Question 1

Nuclear reactor fundamentals: Which reactor type may operate without a moderator (i.e., does not require neutron moderation)?

Accepted Answer

Introduction / Context: Moderators slow down fast neutrons to thermal energies to increase fission probabilities in most power reactors. However, not all reactor concepts need moderation. This question checks understanding of which reactor operates on fast neutrons. Given Data / Assumptions: Standard reactor types: CANDU, fast breeder, homogeneous, and PWR. Definition: a moderator slows neutrons via elastic scattering (e.g., heavy water, light water, graphite). Concept / Approach: Fast breeder reactors are designed to use fast neutrons to breed fissile fuel (e.g., convert fertile U-238 to Pu-239). They therefore do not employ moderators and often use liquid metal coolants (e.g., sodium) to avoid moderation. By contrast, CANDU uses heavy water as moderator, PWRs use light water, and homogeneous reactors also rely on moderation within the fuel-moderator mixture. Step-by-Step Solution: Identify reactor requiring no moderation → fast breeder reactor.Other listed types explicitly employ moderators (heavy or light water).Hence, select the fast breeder reactor. Verification / Alternative check: Reactor physics texts categorize fast reactors as unmoderated, relying on fast neutron spectra for breeding and high conversion ratios. Why Other Options Are Wrong: CANDU: heavy-water moderated. PWR: light-water moderated. Homogeneous reactor: fuel mixed with moderator (historical designs). All require moderation. Common Pitfalls: Confusing coolant type with moderator; some coolants (e.g., sodium) avoid moderation. Assuming “heavy water” reactors are “fast”; they are thermal spectrum. Final Answer: Fast breeder reactor

Question 2

Nuclear units and constants: The symbol MeV (mega–electron-volt) is a unit of which physical quantity?

Accepted Answer

Introduction / Context: In nuclear and particle physics, energies are often quoted in electron-volts and their multiples (keV, MeV, GeV). Recognizing what MeV measures avoids confusion with activity units or electrical units. Given Data / Assumptions: 1 eV is the energy gained by an electron accelerated through a potential difference of 1 volt. 1 MeV = 10^6 eV. Concept / Approach: MeV quantifies energy. Radioactivity is measured in becquerel or curie; potential difference is measured in volts; exposure/dose have different units (gray, sievert). Thus, MeV refers specifically to energy, commonly used for binding energies, decay Q-values, and reaction thresholds. Step-by-Step Solution: Recall definition: eV is an energy unit.Scale to MeV for nuclear magnitudes.Therefore, MeV is a unit of energy. Verification / Alternative check: Mass–energy equivalence frequently uses MeV (e.g., 1 amu ≈ 931 MeV of energy equivalent). Why Other Options Are Wrong: Radioactivity uses Bq or Ci, not MeV. Potential difference uses volt (V). “None of these” is incorrect. Common Pitfalls: Confusing “per nucleon” energies with total reaction energies; both are energy but differ in normalization. Mixing activity (decays per second) with energy scales. Final Answer: Energy

Question 3

Nuclear constants: What is the energy equivalent of one atomic mass unit (1 amu) expressed in MeV (using E = m c^2)?

Accepted Answer

Introduction / Context: Mass–energy equivalence is central to nuclear physics. Converting atomic mass units (amu) into energy (MeV) allows quick estimation of binding energies, reaction Q-values, and decay energies. Given Data / Assumptions: 1 amu ≈ 1.66054 × 10^−27 kg. Speed of light c ≈ 3.00 × 10^8 m/s. E = m * c^2. Concept / Approach: Compute E for 1 amu and convert joules to electron-volts. The widely used constant is 1 amu ≈ 931 MeV of energy equivalent, which underpins tabulated nuclear masses and binding energies per nucleon (~8 MeV for medium-mass nuclei). Step-by-Step Solution: E = m * c^2 = (1.66054 × 10^−27 kg) * (3.00 × 10^8 m/s)^2.E ≈ 1.494 × 10^−10 J.1 eV = 1.602 × 10^−19 J → E ≈ (1.494 × 10^−10) / (1.602 × 10^−19) eV ≈ 9.31 × 10^8 eV.Hence, E ≈ 931 MeV. Verification / Alternative check: Handbooks and nuclear data tables consistently cite 1 amu ≈ 931 MeV (more precisely ~931.5 MeV). Why Other Options Are Wrong: 9.31 and 93.1 MeV are off by powers of ten; 9310 MeV is too large by a factor of 10. Common Pitfalls: Confusing MeV with MeV/c^2 (a mass unit). 1 amu ≈ 931 MeV/c^2 as a mass; energy equivalent is 931 MeV. Rounding inconsistencies; 931 vs. 931.5 MeV are both acceptable approximations. Final Answer: 931 MeV

Question 4

Indian nuclear tests (general knowledge): India’s second underground nuclear test series was conducted at which test site?

Accepted Answer

Introduction / Context: Knowledge of major scientific and strategic events is common in general studies. India’s nuclear tests are historically associated with a specific test range in Rajasthan. Given Data / Assumptions: “Second underground nuclear test” refers to India’s subsequent test campaign after the first underground test. Options list locations known for defence or nuclear facilities. Concept / Approach: Both the 1974 test (Smiling Buddha) and the 1998 test series (Pokhran-II) were conducted at the Pokhran Test Range in Rajasthan. Other listed places are either nuclear power or research sites (Kalpakkam, Narora) or cities without this specific association (Jaisalmer district is nearby but the named test site is Pokhran). Step-by-Step Solution: Identify India’s nuclear test site: Pokhran Test Range, Rajasthan.Match to options: “Pokhran”.Therefore, select Pokhran. Verification / Alternative check: Public records and historical accounts confirm Pokhran as the site for both series of tests. Why Other Options Are Wrong: Jaisalmer: a district/city, but the designated range is Pokhran. Kalpakkam: site of Madras Atomic Power Station and nuclear research facilities, not test site. Narora: nuclear power station in Uttar Pradesh, not a test range. Common Pitfalls: Confusing power-generation sites with test ranges. Associating the broader region (Thar desert) with a specific named site. Final Answer: Pokhran

Question 5

U-235 fission representation: Which balanced representation best describes the fission of 92U235 by a slow (thermal) neutron, showing typical fission products, emitted neutrons, and released energy?

Accepted Answer

Introduction / Context: Fission reactions are written to show target nucleus, incident neutron, fission fragments, emitted neutrons, and the energy released (Q-value). Recognizing a correct schematic equation is essential in nuclear engineering and physics. Given Data / Assumptions: Thermal neutron induces fission of U-235. Representative fission fragments: barium-143 and krypton-90 (one of many possible splits). About 2–3 fast neutrons are typically emitted; 3 is a common depiction. Concept / Approach: The proper schematic includes the incident neutron, two fission products whose mass numbers add up with the emitted neutrons to conserve nucleon number, and the release of energy Q. Many product pairs exist; Ba-143 and Kr-90 are illustrative. The equation should include + Q (energy). Step-by-Step Solution: Write reaction: 92U235 + 0n1 → 56Ba143 + 36Kr90 + 3 0n1 + Q.Check mass numbers: 235 + 1 = 236 on left; right side 143 + 90 + 3*1 = 236 (balanced).Check charges: 92 = 56 + 36 (balanced).Include Q to denote energy release (~200 MeV per fission). Verification / Alternative check: Any valid pair of fission fragments with correct balancing and Q is acceptable; the given pair and three neutrons is standard in texts. Why Other Options Are Wrong: (a) Involves gamma on the reactant side and omits neutrons/Q; incorrect. (b) Omits explicit Q term. (d) Shows formation of U-236 and immediate release of 3 neutrons; this skips explicit fission fragments and is not the usual fission depiction. Common Pitfalls: Forgetting to include emitted neutrons and energy term. Not checking mass and charge conservation. Final Answer: 92U235 + 0n1 → 56Ba143 + 36Kr90 + 3 0n1 + Q (energy)

Question 6

Thorium fuel cycle: Thorium-232 can be converted (bred) into fissile uranium-233 predominantly in which type of nuclear reactor?

Accepted Answer

Introduction / Context: The thorium fuel cycle relies on neutron capture by Th-232 to form U-233, a fissile nuclide. Understanding the neutron spectrum that favors this breeding step is important in reactor design and national nuclear strategies. Given Data / Assumptions: Fertile nuclide: Th-232. Target fissile product: U-233 after neutron capture and beta decays (Th-232 → Th-233 → Pa-233 → U-233). We compare reactor types by neutron spectrum and moderation. Concept / Approach: Th-232 has a relatively high neutron capture cross-section in the thermal (slow) neutron energy range. Consequently, thermal reactors (including heavy-water-moderated designs) are suitable for breeding U-233 from thorium. While fast systems can also produce U-233, the classic thorium cycle development emphasizes thermal spectrum breeding due to favorable capture cross-sections and well-established moderation technologies. Step-by-Step Solution: Identify the fertile-to-fissile route: Th-232 + n → Th-233 → Pa-233 → U-233.Match to spectra: thermal neutrons provide stronger capture probability for Th-232.Therefore, select thermal reactor as the general category that enables effective conversion. Verification / Alternative check: Heavy-water-moderated reactors (a subset of thermal reactors) are often cited in thorium programs because of high neutron economy, reinforcing the broader answer “thermal reactor.” Why Other Options Are Wrong: Fast breeder reactor: primarily optimized for breeding Pu-239 from U-238; thorium breeding is less favorable in fast spectrum. Heavy-water-moderated is indeed a thermal type, but the broader correct category is “thermal reactor.” “Enriched-uranium reactor” describes fuel rather than neutron spectrum and is not specific to thorium breeding. Common Pitfalls: Confusing coolant or fuel enrichment with neutron spectrum classifications. Assuming only heavy-water designs can breed U-233; other thermal systems can as well, given suitable neutron economy. Final Answer: Thermal reactor

Question 7

Nuclear fuels — Which of the following isotopes is NOT a naturally occurring nuclear fuel found in nature (as mined), considering that naturally occurring fuels include uranium-235/238 and thorium-232?

Accepted Answer

Introduction: This question tests basic nuclear fuel-cycle knowledge. Naturally occurring nuclear fuels are those present in mineable ores in appreciable amounts and usable directly as fissile or fertile materials, notably uranium-235, uranium-238, and thorium-232. Identifying which listed isotope is not naturally occurring helps learners avoid common isotope-number mix ups. Given Data / Assumptions: Naturally occurring fuels include U-235/U-238 and Th-232. Isotopes produced only through neutron reactions in reactors (e.g., U-233, Pu-239) are not naturally abundant. The options mention thorium-233, plutonium-239, and uranium-238. Concept / Approach: We separate isotopes into: (1) naturally abundant (ore-grade) vs. (2) reactor-bred. Fertile isotopes (U-238, Th-232) are not fissile at thermal energies but can be converted (bred) to fissile isotopes (Pu-239 from U-238; U-233 from Th-232). Any isotope not present naturally in appreciable quantities is considered not naturally occurring as a fuel. Step-by-Step Solution: Check U-238: It is abundant in nature and is a fertile nuclear fuel. Valid naturally occurring.Check Th-233: Natural thorium is almost entirely Th-232. Thorium-233 is not a natural isotope; it is a short-lived intermediate in the Th-232 breeding chain (Th-232 → Th-233 → Pa-233 → U-233).Check Pu-239: It is predominantly man-made (bred from U-238); traces may occur in nature, but it is not a naturally occurring fuel in mineable form.Therefore, among the given options, thorium-233 clearly is not naturally occurring as a fuel isotope. Verification / Alternative check: Consulting standard nuclear data tables confirms natural thorium is Th-232; Th-233 does not persist in nature. U-238 is the most abundant uranium isotope. Pu-239 is reactor-bred. Why Other Options Are Wrong: Uranium-238: Naturally abundant (fertile) and therefore a naturally occurring nuclear fuel. Plutonium-239: Primarily man-made, but the question asks for which is not naturally occurring; while Pu-239 is also not natural, the isotope specifically wrong as a 'naturally occurring thorium' fuel is Th-233, making it the best single choice here. None of these: Incorrect because Th-233 is not naturally occurring. Common Pitfalls: Confusing Th-233 with Th-232 (the natural thorium). Also, assuming all listed actinides occur naturally in fuel-grade amounts. Final Answer: Thorium-233

Question 8

Coolants in nuclear reactors — Which gas is commonly used as a reactor coolant specifically because of its very low neutron-capture cross-section?

Accepted Answer

Introduction: Reactor coolants must remove heat efficiently while minimally perturbing the neutron economy. A key property is a low neutron-capture cross-section so that the coolant does not parasitically absorb neutrons needed for sustaining fission. Given Data / Assumptions: We compare common gases: H2, N2, He, CO2. Criterion: very low neutron-capture cross-section. Secondary considerations (thermal capacity, chemistry) do not override the stated criterion. Concept / Approach: Helium is a noble gas with exceptionally low neutron absorption and chemical inertness. It does not become radioactive easily and remains single-phase over wide ranges, making it attractive in high-temperature gas-cooled reactor concepts. Step-by-Step Solution: Evaluate H2: Although light and a good moderator, protium absorbs neutrons more than He, and molecular hydrogen poses flammability concerns.Evaluate N2: Nitrogen has higher absorption than He and can form C-14 via activation; thus it is less desirable for neutron economy.Evaluate CO2: Widely used historically (e.g., AGR, Magnox) due to acceptable properties; however, its neutron absorption is higher than He and it can react at high temperatures with graphite.Evaluate He: Extremely low neutron-capture cross-section and chemically inert. Meets the stated criterion best. Verification / Alternative check: High-temperature gas-cooled reactors (HTGRs) adopt helium primarily for its neutronic and chemical advantages. Cross-section compilations consistently place He among the lowest absorbers. Why Other Options Are Wrong: H2: Higher absorption than He and safety concerns. N2: Larger absorption and activation issues (C-14 formation). CO2: Acceptable coolant historically, but not the lowest absorber among the choices. Common Pitfalls: Confusing 'good moderator' with 'low absorber'; moderation effectiveness does not mean minimal capture losses. Final Answer: He

Question 9

Fast breeder reactors — Considering molten sodium as a primary coolant, which statement is correct regarding temperature capability, corrosion, radioactivity, and chemical reactivity?

Accepted Answer

Introduction: Liquid sodium is widely used in fast breeder reactors because it has excellent heat-transfer properties and permits high operating temperatures at near-atmospheric pressure. Understanding its true advantages and hazards is essential for reactor engineering. Given Data / Assumptions: Sodium coolants operate near atmospheric pressure while reaching high temperatures. Sodium is chemically reactive with air and water. Primary-loop sodium can become radioactive mainly by neutron activation (e.g., Na-24), not because of temperature alone. Concept / Approach: We examine each statement for accuracy: temperature capability, corrosion behavior, radioactivity origin, and chemical reactivity. The correct evaluation must reflect that radioactivity arises from neutron activation, not heat; sodium can be corrosive to certain materials; and it certainly can attain high temperatures without high system pressures. Step-by-Step Solution: Check temperature claim: Sodium permits outlet temperatures in the hundreds of degrees Celsius at low pressure. Thus, saying it cannot attain high temperature is false.Check corrosion claim: Sodium can interact with certain alloys and impurities; asserting it is “not at all corrosive” is incorrect.Check radioactivity claim: Primary sodium becomes radioactive by neutron activation (e.g., forming Na-24), not merely by heating; linking radioactivity to temperature is wrong. However, sodium does react vigorously with air/water, potentially causing fires; the explosive hazard stems from chemistry, not temperature-induced radioactivity.Therefore, the correct choice is that none of the given statements accurately capture sodium’s behavior. Verification / Alternative check: Fast reactor literature emphasizes sodium’s low vapor pressure, high thermal conductivity, activation in-core, and strict handling in inert cover gas to prevent air/water reactions. Why Other Options Are Wrong: (a) False: sodium enables high-temperature operation at low pressure. (b) False: corrosion compatibility depends on materials and impurities. (c) False: radioactivity is due to neutron activation, not temperature; the chemical reactivity part alone is insufficient. Common Pitfalls: Confusing temperature effects with neutron activation; underestimating chemical hazards of sodium in air/water. Final Answer: none of these

Question 10

Fusion power — Commercial power generation from a thermonuclear fusion reactor is not yet realized mainly because:

Accepted Answer

Introduction: Fusion promises abundant, low-carbon energy, but commercial plants remain elusive. The central challenge is maintaining a stable, high-temperature plasma long enough (adequate confinement time and density) to achieve net energy gain. Given Data / Assumptions: Fusion fuels (D and T) are available via seawater and breeding (T from lithium). Reactor concepts include magnetic confinement (tokamak, stellarator) and inertial confinement. Question asks for the main reason commercialization is not yet achieved. Concept / Approach: Net-positive fusion requires satisfying the Lawson criterion (sufficient product of density, temperature, and confinement time). Achieving and controlling these conditions without excessive losses or instabilities is the core obstacle. Step-by-Step Solution: Evaluate scarcity claim: Deuterium is abundant; tritium can be bred from lithium. Fuel scarcity is not primary.Evaluate initiation difficulty: Ignition is challenging, but the sustained control and confinement for net gain is the deeper issue.Evaluate fuel quantity: Required quantities are tiny; quantity is not the limiting factor.Therefore, the main barrier is control and confinement stability, making (a) the best answer. Verification / Alternative check: Research highlights include turbulence suppression, disruption avoidance, and materials able to withstand neutron fluxes, aligning with the control/confinement challenge. Why Other Options Are Wrong: (b) Fuel scarcity is not the issue; D is abundant; T can be bred. (c) Initiation alone is not the key barrier; sustaining and controlling the plasma is. (d) Fuel quantity needed is very small, not prohibitively high. Common Pitfalls: Equating ignition with commercialization; ignoring engineering challenges in confinement, materials, and tritium breeding. Final Answer: it is difficult to control fusion reaction (confinement and stability challenges).

Question 11

Spent nuclear fuel — What does the spent fuel discharged from a thermal (fission) reactor contain?

Accepted Answer

Introduction: Upon discharge, spent nuclear fuel is a complex mixture. Understanding its composition is essential for safety, storage, and reprocessing strategies in the back end of the fuel cycle. Given Data / Assumptions: LWR or PHWR spent fuel after typical burnup. U-235 and U-238 initially present; neutron irradiation occurs. Reprocessing (if applied) separates usable isotopes from high-level waste. Concept / Approach: Fission of U-235 (and Pu isotopes once formed) yields a wide distribution of fission products (e.g., Cs-137, Sr-90). Some U-238 captures neutrons to form Pu-239 and higher plutonium isotopes. Not all initial fissile material is consumed; unburnt uranium remains, and plutonium is created. Step-by-Step Solution: Fission products: Present in significant fractions; principal contributors to heat and radioactivity.Plutonium: Bred from U-238 (U-238 → U-239 → Np-239 → Pu-239), with further isotopic evolution to Pu-240/241 etc.Unused fuel: Residual U-235 and abundant U-238 remain after discharge.Hence, spent fuel contains all three categories. Verification / Alternative check: Standard spent-fuel assays confirm approximate compositions: majority U (mostly U-238), a few percent fission products, and a percent-level plutonium inventory depending on burnup. Why Other Options Are Wrong: (a), (b), (c) individually capture only part of the inventory. Common Pitfalls: Assuming spent fuel is 'used up' completely; ignoring plutonium formation during operation. Final Answer: all (a), (b) & (c)

Question 12

Radioactive emissions — The atomic number of a nuclide remains unchanged during the emission of which type of radiation?

Accepted Answer

Introduction: Radioactive decay can change a nuclide’s atomic number Z (proton count) or only its energy state. Recognizing which emissions alter Z is fundamental in nuclear chemistry and reactor physics. Given Data / Assumptions: Alpha (α) emission ejects a helium nucleus (2 protons, 2 neutrons). Beta (β) emission involves a neutron–proton conversion (β− increases Z by 1; β+ or electron capture decreases Z by 1). Gamma (γ) emission is a photon emitted as the nucleus relaxes from an excited state. Concept / Approach: Only emissions that change the number of protons alter the atomic number. Gamma emission removes nuclear excitation energy without changing proton or neutron counts. Step-by-Step Solution: α: Z decreases by 2 → changes atomic number.β−: Z increases by 1; β+: Z decreases by 1 → changes atomic number.γ: Z unchanged; only energy state changes.Therefore, γ emission leaves atomic number unchanged. Verification / Alternative check: Nuclear reaction equations for γ emission show identical nuclide symbols before and after, with an added γ photon. Why Other Options Are Wrong: α and β emissions explicitly change Z. 'α & β' doubles down on changes to Z and is incorrect. Common Pitfalls: Confusing internal conversion electrons (atomic process) with β decay; mixing β− and β+ effects on Z. Final Answer: γ

Question 13

Reactor cooling comparison — Relative to water cooling, which statement about gas cooling of nuclear reactors is correct?

Accepted Answer

Introduction: Nuclear reactors can be cooled by water, heavy water, liquid metals, or gases (e.g., CO2, He). Understanding the thermodynamic and heat-transfer implications of each coolant avoids misconceptions about attainable temperatures and efficiency. Given Data / Assumptions: Gas-cooled reactors include He- and CO2-cooled designs. Water-cooled reactors are the most common globally. Statements compare temperature capability and efficiency. Concept / Approach: Gas coolants allow high outlet temperatures at low pressure (especially helium) because gases have low vapor pressures; this actually aids high-temperature operation. Specific heat per unit mass or volume of gases is not generally higher than water; water’s volumetric heat capacity is far superior. Steam generation mode (saturated or superheated) depends on the secondary cycle design, not strictly on coolant type. Step-by-Step Solution: Assess (a): False. Gas coolants often enable higher temperatures (e.g., HTGRs) than water under similar pressure constraints.Assess (b): False. Water typically has a much higher volumetric heat capacity, giving strong heat-removal capability; gas cooling is not “more efficient” on that basis.Assess (c): False. Plant design can deliver superheated steam or couple to Brayton cycles; it is not limited to saturated steam.Thus, the correct choice is 'none of these'. Verification / Alternative check: HTGR concepts target outlet temperatures well above typical PWR/PHWR secondary temperatures, supporting advanced power cycles. Why Other Options Are Wrong: (a) Inverts the real advantage of gas cooling. (b) Misstates comparative heat capacities. (c) Restricts steam conditions without technical basis. Common Pitfalls: Equating coolant 'efficiency' with specific heat alone, and ignoring system design. Final Answer: none of these.

Question 14

Radioactive decay accounting — In which decay process does the atomic number NOT decrease?

Accepted Answer

Introduction: Tracking atomic number changes in nuclear decay is essential for balancing nuclear equations. Some processes reduce the proton number, while others increase it or leave it unchanged. Given Data / Assumptions: Atomic number Z represents the number of protons. We consider electron capture, β−, α, and β+ decays. Concept / Approach: Determine the sign of change in Z for each decay mode: electron capture and β+ both convert a proton to a neutron (Z decreases by 1). α emission removes two protons (Z decreases by 2). β− converts a neutron to a proton (Z increases by 1), so Z does not decrease. Step-by-Step Solution: Electron capture: p + e− → n + ν → Z decreases by 1.β+ emission: p → n + β+ + ν → Z decreases by 1.α emission: loss of 2 protons → Z decreases by 2.β− emission: n → p + β− + ν̄ → Z increases by 1, so it does not decrease. Verification / Alternative check: Nuclear charts confirm daughter nuclides after β− decay lie one element higher (Z+1). Why Other Options Are Wrong: Electron capture, α, and β+ all reduce Z. Common Pitfalls: Mixing up β− and β+ effects; forgetting that α emission reduces Z by two. Final Answer: β-emission (beta minus)

Question 15

Indian nuclear sites — Identify the location that does NOT host a nuclear power plant among the following:

Accepted Answer

Introduction: Basic awareness of Indian nuclear power sites helps in engineering context, policy studies, and competitive exams. Distinguishing nuclear from major thermal (coal) sites is a common test point. Given Data / Assumptions: Kaiga, Rawatbhata, and Kalpakkam are well-known nuclear sites operated by NPCIL. Talcher in Odisha is known for large coal-based thermal power generation. We assume contemporary, widely recognized site statuses. Concept / Approach: Match each location to its plant type. Nuclear power stations include Kaiga Generating Station (Karnataka), Rajasthan Atomic Power Station at Rawatbhata, and Madras Atomic Power Station at Kalpakkam. Talcher hosts major coal-fired plants (e.g., NTPC Talcher). Step-by-Step Solution: Check Kaiga: Nuclear site (PHWR units).Check Rawatbhata: Nuclear site (RAPS, PHWR units).Check Kalpakkam: Nuclear site (MAPS) and research facilities (IGCAR).Check Talcher: Predominantly coal-fired power; not a nuclear power plant location. Verification / Alternative check: NPCIL public listings confirm operational nuclear units at Kaiga, Rawatbhata, and Kalpakkam; Talcher features thermal stations. Why Other Options Are Wrong: Kaiga, Rawatbhata, Kalpakkam: Each definitely hosts nuclear units. Common Pitfalls: Confusing large thermal hubs with nuclear sites due to similar scale or media coverage. Final Answer: Talcher (Odisha)

Question 16

Artificially produced fissile materials — Which of the following isotopes are primarily man-made (do not occur naturally in useful amounts) and are produced through breeding reactions?

Accepted Answer

Introduction: The nuclear fuel cycle leverages breeding to create fissile isotopes from fertile materials. Recognizing which fissile nuclides are man-made clarifies how reactors extend fuel resources beyond naturally occurring fissiles. Given Data / Assumptions: Natural uranium contains ~0.7% U-235 and ~99.3% U-238. Thorium in nature is almost entirely Th-232. Breeding converts fertile isotopes (U-238, Th-232) into fissile isotopes (Pu-239, U-233). Concept / Approach: Determine which listed isotopes are principally produced in reactors: U-233 (from Th-232) and Pu-239 (from U-238) are classic bred fissile materials. By contrast, U-235 is naturally occurring in significant, mineable quantities. Step-by-Step Solution: U-235: Naturally present; enriched for most power reactors.U-233: Produced via Th-232 → Th-233 → Pa-233 → U-233.Pu-239: Produced via U-238 → U-239 → Np-239 → Pu-239.Hence, the primarily man-made isotopes among the options are U-233 and Pu-239. Verification / Alternative check: Fuel-cycle references document the thorium–uranium and uranium–plutonium breeding chains and their reaction sequences. Why Other Options Are Wrong: U-235 is naturally occurring; while enrichment is industrial, the isotope itself is not primarily man-made. Common Pitfalls: Assuming “all fissile nuclides are man-made” or confusing enrichment with breeding. Final Answer: Both (b) and (c)

Question 17

In nuclear weapons physics, when a nuclear explosion occurs, the released energy is partitioned into blast, thermal radiation, and nuclear radiation. Which form accounts for the largest share and is therefore considered the primary form of released …

Accepted Answer

Introduction / Context: Nuclear explosions release enormous energy over microseconds. That energy is distributed among several modes: blast (mechanical/kinetic energy driving a shock wave), thermal radiation (intense heat and light), and prompt plus residual nuclear radiation. Understanding which form dominates is essential for effects prediction in civil defense, structural design, and hazard modeling. Given Data / Assumptions: A fission-type nuclear detonation in air (typical atmospheric burst). Energy-partitioning averages used in standard effects texts. We consider the initial energy fractions, not long-term fallout energy. Concept / Approach: The blast wave represents kinetic energy imparted to surrounding air, creating an overpressure front that causes the majority of structural damage. Thermal radiation (heat and light) causes burns and ignitions. Nuclear radiation contributes to acute dose and fallout but is a smaller fraction immediately after detonation compared with blast and thermal. Typical partitions reported in effects handbooks are roughly: blast about half, thermal roughly a third, and prompt radiation the remainder (values vary with yield and environment). Hence, the dominant (primary) form is kinetic energy associated with the blast. Step-by-Step Solution: Identify energy modes: blast (kinetic), thermal, radiation.Use canonical partitioning fractions: blast > thermal > prompt radiation (approximate).Conclude the largest share is blast → kinetic energy. Verification / Alternative check: Standard references on nuclear weapon effects consistently show blast as the single largest fraction for typical air bursts, confirming kinetic energy dominance for primary structural damage mechanisms. Why Other Options Are Wrong: thermal: Large but generally second to blast. electrical and potential are not meaningful primary modes here. nuclear binding refers to the source of energy released, not the form in which it propagates in the environment. Common Pitfalls: Assuming the immense fireball implies thermal is always the largest; neglecting that overpressure and dynamic pressure dominate structural damage budgets. Final Answer: kinetic

Question 18

Fast breeder reactors (FBRs) in nuclear engineering: identify what they do NOT do from the list below.

Accepted Answer

Introduction / Context: Fast breeder reactors are advanced nuclear systems designed to produce more fissile material than they consume. They operate with a fast neutron spectrum and a fertile blanket that breeds fissile isotopes. Knowing what FBRs do and do not do is a core concept in nuclear power technology. Given Data / Assumptions: Standard liquid-metal-cooled fast breeder design (e.g., sodium-cooled). Fertile isotopes: U-238, Th-232. Fissile isotopes: U-235, Pu-239, U-233. Fast spectrum implies little or no moderator. Concept / Approach: FBRs deliberately avoid slowing down neutrons; fast neutrons increase breeding gain in blankets containing fertile material. Common fertile-to-fissile conversions include U-238 → Pu-239 and Th-232 → U-233. However, Th-232 itself is not fissile; it is fertile and must be converted to U-233 before it can sustain fission efficiently. FBRs commonly use liquid sodium because of its excellent heat transfer and low neutron moderation properties. Step-by-Step Solution: Identify the role of fertile vs fissile isotopes.Note that Th-232 is fertile, not fissile.Therefore, an FBR does not “use Th-232 as fissile fuel.” Verification / Alternative check: Design literature distinguishes clearly between fissile driver fuel (e.g., Pu-239, U-235) and fertile blankets (e.g., U-238, Th-232) used for breeding. Why Other Options Are Wrong: use fast neutrons: Core principle of FBRs. convert fertile to fissile: The defining feature (breeding). use molten sodium: Common coolant choice in FBR practice. operate without a moderator: True for fast spectrum operation. Common Pitfalls: Confusing fertile with fissile; assuming thorium can directly serve as fissile material without prior conversion to U-233. Final Answer: use Th-232 as fissile fuel

Question 19

Definition check in basic atomic structure: the atomic number (Z) of an element equals the number of which fundamental particle in the atom?

Accepted Answer

Introduction / Context: The atomic number Z uniquely identifies a chemical element on the periodic table. Distinguishing among protons, neutrons, and electrons clarifies how elements differ and how ions form without changing elemental identity. Given Data / Assumptions: Neutral atoms have electrons equal to protons. Ions may gain or lose electrons but retain the same nucleus. Nucleons = protons + neutrons. Concept / Approach: By definition, atomic number Z is the count of protons in the nucleus. While a neutral atom also has Z electrons, electron count can change during ionization without altering elemental identity. Neutron count can vary, producing isotopes (same Z, different mass number A). Thus, Z strictly equals the number of protons. Step-by-Step Solution: Recall: Z = number of protons.Neutral atom → electrons = protons, but this is not the defining criterion.Therefore, the correct defining particle is the proton. Verification / Alternative check: Periodic table ordering by increasing Z confirms definition; isotopes share Z but have different neutron counts. Why Other Options Are Wrong: neutrons: Determine isotopes, not elemental identity. electrons: Vary in ions. either electrons or protons: Misleading; only protons define Z. nucleons: Total of protons plus neutrons defines mass number A, not Z. Common Pitfalls: Equating neutrality with identity; forgetting that ions are still the same element. Final Answer: protons

Question 20

Thermal power in nuclear reactors: by which primary nuclear process is heat actually generated in a typical thermal reactor using uranium fuel?

Accepted Answer

Introduction / Context: In a thermal nuclear reactor (e.g., PWR or BWR), the fuel contains fissile isotopes such as U-235. Understanding the precise mechanism of heat generation is foundational to reactor physics and safety analysis. Given Data / Assumptions: Low-enriched uranium fuel with significant U-235 content. Thermal (moderated) neutron spectrum. Steady-state reactor operation. Concept / Approach: When a U-235 nucleus absorbs a neutron, it becomes unstable and splits (fissions) into two fission fragments, releasing kinetic energy of fragments, prompt gamma radiation, and additional neutrons. The kinetic energy of fragments is deposited as heat in the fuel matrix, which is transferred to the coolant. Chemical “combustion” does not occur, and nuclear “fusion” is not part of thermal reactor operation. Simple neutron capture without fission does not yield the dominant heat output. Step-by-Step Solution: Identify fissile isotope: U-235 in uranium fuel.Process: neutron-induced fission → energy release mainly as fragment kinetic energy.Heat transfer: fuel → cladding → coolant → steam cycle/electricity. Verification / Alternative check: Reactor heat balances and fission energy per event (~200 MeV) confirm fission as the heat source; capture reactions without fission are parasitic and comparatively minor for heat generation. Why Other Options Are Wrong: fusion: Not employed in current power reactors. absorption without fission: Yields little heat, reduces reactivity. combustion: Chemical process, not applicable. alpha decay: Too slow and weak to power a reactor. Common Pitfalls: Confusing neutron capture with fission; assuming “burning” means chemical combustion. Final Answer: fission of U-235 induced by neutrons

Nuclear Power Engineering Questions