Q: Thermal (slow-neutron) reactors: which material is the standard primary fuel used to sustain fission in most thermal reactor designs?

Introduction / Context: Thermal reactors rely on slow (moderated) neutrons to achieve high fission probabilities. Knowing the standard primary fuel clarifies how moderators, enrichment, and core physics interact in mainstream power reactors worldwide. Given Data / Assumptions: “Thermal reactor” means a reactor where neutrons are slowed to thermal energies. Common fuels include uranium-based materials; enrichment level depends on design (natural, low-enriched). Question asks for the standard primary fuel across most thermal designs. Concept / Approach: Uranium—specifically the fissile isotope U-235 within natural or enriched uranium—is the dominant fuel for thermal reactors. Light-water reactors typically use low-enriched uranium (about 2–5% U-235), while heavy-water reactors may use natural uranium due to excellent neutron economy. Plutonium can be used (e.g., MOX fuel) but is not the baseline primary fuel in most introductory contexts; radium is not a reactor fuel; thorium is fertile, requiring breeding to U-233 to become fissile. Step-by-Step Solution: Identify the default fissile isotope for thermal systems: U-235.Recognize uranium fuel forms (UO2 pellets) as industry standard.Eliminate plutonium as a routine primary fuel (used mainly in MOX), radium as non-fuel, and thorium as fertile.Select uranium. Verification / Alternative check: Fuel-fabrication and reactor-operator references consistently cite uranium dioxide as the standard fuel for LWRs and PHWRs (with enrichment varying by design), supporting uranium as the canonical answer. Why Other Options Are Wrong: Plutonium: Used in MOX or fast reactors; not the most common baseline fuel. Radium: Not used as reactor fuel. None of these: Incorrect because uranium is standard. Thorium: Fertile, requires conversion to U-233. Common Pitfalls: Confusing “can be used” (plutonium in MOX) with “is the standard baseline fuel.” Final Answer: Uranium

Q: Nuclear data units: the absorption/scattering cross-section of a nuclide is expressed in the unit “barn”, which is numerically equal to what area?

Introduction / Context: Nuclear engineers quantify the likelihood of neutron–nucleus interactions using an effective area called the microscopic cross-section. The traditional unit is the “barn,” reflecting extremely small length scales in nuclear physics. Given Data / Assumptions: We are comparing units of area and length; only one option expresses an area of the correct magnitude. Conversions use cgs units commonly seen in nuclear data tables. Concept / Approach: The barn is a convenient unit for cross-sections: 1 barn = 10^-24 cm^2. It provides tractable numbers when neutron capture or scattering probabilities are tabulated. While SI usage is increasing (1 barn = 10^-28 m^2), many datasets and exam references still present cross-sections in barns. Step-by-Step Solution: Recall definition: 1 barn = 10^-24 cm^2.Check units: options mixing centimeters and meters must be consistent with area.Select the area expression with cm^2 and the correct exponent, 10^-24. Verification / Alternative check: SI conversion: 10^-24 cm^2 = 10^-28 m^2. Some references use millibarn (mb) = 10^-3 barn; microbarn etc., but the base definition remains 10^-24 cm^2. Why Other Options Are Wrong: 10^-12 cm and 10^-24 cm: lengths, not areas. 1 angstrom: a length (~10^-8 cm), not area. 10^-16 m^2: equals 10^-12 cm^2, not the barn value. Common Pitfalls: Confusing area with length units; mixing SI and cgs without converting exponents properly. Final Answer: 10^-24 cm^2

Q: Thermal (slow) neutrons at 0.025 eV: estimate the typical speed of such neutrons in metres per second for reactor physics calculations.

Introduction / Context: In thermal reactors, neutron speeds at room temperature are often approximated for quick calculations and intuition. A widely used reference energy is 0.025 eV, corresponding to standard “thermal” conditions near 20–25 °C. Given Data / Assumptions: Neutron kinetic energy E ≈ 0.025 eV representative of room-temperature thermal neutrons. We need an order-of-magnitude speed used in hand calculations. Non-relativistic kinetic energy relation applies. Concept / Approach: For non-relativistic particles, E = (1/2) * m * v^2. Using E ≈ 0.025 eV and neutron mass m_n, the commonly memorized result is v ≈ 2200 m/s (more precisely ~2200 m/s at 20 °C). This becomes a standard constant used in cross-section-to-reaction-rate conversions at “2200 m/s” reference conditions in many nuclear data libraries. Step-by-Step Solution: Recall thermal reference: E_ref ≈ 0.025 eV.Use the standard tabulated equivalence: E_ref ↔ v ≈ 2200 m/s.Select 2200 as the typical value. Verification / Alternative check: Detailed calculation with constants (converting eV to joules and inserting neutron mass) yields a velocity close to 2.2 * 10^3 m/s, confirming the memorized figure. Why Other Options Are Wrong: 1100, 3300, 4400, 1500: Not the standard reference velocity for 0.025 eV; 2200 m/s is the accepted benchmark. Common Pitfalls: Forgetting that “thermal” is a temperature-dependent distribution; 2200 m/s is a convenient reference, not a unique speed for all thermal neutrons. Final Answer: 2200

Q: Material properties: the specific gravity (relative density) of uranium and plutonium is approximately equal to which of the following values?

Introduction / Context: Heavy actinide metals like uranium and plutonium are extremely dense. Recognizing their approximate specific gravity helps in estimating fuel mass, shielding considerations, and handling logistics in fuel fabrication and storage. Given Data / Assumptions: Specific gravity is relative to water at 4 °C. We need an approximation commonly used in exams and quick estimates. Concept / Approach: Uranium metal has density ~19.1 g/cm^3 and plutonium (delta phase varies) is about ~19.8 g/cm^3; both round to “about 19.” This contrasts with structural steels (~7.8 g/cm^3) and copper (~8.9 g/cm^3), highlighting the exceptional density of actinide metals. Step-by-Step Solution: Recall densities: U ≈ 19.1 g/cm^3, Pu ≈ 19–20 g/cm^3.Map to specific gravity numbers relative to water ≈ 1 g/cm^3.Select the closest integer: 19. Verification / Alternative check: Engineering handbooks tabulate U and Pu densities in the 19–20 g/cm^3 range; typical practice is to quote “about 19.” Why Other Options Are Wrong: 9 and 13: Far below actinide densities. 27: Too high; few pure metals approach this density. 7.8: Approximate for steel, not U/Pu. Common Pitfalls: Confusing oxide fuel densities (e.g., UO2 pellets ~10–11 g/cm^3 theoretical) with metal densities; mixing specific gravity with atomic weight. Final Answer: 19

Question 1

Half-life dependence: for a radioactive nuclide, does the half-life depend on external conditions such as temperature, pressure, or the initial amount of material?

Accepted Answer

Introduction / Context: Half-life is a cornerstone concept in nuclear science and engineering. It represents the fixed time required for half of the radioactive nuclei in a large population to decay, regardless of how many nuclei remain at any moment. Many learners confuse nuclear decay with chemical reactions and assume that external conditions such as temperature or pressure can alter the decay rate. This item clarifies that nuclear half-life is an intrinsic property of the nuclide and not a function of typical laboratory or environmental conditions. Given Data / Assumptions: The nuclide undergoes spontaneous radioactive decay (no external bombardment). External environment falls within normal ranges of temperature and pressure found in engineering practice. No intense external fields or exotic high-energy environments are applied. Concept / Approach: Radioactive decay is governed by the weak or strong nuclear force depending on the mode (alpha, beta, gamma, etc.). The decay constant lambda determines the probability per unit time that a nucleus will decay. The half-life T is related to the decay constant by T = ln 2 / lambda. Because lambda is a nuclear property, macroscopic parameters like temperature, pressure, sample mass, or dilution do not enter the relation for ordinary conditions. Hence the half-life is effectively independent of these factors for almost all practical purposes. Step-by-Step Solution: Recognize that half-life T = ln 2 / lambda is a function only of the decay constant.External thermal or mechanical variables do not appear in the expression.Therefore, changing temperature, pressure, or the initial amount cannot change T in normal conditions.Select the option that captures this independence: “None of these.” Verification / Alternative check: Empirical measurements across broad ranges of ambient temperature and pressure show no measurable change in half-life for typical nuclides. Minor reported effects apply only to rare cases (e.g., certain electron-capture nuclides at extreme chemical states), which are well outside routine reactor or laboratory practice and do not invalidate the general rule. Why Other Options Are Wrong: Temperature/Pressure: affect chemical rates, not intrinsic nuclear decay. Initial amount: changes activity magnitude, not the half-life value. Humidity: irrelevant to nuclear transformations. Common Pitfalls: Assuming nuclear decay behaves like chemical kinetics; confusing activity A = lambda * N (which scales with amount) with half-life (which does not). Final Answer: None of these

Question 2

Purpose of a thermal (inner) shield in high-power nuclear reactors: what is its primary function inside the reactor vessel region?

Accepted Answer

Introduction / Context: Large power reactors include several shielding systems with distinct purposes. In addition to biological shields that protect personnel outside the core, an inner “thermal shield” (often a thick steel or heavy-metal structure) is positioned to reduce neutron and gamma heating of the reactor vessel and internals. Understanding what this shield does—and what it does not do—helps avoid confusion with moderators, absorbers, and mechanical hardware supports. Given Data / Assumptions: High neutron and gamma flux exists near the core. Excessive radiation heating can raise temperatures and cause embrittlement or swelling in structural materials. Thermal shield is located between core and vessel wall (design-dependent). Concept / Approach: Fast neutrons and prompt gammas deposit energy in surrounding structures. The thermal shield’s mass and composition attenuate this radiation before it reaches the vessel wall, thereby limiting temperature rise and long-term material damage. It does not function as a control system for reactivity, nor as a moderator. Instead, it is a passive protection component designed to intercept and reduce heat deposition and flux at sensitive structures. Step-by-Step Solution: Identify the threat: radiation heating and damage to vessel and internals.Recognize the mitigation: place a dense shield region to attenuate neutrons and gammas.Conclude the purpose is structural protection and thermal load reduction.Choose the statement explicitly describing this role. Verification / Alternative check: Design descriptions show thermal shields lowering vessel wall temperature gradients and fluence, extending component life and inspection intervals. Biological shielding, in contrast, addresses dose outside containment; moderators and absorbers serve different neutronic functions. Why Other Options Are Wrong: Absorb fast neutrons to shut down: that is a control rod/poison role, not the thermal shield’s main purpose. Slow secondary neutrons: that is the moderator’s role. Prevent fuel-coolant contact: a mechanical design and grid/spacer function, not shielding. Steam generator corrosion: unrelated to the in-vessel thermal shield. Common Pitfalls: Confusing inner thermal shielding with biological shielding or control rod functions. Final Answer: Protect the reactor vessel and internal walls from radiation damage and excessive heating

Question 3

Thorium fuel cycle: thorium-232 can be converted (bred) into uranium-233 predominantly in which reactor spectrum/category?

Accepted Answer

Introduction / Context: Thorium-232 is a fertile isotope that can capture a neutron and, through a decay chain, become uranium-233, a fissile material suitable for thermal reactors. Understanding the reactor spectra that favor this conversion is key to appreciating thorium-based fuel cycles explored in several countries. Given Data / Assumptions: Thorium-232 + neutron → Th-233 → Pa-233 → U-233 (after beta decays). Good neutron economy (low parasitic absorption) enhances breeding. We focus on the spectrum/category most associated with Th→U-233 programs. Concept / Approach: While neutron capture is possible in any spectrum, thorium breeding to U-233 is classically associated with thermal-spectrum systems that feature excellent neutron economy, such as heavy-water moderated or molten-salt thermal reactors. These systems minimize neutron losses so that enough captures occur in Th-232 without requiring very high fluxes. Fast reactors typically emphasize U-238→Pu-239 breeding, not Th-232→U-233 as a primary objective. Step-by-Step Solution: Recall the thorium chain leading to U-233 via beta decays.Match reactor types to breeding emphasis: thermal systems for Th→U-233; fast systems for U-238→Pu-239.Select the thermal category as the predominant pathway for thorium utilization. Verification / Alternative check: Historical programs (e.g., molten-salt thermal concepts, heavy-water thermal experiments) document effective Th→U-233 breeding with proper reprocessing or on-line fuel handling. Why Other Options Are Wrong: Fast breeder and other fast-spectrum options concentrate on U-238 breeding. Swimming-pool reactors are a geometry/cooling style; they can irradiate thorium but are not uniquely required. Common Pitfalls: Assuming all breeding must be “fast”; overlooking the role of neutron economy in thermal-spectrum thorium cycles. Final Answer: Thermal (e.g., heavy-water or molten-salt thermal) reactor

Question 4

Isotopes and neutral atoms: for the neutral atoms of an element’s isotopes, which property differs among them?

Accepted Answer

Introduction / Context: Isotopes are atoms of the same element with the same number of protons (same atomic number) but different numbers of neutrons. This leads to different masses while chemical identity, governed by proton count and electron configuration, remains the same for neutral atoms. Distinguishing which properties change or stay constant is essential in nuclear chemistry and analytical applications like mass spectrometry and radiochemistry. Given Data / Assumptions: Isotopes share the same element identity Z (protons). Neutral atoms possess equal numbers of protons and electrons. We compare atomic weight (mass), atomic number, proton count, and electron count. Concept / Approach: Because isotopes differ in neutron number, their mass numbers and atomic weights differ. However, atomic number (Z) is identical for all isotopes of a given element, and neutral atoms have Z electrons to balance Z protons. Therefore, among listed properties, only atomic weight changes from isotope to isotope, while atomic number, number of protons, and number of electrons remain constant for neutral atoms of the same element. Step-by-Step Solution: Define isotope: same Z, different N (neutrons).Link mass to neutron number: more neutrons → higher mass.Conclude that atomic weight is the differing property. Verification / Alternative check: Periodic table listings often show weighted average atomic masses reflecting natural isotopic abundances, confirming that individual isotopes have distinct masses that contribute to the average value. Why Other Options Are Wrong: Atomic number/number of protons: fixed for the element; changing these changes the element. Number of electrons (neutral atoms): equals Z for neutrality. Nuclear charge state: determined by Z, same across isotopes of the same element. Common Pitfalls: Confusing ionic charge (electron gain/loss) with isotopic differences; thinking isotopes alter chemical valence dramatically—most chemistry is similar despite mass differences. Final Answer: Atomic weights

Question 5

Counting atoms: how many atoms are present in one gram-atom (i.e., one mole) of an element?

Accepted Answer

Introduction / Context: The concept of the mole (historically “gram-atom” for elemental substances) is central to stoichiometry, nuclear accounting, and materials balances. It provides a bridge between microscopic particle counts and macroscopic masses measured in the laboratory. Given Data / Assumptions: One mole corresponds to Avogadro’s number of entities. Entities can be atoms, molecules, ions, or other specified particles. Concept / Approach: Avogadro’s number, approximately 6.022 × 10^23 per mole, defines how many atoms are in one gram-atom of an element. In educational MCQs, it is often rounded to 6 × 10^23 for simplicity. This universal constant allows direct conversion between mass (using atomic or molar mass) and count of particles, which is essential in reaction stoichiometry and nuclear fuel calculations where atom counts determine reaction rates and activities. Step-by-Step Solution: Recall the definition: 1 mole = 6.022 × 10^23 particles.Recognize that “gram-atom” = 1 mole of atoms for an element.Select the nearest conventional rounded value: 6 × 10^23. Verification / Alternative check: By definition in the SI system, one mole is the amount of substance containing exactly 6.02214076 × 10^23 specified entities. Rounding to 6 × 10^23 is acceptable for MCQ contexts. Why Other Options Are Wrong: 2 × 10^23 and 3 × 10^21: incorrect magnitudes. 6 × 10^32 and 5 × 10^5: wildly off by many orders of magnitude. Common Pitfalls: Misreading exponents; confusing gram-atom with gram-mole of compounds (the concept extends but the count per mole remains the same). Final Answer: 6 × 10^23

Question 6

Role of graphite in a nuclear reactor: in standard terminology, graphite primarily serves as which component?

Accepted Answer

Introduction / Context: Reactor cores use moderators to slow down fast neutrons to thermal energies where fission in certain fuels (e.g., U-235) is more probable. Graphite is one of the classic moderators used in historical and some modern designs due to its favorable scattering properties and low neutron absorption. Given Data / Assumptions: Graphite is crystalline carbon with low thermal-neutron capture cross-section. The purpose of a moderator is to reduce neutron energies by elastic scattering. Other core components include fuel, coolant, reflectors, absorbers, and structural materials. Concept / Approach: Effective moderation requires many collisions with minimal neutron loss by absorption. Graphite’s atomic mass and scattering characteristics allow efficient neutron slowing while preserving the neutron population. This enhances the probability that thermal neutrons will induce fission in fissile isotopes, improving neutron economy in thermal-spectrum reactors. Step-by-Step Solution: Identify the function “retarder of neutron velocity”.Match to the reactor component category: moderator.Select graphite’s role accordingly. Verification / Alternative check: Design literature for graphite-moderated reactors (e.g., certain gas-cooled designs) documents its use specifically for moderation with helium or CO2 as coolants. Why Other Options Are Wrong: Insulation lining/gamma absorber: not graphite’s primary engineered role. Fuel: graphite is not a fissile/fertile fuel. Lubricant: graphite can lubricate mechanically, but that is not its nuclear function. Common Pitfalls: Confusing moderator with absorber (control rods/poisons); assuming any carbon-based material absorbs gammas strongly (high-Z materials are better gamma shields). Final Answer: Retarder of neutron velocity (moderator)

Question 7

Fuel choice for BWRs: which nuclear fuel is most commonly used in a boiling water reactor (BWR)?

Accepted Answer

Introduction / Context: Boiling water reactors (BWRs) are light-water reactors in which the reactor coolant water boils in the core to produce steam directly for the turbine. Fuel selection affects core design, reactivity control, and refueling strategies. Knowing the standard fuel type is basic industry literacy. Given Data / Assumptions: BWRs use ordinary light water as both coolant and moderator. Light-water moderation has higher neutron absorption than heavy water. Commercial BWRs operate at large thermal powers with on-cycle boiling. Concept / Approach: Because light water absorbs neutrons more than heavy water, achieving criticality with natural uranium is impractical in a BWR. Therefore, BWR cores use low-enriched uranium (LEU), typically 3–5% U-235, assembled into fuel rods and bundles. While plutonium may be present in mixed-oxide (MOX) fuel in some reactors, the main and most common fuel specification remains LEU uranium dioxide. Step-by-Step Solution: Identify moderator/coolant: light water → demands enrichment.Recall typical LEU range: about 3–5% U-235.Select “Enriched uranium” as the common BWR fuel. Verification / Alternative check: Vendor documentation and operating histories for BWR fleets show widespread use of LEU UO2 fuel, with some plants licensed for partial MOX loads under strict controls. Why Other Options Are Wrong: Natural uranium: feasible with heavy-water moderators, not standard for BWRs. Plutonium metal/monazite sand/thorium metal: not the mainstream BWR fuel form or composition. Common Pitfalls: Assuming all reactor types can run on natural uranium; confusing BWRs with heavy-water designs like PHWR/CANDU. Final Answer: Enriched uranium

Question 8

Half-life and sample size: a 4 g sample of a radioisotope has a half-life of 10 days. What is the half-life if we instead take a 2 g sample of the same isotope?

Accepted Answer

Introduction / Context: This question tests the key principle that half-life is an intrinsic nuclear property of a nuclide and does not depend on how much of the substance you start with. While the activity (decays per second) scales with the number of atoms, the half-life remains constant for a given isotope under ordinary conditions. Given Data / Assumptions: Isotope half-life T_1/2 = 10 days. Two different initial masses: 4 g and 2 g. Same physical and chemical environment (no exotic conditions). Concept / Approach: The decay law N(t) = N0 * (1/2)^(t/T_1/2) shows that the characteristic time T_1/2 depends only on the decay constant lambda via T_1/2 = ln 2 / lambda. Changing N0 (initial amount) alters only the starting activity A0 = lambda * N0, but it does not alter lambda or T_1/2. Therefore, halving the sample mass halves the initial activity but leaves the half-life unchanged. Step-by-Step Solution: Recognize T_1/2 is related to lambda only: T_1/2 = ln 2 / lambda.Sample mass affects N0 and thus A0, not lambda.Hence T_1/2 for 2 g remains 10 days, identical to that for 4 g. Verification / Alternative check: Compute activity ratio: A0(2 g) = (2/4) * A0(4 g) = 0.5 * A0(4 g). Despite different A0 values, the time to halve each sample is the same (10 days), confirming independence from the amount. Why Other Options Are Wrong: 5, 20, 30 days: suggest mass-dependent half-lives, which is incorrect. “Depends on geometry”: ordinary geometry has no effect on nuclear decay constants. Common Pitfalls: Confusing activity (amount-dependent) with half-life (amount-independent); assuming dilution or concentration changes the nuclear decay rate. Final Answer: 10 days

Question 9

Energy of thermal neutrons used to fission U-235: which range best represents their kinetic energy at room temperature?

Accepted Answer

Introduction / Context: In thermal reactors, neutrons are moderated to low energies where the fission cross-section of U-235 is high. Recognizing the approximate energy scale of “thermal” neutrons at room temperature is essential for interpreting cross-section data, diffusion lengths, and reactor kinetics. Given Data / Assumptions: Room temperature near 293 K is assumed. Thermal equilibrium of neutrons with the moderator. Energy distribution is approximately Maxwellian. Concept / Approach: The characteristic energy of thermal neutrons at 293 K corresponds to about kT, where k is Boltzmann’s constant. Numerically, kT ≈ 0.025 eV at room temperature. Therefore, typical thermal neutrons have energies around 0.025 eV or less, much lower than epithermal (eV to keV) or fast (~MeV) neutron energies. Step-by-Step Solution: Estimate kT at 293 K: kT ≈ 0.025 eV.Thermal neutrons populate energies around and below this value.Select the option indicating energies below about 0.025 eV. Verification / Alternative check: Cross-section charts for U-235 show a pronounced 1/v dependence at low energies, peaking effectiveness in the thermal range near 0.025 eV at room temperature. Why Other Options Are Wrong: 1–25 eV and > 1 eV: epithermal energies, not thermal. > 200 eV and ~1 MeV: fast neutron ranges, orders of magnitude too high for thermal neutrons. Common Pitfalls: Confusing average energy (≈0.025 eV) with a sharp value; thermal spectra are distributions centered around kT. Final Answer: < 0.025 eV

Question 10

Maximum U-235 content in enriched-uranium fast breeder reactors: up to what percentage of U-235 may be used in such cores?

Accepted Answer

Introduction / Context: Fast breeder reactors (FBRs) are designed to operate with a fast neutron spectrum and to convert fertile material (often U-238) into fissile material (such as Pu-239). Core fuel compositions vary by design and purpose (experimental, prototype, or commercial), but an understanding of the upper bounds used in enriched-uranium fast cores is a common exam topic. Given Data / Assumptions: We refer to fast reactors that employ enriched uranium as fissile driver material. Enrichment denotes the percentage of U-235 in uranium. The question asks for an upper-end figure sometimes cited for such designs. Concept / Approach: Fast-spectrum reactors do not require moderation, and their neutronics can accommodate a wide range of fissile assays. While many practical fast reactors use lower enrichments (e.g., tens of percent or MOX fuels), educational references and MCQ compendia often cite that fast reactors using enriched uranium may contain up to very high enrichments on the order of several tens of percent and even approaching weapons-grade levels in certain specialized cores. The extreme upper bound commonly quoted in exam contexts is about 85% U-235. Step-by-Step Solution: Recognize that the question seeks a maximum typical exam-cited value, not a standard operating assay.Match to the provided options: 85% represents the highest listed enrichment and aligns with the “up to” phrasing.Select 85% as the answer consistent with traditional MCQ references. Verification / Alternative check: While many fast reactor cores today prefer MOX or lower-assay fuels, historical and theoretical treatments acknowledge designs that could utilize very high U-235 enrichments; hence, 85% is used as an accepted “maximum up to” figure in numerous exam banks. Why Other Options Are Wrong: 5% and 15%: typical of light-water or some experimental ranges, not an upper bound for fast cores. 45% and 65%: possible in some contexts but do not represent the quoted “maximum up to” figure. Common Pitfalls: Confusing typical operating enrichments with exam-style maximum values; assuming the value implies common practice rather than a permissible upper bound. Final Answer: 85%

Question 11

Indian uranium resources: uranium ore is presently mined and concentrated at which of the following locations in India? (Select the best-known operating mine/processing site.)

Accepted Answer

Introduction / Context: This question checks your knowledge of India’s primary uranium mining and milling locations. Identifying the correct operating site helps learners connect nuclear fuel-cycle theory with real-world Indian geography and resource management. Given Data / Assumptions: The focus is on the best-known, long-operating uranium mine and associated processing center in India. We consider widely cited textbook and exam references rather than every satellite deposit. Choice names may refer to towns near deposits; select the prominent uranium site most associated with mining and concentration. Concept / Approach: Uranium Corporation of India Limited (UCIL) has historically operated in the Singhbhum Shear Zone (Jharkhand). The Jadugoda mine and mill are the classic exam-cited locations where ore is mined and processed to a concentrate (commonly U3O8, often called yellowcake). Nearby towns like Ghatsila are geographically close but are not the canonical mine name used in foundational questions. Other options listed are known for different minerals or processes (e.g., copper or monazite/rare earths), not the flagship underground uranium mining operation identified in Indian nuclear engineering exams. Step-by-Step Solution: List sites and their typical associations.Jadugoda (Jharkhand) is India’s most cited uranium mine and milling site.Ghatsila is a nearby town but not the canonical mine name in exams.Khetri is known for copper mining; Alwaye (Aluva) is associated with monazite processing/rare earths.Therefore, select Jadugoda. Verification / Alternative check: Introductory nuclear fuel-cycle overviews for India routinely mention Jadugoda as the pioneering uranium mine and processing location, reinforcing its prominence in exam questions and industry history. Why Other Options Are Wrong: Ghatsila: Nearby region, not the exam-standard mine/mill name. Khetri: Copper mining district, not uranium. Alwaye: Known for monazite and rare earth processing, not primary uranium mining. Tumkur: Not India’s flagship uranium mining site in standard MCQs. Common Pitfalls: Confusing nearby towns with the named mine; mixing monazite/rare earth processing centers with uranium mining locations. Final Answer: Jadugoda

Question 12

Thermal (slow-neutron) reactors: which material is the standard primary fuel used to sustain fission in most thermal reactor designs?

Accepted Answer

Introduction / Context: Thermal reactors rely on slow (moderated) neutrons to achieve high fission probabilities. Knowing the standard primary fuel clarifies how moderators, enrichment, and core physics interact in mainstream power reactors worldwide. Given Data / Assumptions: “Thermal reactor” means a reactor where neutrons are slowed to thermal energies. Common fuels include uranium-based materials; enrichment level depends on design (natural, low-enriched). Question asks for the standard primary fuel across most thermal designs. Concept / Approach: Uranium—specifically the fissile isotope U-235 within natural or enriched uranium—is the dominant fuel for thermal reactors. Light-water reactors typically use low-enriched uranium (about 2–5% U-235), while heavy-water reactors may use natural uranium due to excellent neutron economy. Plutonium can be used (e.g., MOX fuel) but is not the baseline primary fuel in most introductory contexts; radium is not a reactor fuel; thorium is fertile, requiring breeding to U-233 to become fissile. Step-by-Step Solution: Identify the default fissile isotope for thermal systems: U-235.Recognize uranium fuel forms (UO2 pellets) as industry standard.Eliminate plutonium as a routine primary fuel (used mainly in MOX), radium as non-fuel, and thorium as fertile.Select uranium. Verification / Alternative check: Fuel-fabrication and reactor-operator references consistently cite uranium dioxide as the standard fuel for LWRs and PHWRs (with enrichment varying by design), supporting uranium as the canonical answer. Why Other Options Are Wrong: Plutonium: Used in MOX or fast reactors; not the most common baseline fuel. Radium: Not used as reactor fuel. None of these: Incorrect because uranium is standard. Thorium: Fertile, requires conversion to U-233. Common Pitfalls: Confusing “can be used” (plutonium in MOX) with “is the standard baseline fuel.” Final Answer: Uranium

Question 13

Nuclear data units: the absorption/scattering cross-section of a nuclide is expressed in the unit “barn”, which is numerically equal to what area?

Accepted Answer

Introduction / Context: Nuclear engineers quantify the likelihood of neutron–nucleus interactions using an effective area called the microscopic cross-section. The traditional unit is the “barn,” reflecting extremely small length scales in nuclear physics. Given Data / Assumptions: We are comparing units of area and length; only one option expresses an area of the correct magnitude. Conversions use cgs units commonly seen in nuclear data tables. Concept / Approach: The barn is a convenient unit for cross-sections: 1 barn = 10^-24 cm^2. It provides tractable numbers when neutron capture or scattering probabilities are tabulated. While SI usage is increasing (1 barn = 10^-28 m^2), many datasets and exam references still present cross-sections in barns. Step-by-Step Solution: Recall definition: 1 barn = 10^-24 cm^2.Check units: options mixing centimeters and meters must be consistent with area.Select the area expression with cm^2 and the correct exponent, 10^-24. Verification / Alternative check: SI conversion: 10^-24 cm^2 = 10^-28 m^2. Some references use millibarn (mb) = 10^-3 barn; microbarn etc., but the base definition remains 10^-24 cm^2. Why Other Options Are Wrong: 10^-12 cm and 10^-24 cm: lengths, not areas. 1 angstrom: a length (~10^-8 cm), not area. 10^-16 m^2: equals 10^-12 cm^2, not the barn value. Common Pitfalls: Confusing area with length units; mixing SI and cgs without converting exponents properly. Final Answer: 10^-24 cm^2

Question 14

Thermal (slow) neutrons at 0.025 eV: estimate the typical speed of such neutrons in metres per second for reactor physics calculations.

Accepted Answer

Introduction / Context: In thermal reactors, neutron speeds at room temperature are often approximated for quick calculations and intuition. A widely used reference energy is 0.025 eV, corresponding to standard “thermal” conditions near 20–25 °C. Given Data / Assumptions: Neutron kinetic energy E ≈ 0.025 eV representative of room-temperature thermal neutrons. We need an order-of-magnitude speed used in hand calculations. Non-relativistic kinetic energy relation applies. Concept / Approach: For non-relativistic particles, E = (1/2) * m * v^2. Using E ≈ 0.025 eV and neutron mass m_n, the commonly memorized result is v ≈ 2200 m/s (more precisely ~2200 m/s at 20 °C). This becomes a standard constant used in cross-section-to-reaction-rate conversions at “2200 m/s” reference conditions in many nuclear data libraries. Step-by-Step Solution: Recall thermal reference: E_ref ≈ 0.025 eV.Use the standard tabulated equivalence: E_ref ↔ v ≈ 2200 m/s.Select 2200 as the typical value. Verification / Alternative check: Detailed calculation with constants (converting eV to joules and inserting neutron mass) yields a velocity close to 2.2 * 10^3 m/s, confirming the memorized figure. Why Other Options Are Wrong: 1100, 3300, 4400, 1500: Not the standard reference velocity for 0.025 eV; 2200 m/s is the accepted benchmark. Common Pitfalls: Forgetting that “thermal” is a temperature-dependent distribution; 2200 m/s is a convenient reference, not a unique speed for all thermal neutrons. Final Answer: 2200

Question 15

Material properties: the specific gravity (relative density) of uranium and plutonium is approximately equal to which of the following values?

Accepted Answer

Introduction / Context: Heavy actinide metals like uranium and plutonium are extremely dense. Recognizing their approximate specific gravity helps in estimating fuel mass, shielding considerations, and handling logistics in fuel fabrication and storage. Given Data / Assumptions: Specific gravity is relative to water at 4 °C. We need an approximation commonly used in exams and quick estimates. Concept / Approach: Uranium metal has density ~19.1 g/cm^3 and plutonium (delta phase varies) is about ~19.8 g/cm^3; both round to “about 19.” This contrasts with structural steels (~7.8 g/cm^3) and copper (~8.9 g/cm^3), highlighting the exceptional density of actinide metals. Step-by-Step Solution: Recall densities: U ≈ 19.1 g/cm^3, Pu ≈ 19–20 g/cm^3.Map to specific gravity numbers relative to water ≈ 1 g/cm^3.Select the closest integer: 19. Verification / Alternative check: Engineering handbooks tabulate U and Pu densities in the 19–20 g/cm^3 range; typical practice is to quote “about 19.” Why Other Options Are Wrong: 9 and 13: Far below actinide densities. 27: Too high; few pure metals approach this density. 7.8: Approximate for steel, not U/Pu. Common Pitfalls: Confusing oxide fuel densities (e.g., UO2 pellets ~10–11 g/cm^3 theoretical) with metal densities; mixing specific gravity with atomic weight. Final Answer: 19

Question 16

Nature of transformations: radioactive decay should be classified as which type of change at the atomic scale?

Accepted Answer

Introduction / Context: Understanding what distinguishes nuclear changes from chemical or physical changes is fundamental in radiochemistry and nuclear engineering. Radioactive decay alters the nucleus, not merely electron arrangements or phases of matter. Given Data / Assumptions: Chemical changes involve electron-shell interactions forming/breaking bonds. Physical changes involve state, shape, or other macroscopic properties without altering atomic identities. Radioactive decay originates within the nucleus. Concept / Approach: In radioactive decay (alpha, beta, gamma), the composition or energy of the nucleus changes. The element may transmute (change Z), which cannot be achieved by purely chemical means. Thus, decay is a nuclear change; it may accompany emitted radiation and energy release but does not principally involve chemical bonding mechanisms. Step-by-Step Solution: Identify the locus of the transformation: nucleus vs electron cloud.Alpha/beta change the nucleus; gamma releases nuclear energy without changing Z or A.Therefore, classify as nuclear change. Verification / Alternative check: Transmutation examples (e.g., beta decay increasing atomic number by 1) confirm nuclear identity changes that are independent of chemical environment, pressure, or temperature. Why Other Options Are Wrong: Chemical change: Involves electron-shell rearrangements, not nuclear transformations. Physical change only: Does not alter atomic identity. None of these/Electrochemical change: Do not capture nuclear processes. Common Pitfalls: Assuming decay rates can be significantly altered by chemical state; while tiny effects exist in special cases, decay is fundamentally nuclear. Final Answer: Nuclear change

Question 17

Fuel cladding selection: which of the following materials is NOT used as a nuclear fuel cladding in power reactors?

Accepted Answer

Introduction / Context: Fuel cladding provides the first barrier against fission product release and must combine low neutron absorption with corrosion resistance and mechanical integrity. Distinguishing cladding materials from control-rod absorbers prevents dangerous misconceptions in reactor materials engineering. Given Data / Assumptions: Commercial power reactors historically use zirconium alloys or stainless steels as cladding. Advanced concepts explore ceramic composites, but conventional fleet practice is metal cladding. Cadmium is known as a strong neutron absorber for control rods and shields. Concept / Approach: Zircaloy and zirconium–niobium alloys are standard LWR/PHWR claddings because of low neutron absorption and good corrosion resistance in water. Stainless steel has been used in certain designs historically. Cadmium, by contrast, has a very high thermal neutron capture cross-section and is employed in control rods or as neutron poison—not as cladding surrounding fuel where low absorption is crucial. Step-by-Step Solution: List typical cladding: zirconium-based alloys; some designs used stainless steel.Consider ceramics: explored for accident-tolerant cladding but not traditional baseline.Identify cadmium as a neutron absorber material for control rods.Therefore, cadmium is not used as fuel cladding. Verification / Alternative check: Material specifications from reactor vendors consistently show zirconium alloys for cladding; cadmium appears in absorber components, confirming the role separation. Why Other Options Are Wrong: Zircaloy and Zr–Nb: Canonical cladding materials. Stainless steel: Used in some earlier or specific designs. Ceramics: Investigational/advanced concepts; not the “not used” distractor compared to cadmium. Common Pitfalls: Assuming any corrosion-resistant metal could serve as cladding without considering neutron absorption; confusing absorber materials with cladding. Final Answer: Cadmium

Question 18

Isotopes and radioactivity: which of the following hydrogen isotopes is intrinsically radioactive by nature?

Accepted Answer

Introduction / Context: Hydrogen has three isotopes of interest: protium (H-1), deuterium (H-2), and tritium (H-3). Recognizing which isotope is radioactive is important in fusion research, radiolabeling, and environmental monitoring. Given Data / Assumptions: We compare stability of hydrogen isotopes. Terminology: “heavy hydrogen” is deuterium. Helium is included as a distractor and is a different element entirely. Concept / Approach: Protium (H-1) and deuterium (H-2) are stable. Tritium (H-3) is radioactive, undergoing beta decay to helium-3 with a half-life of about 12.3 years. Therefore, among the options, only tritium is intrinsically radioactive. Step-by-Step Solution: List isotopes: H-1 (stable), H-2 = deuterium/heavy hydrogen (stable), H-3 = tritium (radioactive).Identify the radioactive one: tritium.Eliminate helium and duplicate naming of deuterium (“heavy hydrogen”). Verification / Alternative check: Standard nuclear data handbooks confirm tritium's beta decay to He-3, widely used in fusion research and self-luminous devices (with strict safety controls). Why Other Options Are Wrong: Helium: Different element; stable isotopes He-3 and He-4. Deuterium/heavy hydrogen: Stable. Protium: Stable. Common Pitfalls: Confusing “heavy hydrogen” with tritium; heavy hydrogen is deuterium, which is stable. Final Answer: Tritium

Question 19

Thorium metal characteristics: which set of statements correctly describes thorium’s macroscopic appearance and mechanical behavior?

Accepted Answer

Introduction / Context: Thorium is a silvery, actinide metal of interest for alternative nuclear fuel cycles. Recognizing its basic physical appearance and mechanical traits helps distinguish it from other structural and fuel materials in nuclear engineering contexts. Given Data / Assumptions: We consider pure thorium metal under normal ambient conditions. Comparisons like “similar to silver” refer to hardness feel among common metals. Surface tarnish can occur, but intrinsic properties are the focus. Concept / Approach: Thorium is typically described as silvery and steel-like in luster when freshly cut. It is relatively soft (less hard than steels; closer to silver in hardness) and workable, demonstrating good ductility. These traits align with textbook descriptions and general metallurgy references, especially when unoxidized and properly handled. Step-by-Step Solution: Assess appearance: fresh thorium is bright, metallic, steel-like.Assess hardness: relatively soft, comparable to silver rather than hardened steels.Assess ductility: thorium can be drawn/formed, indicating notable ductility.Combine: all three statements hold—select “All (a), (b) and (c).” Verification / Alternative check: Standard materials descriptions categorize thorium as a soft, ductile, silvery metal that tarnishes in air—consistent with the selected combined statement. Why Other Options Are Wrong: Any single statement alone is incomplete; the set of all three captures the common description. “Brittle and black” mischaracterizes fresh thorium; darkening is due to surface oxidation, not intrinsic brittleness. Common Pitfalls: Confusing surface oxidation color with bulk appearance; equating “actinide” with brittle behavior—thorium is comparatively ductile. Final Answer: All (a), (b) and (c).

Question 20

Fertile versus fissile: identify the nuclide that is fertile (can be converted into a fissile isotope upon neutron absorption and decay) rather than directly fissile.

Accepted Answer

Introduction / Context: Nuclear fuel-cycle discussions hinge on the distinction between fissile materials (which readily fission with thermal neutrons) and fertile materials (which convert into fissile isotopes after neutron absorption and subsequent decay). Correct identification guides fuel design and reactor strategy. Given Data / Assumptions: Fissile examples: U-235, Pu-239, U-233, Pu-241. Fertile examples: Th-232, U-238. We compare the listed nuclides and select the fertile one. Concept / Approach: Th-232 is fertile; it does not fission easily with thermal neutrons. Instead, it absorbs a neutron to become Th-233, which beta-decays to Pa-233 and then to fissile U-233. In contrast, U-235, Pu-239, and Pu-241 are fissile with thermal neutrons. U-233 is itself the fissile endpoint of the thorium cycle. Step-by-Step Solution: Classify each isotope: U-233 (fissile), U-235 (fissile), Pu-239 (fissile), Pu-241 (fissile), Th-232 (fertile).Identify the single fertile nuclide in the list.Select Th-232. Verification / Alternative check: Thorium fuel-cycle literature consistently presents Th-232 → U-233 breeding via Pa-233, validating Th-232 as the fertile starting point. Why Other Options Are Wrong: U-233: Fissile product of thorium cycle. U-235: Canonical fissile isotope in uranium fuels. Pu-239: Fissile, widely used in MOX and fast reactors. Pu-241: Also fissile. Common Pitfalls: Confusing fertile Th-232 with fissile U-233; assuming all actinides listed are fissile without checking their neutron-induced behavior. Final Answer: Th-232

Nuclear Power Engineering Questions