperature effects that swing a few percentage points. Final Answer: 32

Question 4

Coolant selection: liquid metals such as molten sodium are preferred in which reactor type due to neutronic and thermal-hydraulic requirements?

Accepted Answer

Introduction / Context: Reactor coolant choice must align with the desired neutron energy spectrum. Fast breeder reactors (FBRs) rely on a fast spectrum to convert fertile isotopes (e.g., U-238) into fissile material (e.g., Pu-239) while producing power. Any coolant or moderator that strongly slows neutrons is counterproductive in an FBR. Given Data / Assumptions: Objective: maintain a fast neutron spectrum. Need high thermal conductivity and low neutron moderation. Coolant must operate at high temperature at near-atmospheric pressure. Concept / Approach: Liquid sodium has excellent thermal conductivity and heat capacity per unit volume, enabling compact cores with high power density. Crucially, sodium has a low moderating power and low neutron absorption (for Na-23), preserving the fast spectrum needed for breeding. Operating near atmospheric pressure reduces mechanical loading on the primary system compared with high-pressure water. Step-by-Step Solution: 1) Identify spectrum need: fast → avoid strong moderators (e.g., light water).2) Select coolant with high heat-transfer capability and minimal moderation → liquid sodium fits.3) Conclude that FBRs prefer sodium or other liquid metals for neutronics and thermal reasons. Verification / Alternative check: Historical and modern fast reactors (e.g., EBR-II, BN series) have used sodium; some designs use lead or lead-bismuth for similar spectral reasons, trading chemistry and activation differences. Why Other Options Are Wrong: Homogeneous or graphite-moderated reactors are thermal designs; sodium would not add benefit.Enriched-uranium LWRs depend on strong moderation by water, not liquid metal. Common Pitfalls: Overlooking sodium–water reaction hazards in steam generators; confusing breeding (fuel cycle) with simple power production. Final Answer: Fast breeder reactor

Question 5

India locations check: among the following sites, which one is NOT a location of a Heavy Water Plant (HWP) operated in the national programme?

Accepted Answer

Introduction / Context: Heavy water (D2O) is used as a moderator and, in some designs, as a coolant in thermal reactors, especially in pressurised heavy-water reactors. India established multiple Heavy Water Plants (HWPs) to ensure domestic supply, and exam questions often test recognition of authentic sites. Given Data / Assumptions: Options list Indian industrial cities associated with energy/chemical facilities. We are identifying the site that does not host an HWP. Concept / Approach: Historic and current HWP sites have included locations such as Baroda (Vadodara), Talcher, Tuticorin, Kota, Manuguru, Nangal, Thal (Vaishet), and Hazira, among others in the national programme. Some industrial hubs are closely associated with power projects but not with an HWP. Step-by-Step Solution: 1) Check each option against known HWP sites.2) Baroda (Vadodara), Talcher, and Tuticorin are established HWP locations.3) Ramagundam is a major thermal-power centre; however, the dedicated Heavy Water Plant in that region is at Manuguru, not Ramagundam.4) Therefore, Ramagundam is the option that is not an HWP site. Verification / Alternative check: Cross-referencing typical exam lists shows Manuguru (near the Godavari valley) as the HWP location in Telangana, distinct from the Ramagundam power complex. Why Other Options Are Wrong: Baroda: recognised HWP location in Gujarat.Talcher: HWP site in Odisha.Tuticorin: HWP site in Tamil Nadu. Common Pitfalls: Confusing Manuguru with Ramagundam because both are large energy hubs in the same broader region; assuming any city with a big power station also hosts an HWP. Final Answer: Ramagundam

Question 6

Boiling Water Reactor (BWR) basics: what working fluid serves as the reactor coolant in the core?

Accepted Answer

Introduction / Context: In a Boiling Water Reactor (BWR), the reactor coolant removes fission heat from fuel rods and also acts as the working fluid for the turbine. Understanding the state of the coolant in the core and how the cycle is arranged is a basic reactor-engineering concept. Given Data / Assumptions: The BWR uses light water for both moderation and cooling. Core operation allows nucleate boiling within fuel channels. Steam is separated from liquid before the turbine. Concept / Approach: By design, the coolant entering the core is liquid water. As it flows upward, controlled boiling occurs, generating a two-phase mixture. Steam is then separated and dried; the resulting dry steam drives the turbine directly. The term “coolant” refers to the fluid introduced to remove heat from the core—liquid water—not the downstream phase after separation. Step-by-Step Solution: 1) Identify the coolant at core inlet: subcooled or saturated liquid water.2) Recognise that within the core, a mixture forms due to boiling.3) After separators/dryers, steam goes to the turbine while liquid recirculates. Verification / Alternative check: System schematics show recirculation pumps handling liquid water; steam lines branch only after moisture separators in the reactor vessel. Why Other Options Are Wrong: Hydrogen is not used as a BWR coolant.Steam alone is not injected as coolant; it is produced in situ.“Mixture” describes a state inside the core, but the coolant as a working fluid is water. Common Pitfalls: Equating “coolant” with whatever exits the core; forgetting the functional definition tied to what is pumped into the core to remove heat. Final Answer: Water

Question 7

Mass–energy check in fission: compared with the original heavy nucleus, the sum of the rest masses of the two fission product nuclei is

Accepted Answer

Introduction / Context: Nuclear fission converts a small amount of mass into a large amount of energy, manifested primarily as kinetic energy of fission fragments plus neutron and gamma energies. Understanding the mass balance qualitatively is central to reactor physics. Given Data / Assumptions: Heavy nucleus such as U-235 absorbs a neutron and splits. Two fission product nuclei and several neutrons are produced. Binding energy considerations determine the mass defect. Concept / Approach: The mass of the products (two nuclei + emitted neutrons) is less than the mass of the original nucleus plus the captured neutron by an amount Δm. The corresponding energy release is E = Δm * c^2, about 200 MeV per fission. Thus, the sum of the two product nuclei alone is necessarily less than the original heavy nucleus mass. Step-by-Step Solution: 1) Write qualitative balance: m_initial > m_products_total.2) Attribute the difference to increased binding energy of medium-mass nuclei and released kinetic/gamma energy.3) Conclude “Less”. Verification / Alternative check: Binding energy per nucleon peaks around iron; splitting heavier nuclei increases total binding energy, forcing a reduction in rest mass. Why Other Options Are Wrong: “More” or “much more” would violate energy accounting for exoergic fission.“Same” would imply zero energy release. Common Pitfalls: Ignoring the masses of emitted neutrons in casual statements; mixing up rest mass with relativistic kinetic energy of fragments. Final Answer: Less

Question 8

Uranium extraction from pitchblende (uraninite): which primary process step is used to bring uranium into solution for subsequent recovery?

Accepted Answer

Introduction / Context: Processing uranium ores such as pitchblende (uraninite) involves crushing, grinding, and hydrometallurgical steps to solubilise uranium and separate it from gangue. Recognising the primary leaching chemistry is important for materials and nuclear fuel engineers. Given Data / Assumptions: Host ore contains uranium oxides with associated minerals. Conventional acid leach flowsheets are considered. Concept / Approach: In acid leach circuits, sulphuric acid (often with an oxidant like MnO2, ferric iron, or oxygen) converts uranium oxides to soluble uranyl sulphate complexes. The pregnant leach solution is then clarified and uranium is recovered by solvent extraction or ion exchange, followed by precipitation (e.g., as ammonium diuranate) and calcination to U3O8. Step-by-Step Solution: 1) Crush and grind ore to liberate uranium minerals.2) Leach with H2SO4 and oxidant to dissolve uranium as uranyl complexes.3) Separate solids/liquids; purify via SX or IX; precipitate and calcine to product. Verification / Alternative check: Some ores use alkaline carbonate leaching when acid-consuming gangue is high; the question points to sulphuric acid as the standard for many pitchblende deposits. Why Other Options Are Wrong: Froth flotation may upgrade ore but does not itself dissolve uranium.Smelting is not the common primary recovery route for uranium oxides.Uranium oxides do not dissolve in plain water. Common Pitfalls: Assuming a single universal flowsheet; failing to note the role of oxidants in acid leaching. Final Answer: Leaching with sulphuric acid

Question 9

Pressurised Water Reactor (PWR): what is the specific function of the pressuriser and primary-side pressure control?

Accepted Answer

Introduction / Context: In a Pressurised Water Reactor, the primary circuit carries heat from the core to steam generators. Boiling in the primary side must be avoided to maintain stable heat transfer and fuel cladding margins. The pressuriser is a dedicated vessel connected to the primary loop to control pressure. Given Data / Assumptions: Closed primary loop with reactor vessel, pumps, and steam generators. Dedicated pressuriser provides a steam-water cushion. Load changes disturb temperature and volume in the loop. Concept / Approach: The pressuriser maintains the primary pressure at a high setpoint so that the coolant remains subcooled (no bulk boiling) throughout normal operation. Electric heaters and spray systems inside the pressuriser add or remove energy to adjust the steam-water inventory and, hence, the loop pressure. Step-by-Step Solution: 1) Set a pressure high enough that core outlet remains below saturation temperature.2) Use heaters to increase pressure when it tends to drop; use sprays to condense steam and reduce pressure when it tends to rise.3) Keep the primary loop single-phase liquid to ensure predictable heat transfer and pump performance. Verification / Alternative check: Plant schematics show the pressuriser connected to the hot leg; control logic references primary pressure, not secondary steam superheat (which is produced on the secondary side). Why Other Options Are Wrong: Maintaining “constant” pressure is a control outcome, but the core purpose is to keep the primary coolant pressurised to prevent boiling.Superheating steam occurs on the secondary side in some designs, not via the pressuriser.Hence “None of these” is incorrect. Common Pitfalls: Confusing primary and secondary systems; assuming the pressuriser affects turbine steam conditions directly. Final Answer: Pressurise the primary coolant water to prevent boiling

Question 10

Fuel choice for a Boiling Water Reactor (BWR): which nuclear fuel is typically loaded in the core?

Accepted Answer

Introduction / Context: Commercial BWRs are thermal reactors using light water as both moderator and coolant. The neutron economy of light water requires fuel with higher fissile content than naturally occurring uranium to achieve practical core sizes and cycle lengths. Given Data / Assumptions: Moderator and coolant: ordinary light water. Standard commercial fuel form: UO2 pellets in zirconium alloy cladding. Typical enrichment: a few weight percent U-235. Concept / Approach: Because hydrogen in light water captures neutrons and moderates strongly, natural uranium (0.711% U-235) is usually insufficient for steady criticality with practical core geometry. Therefore, BWRs, like PWRs, use enriched uranium—commonly in the 2–5% U-235 range—fabricated as ceramic oxide fuel. Step-by-Step Solution: 1) Identify reactor type: thermal, light-water moderated.2) Evaluate neutronic requirements: light water absorptivity demands higher fissile content.3) Conclude that enriched uranium fuel is standard for BWR cores. Verification / Alternative check: Vendor data sheets and generic BWR core design descriptions specify low-enriched uranium fuel assemblies; MOX loading is possible in some programmes but is not the baseline. Why Other Options Are Wrong: Plutonium metal is not the typical commercial BWR fuel form.Natural uranium is used in heavy-water-moderated reactors, not standard BWRs.Thorium requires a breeding strategy (e.g., Th-U fuel cycle) and is not the default BWR fuel. Common Pitfalls: Assuming “natural uranium” works in all thermal reactors; overlooking the role of moderator absorption in fuel enrichment needs. Final Answer: Enriched uranium

Question 11

Nuclear reactor coolants: which gas is normally avoided as a primary coolant because it has a relatively high neutron-capture cross-section (leading to parasitic absorption and activation)?

Accepted Answer

Introduction / Context: Gas-cooled nuclear reactors select coolants that remove heat efficiently while minimizing parasitic neutron losses. A good coolant should have a small neutron-capture cross-section so that neutrons remain available to sustain the chain reaction and to achieve high neutron economy. Given Data / Assumptions: Candidate gases include nitrogen, hydrogen, helium, carbon dioxide, and others. We focus on thermal-neutron absorption (capture) tendencies and operational practicality. The question asks which gas is normally not used as a reactor coolant because of high neutron capture. Concept / Approach: Helium and carbon dioxide are classic gaseous coolants. Helium is chemically inert, single-phase, and has a very low thermal-neutron absorption cross-section, making it excellent for high-temperature gas-cooled reactors. Carbon dioxide has been used historically (e.g., AGR/Magnox) due to reasonable neutronic and thermal properties. Nitrogen, by contrast, has a significantly higher neutron-capture cross-section and also undergoes activation (e.g., formation of C-14 via N-14(n,p)C-14 in certain conditions), which is undesirable for both neutron economy and radiological considerations. Hydrogen gas itself is rarely used as a standalone reactor coolant despite hydrogen in light water being the primary moderator/coolant in LWRs; the molecular gas introduces other design challenges and safety concerns, but the canonical 'high capture' reason points to nitrogen. Step-by-Step Solution: 1) Identify the gases commonly adopted as coolants: helium (preferred), carbon dioxide (historical use).2) Compare neutron-capture tendencies: helium is very low; CO2 moderate; nitrogen comparatively higher and leads to activation products.3) Conclude that nitrogen is normally avoided primarily for neutronic reasons (parasitic absorption) and activation concerns. Verification / Alternative check: Standard reactor physics texts list helium among the best gas coolants for neutron economy; nitrogen is not standard due to capture/activation and inferior neutronic characteristics compared with helium and even CO2. Why Other Options Are Wrong: Hydrogen: not used as a gaseous coolant, but the reason traditionally cited for 'not used' in this context is not mainly high capture relative to nitrogen. Helium: extremely low capture, ideal coolant. Carbon dioxide: historically used successfully. Neon: not typical industrial choice, but the question targets the high-capture culprit, i.e., nitrogen. Common Pitfalls: Conflating hydrogen in water (effective moderator/coolant as light water) with using H2 gas directly; overlooking activation issues. Final Answer: Nitrogen

Question 12

Indian nuclear power geography: which of the following locations does not host a nuclear power plant?

Accepted Answer

Introduction / Context: Recognizing the locations of major Indian nuclear power stations is a common general knowledge and engineering awareness topic. This question checks whether you can distinguish nuclear sites from thermal or other power stations. Given Data / Assumptions: Options include well-known Indian power locations. Only one option does not correspond to a nuclear power station site. Concept / Approach: Rawatbhata (Rajasthan), Kalpakkam (Tamil Nadu), Kaiga (Karnataka), and Tarapur (Maharashtra) are recognized nuclear power plant sites. Talcher, in Odisha, is famous for large coal-fired (thermal) power generation and related industrial energy infrastructure, not for a nuclear plant. Step-by-Step Solution: 1) List Indian nuclear sites: Rawatbhata (RAPS), Kalpakkam (MAPS & research facilities), Kaiga (KGS), Tarapur (TAPS).2) Identify Talcher’s primary association: major coal-based thermal power projects.3) Therefore, the sole non-nuclear site in the list is Talcher. Verification / Alternative check: Public domain records and India’s Department of Atomic Energy lists confirm the nuclear status of Rawatbhata, Kalpakkam, Kaiga, and Tarapur. Why Other Options Are Wrong: Rawatbhata: hosts Rajasthan Atomic Power Station units. Kalpakkam: hosts Madras Atomic Power Station and research facilities. Kaiga: hosts Kaiga Generating Station. Tarapur: hosts Tarapur Atomic Power Station. Common Pitfalls: Mixing up thermal and nuclear sites; assuming every large power hub is nuclear. Final Answer: Talcher (Odisha)

Question 13

Neutron moderation: which of the following is not a good moderating material for thermal reactors?

Accepted Answer

Introduction / Context: Moderators slow down fast neutrons to thermal energies to sustain fission in thermal reactors. An effective moderator has high scattering effectiveness per absorption (low capture cross-section and suitable atomic mass) so that neutrons thermalize without being lost. Given Data / Assumptions: Materials under consideration: concrete, boron, and 18/8 stainless steel. Comparison is with standard moderators such as light water, heavy water, graphite, and beryllium. Concept / Approach: Concrete is primarily a structural/shielding material; while it can slow neutrons, it also contains elements that absorb neutrons and is not an efficient, controllable moderator inside the core. Boron (notably B-10) has a very high neutron absorption cross-section and is deliberately used as a poison or control material, not as a moderator. Austenitic stainless steel (18/8) has relatively high absorption compared with classic moderators and is used for structural components, not for moderation. Step-by-Step Solution: 1) Evaluate boron: strong neutron absorber → unsuitable as moderator.2) Evaluate stainless steel: structural alloy, higher capture than desired for moderation → unsuitable.3) Evaluate concrete: principally shielding; moderation is incidental and inefficient → not a good core moderator.4) Hence, all three are not good moderating materials. Verification / Alternative check: Standard reactor design texts list graphite, heavy water, and light water as preferred moderators; boron is explicitly used to absorb neutrons, not to moderate; steels are avoided for moderation roles. Why Other Options Are Wrong: Graphite (option e) is an established high-quality moderator. Common Pitfalls: Assuming anything that slows neutrons suffices; ignoring absorption penalties that undermine neutron economy. Final Answer: All of the above

Question 14

Nuclear measurements: the radioactivity (decay rate) of an isotope is commonly expressed in which unit?

Accepted Answer

Introduction / Context: Radioactivity quantifies the number of nuclear disintegrations per unit time. Historically and in many references, the Curie (Ci) is used, while the SI unit is the Becquerel (Bq). Given Data / Assumptions: We are asked for the commonly used unit for activity (decays per second). Options include units for cross-section, energy, and activity. Concept / Approach: The Curie is defined as 3.7 × 10^10 disintegrations per second, originally based on the activity of 1 gram of radium-226. The SI unit Becquerel equals 1 disintegration per second. A barn measures nuclear reaction cross-sections; MeV and erg are energy units. Step-by-Step Solution: 1) Identify the property: activity = decay rate.2) Match activity to unit: Curie (traditional) and Becquerel (SI).3) From the listed options, Curie is the correct match. Verification / Alternative check: Dimensional analysis and definitions confirm that Bq and Ci measure events per time, whereas barns and MeV/erg do not. Why Other Options Are Wrong: Barn: area unit for cross-sections. MeV/erg: energy units. Becquerel: correct SI unit, but the specific correct choice here is Curie as per the option list focus. Common Pitfalls: Confusing cross-section (barn) or energy (MeV) with activity. Final Answer: Curie

Question 15

Radiation protection: in a power reactor, the biological shield is provided primarily to protect which group from radiation exposure?

Accepted Answer

Introduction / Context: A biological shield is a mass of material (e.g., heavy concrete, water tanks, steel) designed to attenuate neutron and gamma radiation to safe levels outside the reactor core region. Its primary purpose is occupational safety. Given Data / Assumptions: Shield design focuses on protecting people working around the reactor. Materials are selected for attenuation of both fast/thermal neutrons and photons. Concept / Approach: While fuel elements and structural walls must also withstand radiation, the explicit term ”biological” emphasizes protection of living beings—namely operating and maintenance personnel, and the surrounding public. Shielding reduces dose rates in accessible areas to below regulatory limits. Step-by-Step Solution: 1) Identify the object of protection in biological shielding: humans.2) Evaluate options: the direct beneficiaries are operating personnel.3) Hence, the correct answer is operating personnel. Verification / Alternative check: Facility layouts and ALARA (As Low As Reasonably Achievable) practices prioritize dose reduction for workers through fixed shields and distance/time controls. Why Other Options Are Wrong: Fuel elements: are the source; shielding is designed around them. Reactor building walls/control rods: not the principal target of biological dose limitation. None of these: incorrect because personnel protection is the aim. Common Pitfalls: Misreading ”biological” as structural; overlooking neutron vs gamma attenuation requirements. Final Answer: Operating personnel

Question 16

Fissile vs fertile: which of the following is not a fissile nuclear material under thermal neutrons?

Accepted Answer

Introduction / Context: Fissile isotopes can undergo fission with low-energy (thermal) neutrons. Fertile isotopes do not fission readily with thermal neutrons but can transmute into fissile materials upon neutron capture and subsequent decay. Given Data / Assumptions: We compare common reactor-relevant nuclides. Standard behavior under thermal neutron spectrum is implied. Concept / Approach: U-233, U-235, Pu-239, and Pu-241 are classic fissile isotopes in thermal reactors. U-238 is fertile: it captures a neutron to form U-239, which beta-decays to Np-239 and then to Pu-239, a fissile nuclide. Therefore, U-238 itself is not fissile under thermal neutrons. Step-by-Step Solution: 1) List fissile examples: U-233, U-235, Pu-239, Pu-241.2) Identify U-238 behavior: primarily fertile, not fissile with thermal neutrons.3) Conclude: U-238 is the non-fissile option. Verification / Alternative check: Reactor physics references consistently classify U-238 as fertile and the others as fissile for thermal spectra. Why Other Options Are Wrong: U-233/U-235/Pu-239/Pu-241: all fissile with thermal neutrons. Common Pitfalls: Confusing ”fissile” with ”fissionable” at high energies; U-238 can fission with fast neutrons but not efficiently with thermal ones. Final Answer: U-238

Question 17

Fusion fuels: which of the following is a practical fuel pair for thermonuclear (fusion) reactors?

Accepted Answer

Introduction / Context: Fusion energy seeks to combine light nuclei to release energy, as in D-T (deuterium-tritium) fusion. The choice of fuel affects reaction cross-section, ignition temperature, and radiation output (e.g., neutron energy). Given Data / Assumptions: We are comparing fuels for fusion (not fission). Practical consideration emphasizes achievable cross-sections at attainable plasma temperatures. Concept / Approach: The D-T reaction has the largest cross-section at relatively ”low” fusion temperatures (order of 10^8 K compared with alternatives). It produces a 14.1 MeV neutron and a 3.5 MeV alpha, making it the leading candidate in present experimental devices (e.g., tokamaks and inertial systems). U-233, thorium, and Pu-239 are fission fuels; heavy water is a moderator, not a fuel. Step-by-Step Solution: 1) Identify the only fusion fuel pair listed: D and T.2) Recognize others as fission fuels or moderators.3) Therefore, select deuterium and tritium. Verification / Alternative check: Fusion literature consistently treats D-T as the near-term fuel due to favorable reactivity compared with D-D or D-^3He, which require higher temperatures or have lower cross-sections. Why Other Options Are Wrong: U-233, Pu-239: fissile for fission, not fusion fuel. Thorium: fertile material for breeding, not a fusion fuel. Heavy water: D2O, used as moderator and coolant in some fission reactors. Common Pitfalls: Conflating fusion and fission fuel cycles; assuming any ”nuclear” material is a fusion fuel. Final Answer: Deuterium and tritium

Question 18

Nuclear facts: pick the wrong statement about radioactivity and nuclear particles.

Accepted Answer

Introduction / Context: This item mixes fundamental statements from nuclear physics and chemistry. The task is to identify the one incorrect statement among generally correct facts about decay, particle properties, and neutron absorption. Given Data / Assumptions: Thermal effects on decay rates are negligible (radioactive decay is a nuclear, not chemical, process). Elementary particle properties are at textbook values. Typical neutron absorbers (like cadmium, boron) are well known. Concept / Approach: Deuterium (hydrogen-2) has one proton and one neutron in the nucleus. Therefore, any statement asserting ”one proton and two neutrons” describes tritium (hydrogen-3), not deuterium. Heating has no practical effect on decay constant; electrons carry −1 elementary charge and very small rest mass; cadmium is a strong thermal-neutron absorber used in control rods and shielding. Step-by-Step Solution: 1) Evaluate (a): correct—decay constant is temperature-independent within normal conditions.2) Evaluate (b): correct—electron has unit negative charge and negligible rest mass relative to nucleons.3) Evaluate (c): incorrect—deuterium is 1p+1n; 1p+2n corresponds to tritium.4) Evaluate (d): correct—cadmium strongly absorbs thermal neutrons. Verification / Alternative check: Standard nuclear data tables confirm deuterium's composition as 1 proton + 1 neutron. Why Other Options Are Wrong: (a) True: decay is governed by nuclear forces; macroscopic heating doesn't alter λ. (b) True: electron charge is −1e; mass ≈ 9.11×10^−31 kg. (d) True: cadmium is a classic neutron absorber (e.g., Cd control blades). Common Pitfalls: Confusing deuterium with tritium; overinterpreting minor environmental effects on decay rates. Final Answer: A deuterium atom has one proton and two neutrons in its nucleus.

Question 19

Heavy water basics: what is the molecular weight (molar mass) of heavy water (D2O)?

Accepted Answer

Introduction / Context: Heavy water (D2O) is used as a moderator and sometimes coolant in certain reactor types. Knowing its composition and molar mass is basic nuclear-engineering literacy. Given Data / Assumptions: D denotes deuterium (hydrogen-2), atomic mass ≈ 2. Oxygen in water is the O-16 isotope (dominant), atomic mass ≈ 16. Concept / Approach: The molecular weight is computed by summing atomic masses of the constituent atoms: two deuterium atoms (2 × 2 = 4) plus one oxygen (16) equals 20 g/mol for D2O. Step-by-Step Solution: 1) Count deuterium atoms: 2 × 2 = 4.2) Add oxygen: 4 + 16 = 20.3) Therefore, molar mass(D2O) ≈ 20 g/mol. Verification / Alternative check: Isotopic mass tables confirm D ≈ 2.014 and O ≈ 15.999 → sum ≈ 20.027, which rounds to 20 for standard choices. Why Other Options Are Wrong: 18: that is ordinary water (H2O). 10, 22, 36: do not match the isotopic composition of D2O. Common Pitfalls: Using H instead of D; forgetting isotopic mass differences. Final Answer: 20

Question 20

Alpha decay effect: the emission of an alpha (α) particle from a nucleus changes the daughter nucleus in what way?

Accepted Answer

Introduction / Context: Alpha decay is a common mode of radioactive decay among heavy nuclei. An alpha particle is a helium-4 nucleus consisting of 2 protons and 2 neutrons, so its emission alters both atomic number (Z) and mass number (A) of the parent nucleus. Given Data / Assumptions: α particle = ^4He nucleus = 2p + 2n. Mass number A decreases by 4; atomic number Z decreases by 2 with each α emission. Concept / Approach: Conservation of nucleon number and charge dictates the daughter’s numbers: A_daughter = A_parent − 4; Z_daughter = Z_parent − 2. Hence both the atomic mass (proportional to A) and atomic number decrease. Step-by-Step Solution: 1) Subtract the α particle nucleons: A → A − 4.2) Subtract the α particle protons: Z → Z − 2.3) Therefore, the daughter has less atomic weight and less atomic number. Verification / Alternative check: Example: ^238U → ^234Th + α. A: 238 → 234 (−4), Z: 92 → 90 (−2), consistent with the rule. Why Other Options Are Wrong: More weight with less Z or less weight with more Z contradict conservation with α emission. Unchanged values or ”none of these” do not represent α decay. Common Pitfalls: Confusing α with β decay (β changes Z by ±1 with nearly unchanged A); overlooking that α removes two protons and two neutrons. Final Answer: Less atomic weight and less atomic number

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