Question 1

Modern Physics – Mesons are predominantly found in which natural/high-energy source?

Accepted Answer

Introduction / Context: Mesons are hadronic particles (such as pions and kaons) composed of a quark–antiquark pair. Historically, mesons were discovered in secondary particle showers produced when high-energy cosmic rays strike the Earth’s atmosphere. Recognizing where mesons are commonly observed links particle physics to astrophysical phenomena. Given Data / Assumptions: The question asks for a natural or high-energy source where mesons are found. Options include electromagnetic radiation (laser, X-rays, gamma rays) and cosmic rays (high-energy charged particles). Mesons are massive particles, not photons. Concept / Approach: Cosmic rays are primarily protons and heavier nuclei traveling at relativistic speeds. When they collide with atmospheric nuclei, they produce showers containing pions (pi-mesons) and kaons, which then decay into muons and neutrinos. Electromagnetic radiation (laser/X/gamma) consists of photons and does not by itself constitute mesons. Step-by-Step Solution: Identify meson nature: hadrons, not light.Select the source that generates particle cascades: cosmic rays in the atmosphere.Eliminate purely photonic options. Verification / Alternative check: Particle detectors at mountain altitudes and collider experiments both record meson production from high-energy collisions—consistent with cosmic-ray air showers. Why Other Options Are Wrong: Laser beam: coherent light (photons), not hadrons. X-rays: high-energy photons; do not “contain” mesons. Gamma rays: also photons; not mesons. Common Pitfalls: Assuming any “high-energy radiation” contains all particle types. Only collisions producing hadronic showers yield mesons. Final Answer: Cosmic rays

Question 2

Nuclear Engineering – Which of the following is used as a moderator in a nuclear reactor?

Accepted Answer

Introduction / Context: In thermal nuclear reactors, a moderator slows down fast neutrons to thermal energies so that they have a higher probability of causing fission in fuels like U-235. Common moderators include graphite, heavy water (D2O), and light (ordinary) water in specific reactor types. Given Data / Assumptions: We must choose a substance known and widely cited as a reactor moderator. Options include graphite and ordinary water (light water), along with two inappropriate radioisotopes. The question expects a single best answer aligned with classic examples. Concept / Approach: Graphite (carbon) is a well-established moderator (e.g., RBMK, early gas-cooled reactors). Light water (ordinary H2O) also acts as a moderator in many pressurized and boiling water reactors; however, in many exam contexts, “graphite” is the prototypical textbook answer when a single choice is expected. Thorium and radium are not moderators (thorium is fertile material; radium is a radioactive element not used for moderation). Step-by-Step Solution: Recall definition: moderator = slows down neutrons without capturing too many.Recognize classic moderator materials: graphite, heavy water; light water in LWRs.Select the prototypical answer: Graphite. Verification / Alternative check: Moderator quality depends on scattering-to-absorption ratio. Graphite has favorable scattering with low absorption, making it a historically important moderator material. Why Other Options Are Wrong: Thorium: acts as fertile material (can breed U-233), not a moderator. Radium: not used as a moderator; high radioactivity and unsuitable properties. Ordinary water: while it does serve as a moderator in LWRs, many single-answer exams expect “graphite” as the canonical choice here; the question format allows only one correct option. Common Pitfalls: Forgetting that several materials can moderate; when only one option is permitted, choose the canonical textbook moderator if multiple correct materials appear. Final Answer: Graphite

Question 3

Nuclear Physics – Nuclear fission in fissile materials is caused by the impact of which particle?

Accepted Answer

Introduction / Context: Nuclear fission is the process where a heavy nucleus (like U-235 or Pu-239) splits into two lighter nuclei with the release of energy and more neutrons. Understanding which incident particle typically triggers fission is fundamental to reactor physics and nuclear engineering. Given Data / Assumptions: We seek the projectile that most effectively induces fission in fissile isotopes. Common candidates: neutron, proton, deuteron, electron. Thermal neutrons are especially effective for U-235. Concept / Approach: Neutrons have no electric charge, so they can penetrate the Coulomb barrier of a nucleus more readily than charged particles (protons, deuterons). Once absorbed, the compound nucleus becomes unstable and may fission. Electrons interact via electromagnetic processes and do not induce fission in the way neutrons do. Step-by-Step Solution: Identify the Coulomb barrier issue for charged particles.Note that neutrons avoid electrostatic repulsion and are readily captured.Recognize that thermal (slow) neutrons are ideal for many fissile isotopes.Select “neutron”. Verification / Alternative check: Chain reactions rely on fission neutrons producing further fissions; this feedback would not work with charged projectiles in conventional reactors. Why Other Options Are Wrong: proton: repelled by positive nucleus; inefficient for inducing fission. deuteron: also charged; used in other nuclear reactions, not standard fission initiation. electron: interacts weakly with nuclei for fission purposes; mainly causes ionization. Common Pitfalls: Assuming any subatomic particle can cause fission; in practice, neutrons are uniquely suited for initiating and sustaining chain reactions. Final Answer: neutron

Question 4

Atomic Structure – Atoms of one element differ from atoms of all other elements primarily in which characteristics?

Accepted Answer

Introduction / Context: Each chemical element is uniquely identified by its atomic number (Z), which equals the number of protons in the nucleus. This determines the number of electrons in a neutral atom and thus drives its electronic configuration and chemical properties. Given Data / Assumptions: We compare features that could distinguish one element from another. Atomic number (Z) uniquely defines the element’s identity. Electronic configuration shapes chemical behavior for the neutral atom. Concept / Approach: Different elements must have different atomic numbers. Consequently, their full electronic configurations also differ for neutral atoms (though elements in the same group may share valence patterns). Neutron count can vary within the same element (isotopes), so it does not uniquely distinguish elements. Step-by-Step Solution: Identify the invariant: atomic number is unique to each element.Link atomic number to electron count in neutral atoms, which sets electronic configuration.Eliminate choices that rely on neutron number or only valence electrons.Choose “atomic number and electronic configuration”. Verification / Alternative check: Elements in the same group (e.g., alkali metals) share valence electron counts but still have different complete configurations and different atomic numbers. Isotopes show the same Z with different neutron numbers, confirming neutrons are not decisive for element identity. Why Other Options Are Wrong: number of neutrons and number of valence electrons: neutron number is not unique to different elements; valence electron counts can repeat across groups. atomic number and number of valence electrons: valence counts can be identical for several elements. number of neutrons and electronic configuration: neutron number varies within an element; configuration is derivative of Z. Common Pitfalls: Confusing isotopes (same element, different neutron counts) with different elements. Only atomic number fixes the elemental identity. Final Answer: atomic number and electronic configuration

Question 5

Environmental Health – Which radioactive pollutant has recently drawn public attention due to its occurrence in building materials?

Accepted Answer

Introduction / Context: Indoor air quality can be affected by naturally occurring radioactive materials. Radon (Rn), a noble gas produced from the decay of radium in the uranium-238 series, can seep from soil and certain building materials (like some granites, phosphogypsum, or fly-ash based products) into homes, contributing to long-term radiation exposure. Given Data / Assumptions: The focus is on a pollutant linked specifically to building materials and indoor environments. Public awareness has risen due to measurements of radon exhalation from materials and entry through foundations. We must select the most relevant pollutant. Concept / Approach: Radon is colorless, odorless, and radioactive. It can accumulate indoors, especially in poorly ventilated spaces. Unlike metals like plutonium (industrial/weapon-related) or thorium/radium in bulk form, radon as a gas readily migrates and is directly inhaled, making it a primary indoor radiation concern. Step-by-Step Solution: Identify pollutant with indoor relevance and linkage to building materials: radon gas.Recognize source: decay of radium/uranium in rocks, soil, and certain construction materials.Choose the option that names the gas directly affecting indoor air. Verification / Alternative check: Health advisories often recommend radon testing in basements/ground floors and improving ventilation or sealing entry pathways where levels are high. Why Other Options Are Wrong: Thorium: present in monazite sands and some materials but not the main indoor gas hazard. Radium: parent of radon; the gas is the immediate indoor air concern. Plutonium: not a typical residential building-material contaminant. Common Pitfalls: Confusing parent radionuclides with the airborne daughter product. Radon’s gaseous nature makes it uniquely problematic indoors. Final Answer: Radon

Question 6

Radioactivity – Which of the following is a positively charged particle (radiation) emitted by a radioactive element?

Accepted Answer

Introduction / Context: Radioactive emissions include alpha, beta, and gamma radiations, each with different charge, mass, and penetrating power. Identifying the charge helps infer shielding needs and biological impact. Given Data / Assumptions: We must choose the radiation that carries a positive electric charge. Options mix particle and electromagnetic emissions. Classical definitions apply: alpha = He nuclei; beta often means electrons; gamma = photons. Concept / Approach: An alpha particle is a helium-4 nucleus containing 2 protons and 2 neutrons; its net charge is +2e. Beta radiation typically refers to beta-minus (electrons, charge −e), though beta-plus (positrons, +e) exists; exam convention usually treats “beta rays” as negative unless specified as positrons. Gamma radiation is neutral electromagnetic radiation, and cathode rays are streams of electrons (negative). Step-by-Step Solution: Recall: alpha = +2 charged, heavy; low penetration, high ionization.Beta (classical) = electrons, −1 charge; can also be positrons if stated as beta-plus.Gamma = photons, neutral.Cathode rays = electrons, −1 charge.Select the only clearly positive option: alpha ray. Verification / Alternative check: Deflection in electric/magnetic fields: alpha deflects towards negative plate; beta (−) towards positive plate; gamma unaffected, which corroborates charge assignments. Why Other Options Are Wrong: Beta ray: conventionally negative electrons unless specified as positrons. Cathode ray: electrons emitted from a cathode; negative. Gamma ray: neutral photons; no charge. Common Pitfalls: Forgetting that “beta” can be either electron or positron; unless specified, exams typically mean beta-minus, which is negative. “Alpha” is unambiguously positive. Final Answer: Alpha ray

Question 7

Physics – Electromagnetic Spectrum (X-rays) What is the typical order of magnitude of the wavelength of X-rays used in physics and medical imaging? Choose the closest standard value.

Accepted Answer

Introduction / Context: X-rays are high-energy electromagnetic waves widely used in medical radiography, crystallography, and materials testing. This question checks whether you know the correct order of magnitude of their wavelength, which helps distinguish X-rays from ultraviolet light and gamma rays in the electromagnetic spectrum. Given Data / Assumptions: X-rays lie between ultraviolet and gamma rays in frequency/energy. Typical diagnostic and diffraction X-rays have very short wavelengths. We must select the closest standard order of magnitude. Concept / Approach: Electromagnetic spectrum regions are often identified by wavelength bands. X-rays commonly used in crystallography (Cu Kα, Mo Kα) have wavelengths on the order of 1 angstrom (1 Å = 10^-10 m). Medical X-ray tubes produce a spectrum with effective wavelengths in the sub-nanometre range, again of angstrom order. Values like centimetres or metres belong to radio/microwave, and microns (micrometres) correspond to infrared, not X-rays. Step-by-Step Solution: Recall: 1 angstrom = 10^-10 m; visible light is ~4000–7000 Å; X-rays are far shorter.Typical X-ray wavelengths for diffraction: about 0.5–2.5 Å.Compare options: 10 micron = 10^-5 m (infrared); 1 cm = 10^-2 m (microwave); 1 m (radio); 1 angstrom = 10^-10 m (X-ray region).Therefore, the best order of magnitude is 1 angstrom. Verification / Alternative check: Energy–wavelength relation: E = hc / λ. X-ray photon energies in keV imply λ in Å. For example, 12.4 keV corresponds roughly to 1 Å using hc ≈ 12.4 keV*Å. This matches the angstrom scale and validates the choice. Why Other Options Are Wrong: 10 micron: Infrared radiation, much longer than X-rays. 1 cm: Microwave band. 1 m: Radio waves. Common Pitfalls: Confusing nanometres and angstroms. Note that 0.1 nm = 1 Å; many X-ray problems quote λ in Å because crystal interplanar spacings are of similar size, enabling diffraction (Bragg’s law). Final Answer: 1 angstrom

Question 8

Atomic Structure – Quantum Numbers Identify the principle stated by the rule: ‘‘No two electrons in an atom can have the same set of four quantum numbers.’’

Accepted Answer

Introduction / Context: Quantum numbers (n, l, m_l, m_s) uniquely describe the state of an electron in an atom. This question targets a foundational postulate that governs electron occupancy of orbitals and explains the structure of the periodic table and electron configurations. Given Data / Assumptions: Statement: No two electrons in an atom can share all four quantum numbers. Four quantum numbers: principal n, azimuthal l, magnetic m_l, and spin m_s. We must name the principle encapsulating the statement. Concept / Approach: The statement is the essence of Pauli’s exclusion principle. It restricts occupancy to at most two electrons per orbital with opposite spins (since n, l, m_l are the same for an orbital, m_s must differ). Aufbau principle is a filling order guideline (lowest energy first), while Newton’s and Bohr’s laws do not address quantum number uniqueness across electrons. Step-by-Step Solution: Recognize the uniqueness clause → exclusion among identical fermions (electrons).Map to principle → Pauli’s exclusion principle.Check consistency with observed electron configurations and the periodic table.Eliminate distractors: Aufbau (filling order), Bohr (quantized orbits), Newton (classical mechanics). Verification / Alternative check: Consider the 1s orbital: it holds two electrons only if spins are +1/2 and −1/2. A third electron cannot have a unique set of four quantum numbers in that orbital, so it must occupy a different state—exactly what Pauli’s principle dictates. Why Other Options Are Wrong: Aufbau principle: Addresses order of filling, not uniqueness of quantum states. Bohr's law: Concerns energy levels and spectral lines, not fermion exclusion. Newton's law: Classical motion; not applicable to quantum electron states. Common Pitfalls: Confusing “how many can fit” (Pauli) with “which fills first” (Aufbau). Both are used together in configuration problems but answer different questions. Final Answer: Pauli's exclusion principle

Question 9

Particle Physics – Relative Mass Scales Which ordering lists the rest masses of the three elementary particle families in decreasing order (heaviest to lightest)?

Accepted Answer

Introduction / Context: In high-energy physics, hadrons (baryons and mesons) are composite particles built from quarks, while leptons (such as electrons and neutrinos) are elementary and generally lighter. Knowing typical mass scales helps you quickly sanity-check reaction energetics and decay possibilities. Given Data / Assumptions: Baryons are three-quark hadrons (e.g., proton, neutron). Mesons are quark–antiquark pairs (e.g., pion, kaon). Leptons include electron, muon, tau, and neutrinos. Concept / Approach: Typical masses: baryons are around 1 GeV/c^2 (proton 938 MeV/c^2); light mesons such as pions are ~140 MeV/c^2; electrons are ~0.511 MeV/c^2, and neutrinos are far lighter. Although some heavy mesons (charmonium, bottomonium) exceed proton mass, the standard comparative order taught in introductory courses is baryons > mesons > leptons for representative species. Step-by-Step Solution: Recall representative masses: proton ~938 MeV, pion ~140 MeV, electron ~0.5 MeV.Arrange decreasing: ~938 > ~140 > ~0.5 → Baryons > Mesons > Leptons.Match to options: choose “Baryons, Mesons, Leptons.”Verify that alternatives flip the correct order. Verification / Alternative check: Cross-check with decay kinematics: baryons can decay into mesons and leptons when allowed; mesons frequently decay into lepton pairs, consistent with mass hierarchies. Why Other Options Are Wrong: Leptons first: contradicts typical mass scales (electron and neutrinos are light). Mesons first: lighter than baryons in standard examples. Disordered listings mix heavy and light families incorrectly. Common Pitfalls: Overgeneralising from heavy mesons (e.g., B mesons) that can exceed proton mass. The question concerns the broad, representative trend taught at entry level. Final Answer: Baryons, Mesons, Leptons

Question 10

Chemistry – Avogadro’s Hypothesis (Terminology) According to Avogadro’s hypothesis, the smallest particle of an element or a compound that can exist independently is called _____.

Accepted Answer

Introduction / Context: Avogadro’s hypothesis states that equal volumes of gases, at the same temperature and pressure, contain equal numbers of particles. The related terminology emphasises that for many substances the independently existing particle is not a single atom but a molecule (e.g., O2, N2). Given Data / Assumptions: We refer to the tiniest independently existing particle of an element or compound. Examples: oxygen gas exists as O2 molecules; water as H2O molecules. Cations and anions are charged species, not the universal form of independent existence. Concept / Approach: In gaseous state, Avogadro treated the basic countable entities as “molecules,” whether the substance is an element (often diatomic) or a compound. Atoms are subunits of molecules for many substances, though monoatomic gases (e.g., noble gases) are special cases where the molecule equals a single atom. Step-by-Step Solution: Identify the key: “exist independently” in ordinary conditions.Recall that many elemental gases are diatomic; compounds are molecular.Therefore, the general term is “molecule.”Eliminate charged species (cation/anion) and the narrower “atom.” Verification / Alternative check: For helium (He), the “molecule” is one atom; for oxygen (O2), the molecule has two atoms. The umbrella term remains molecule across substances, aligning with Avogadro’s counting entity. Why Other Options Are Wrong: a cation / an anion: Ions; not universally representative of all substances. an atom: Too narrow; many substances exist naturally as multi-atom molecules. Common Pitfalls: Equating “smallest particle” with “atom” in all contexts. Chemistry distinguishes atoms (building blocks) from molecules (independent chemical entities). Final Answer: a molecule

Question 11

Physics – Visible Light Range What is the approximate wavelength range of the visible spectrum perceived by the human eye (express your choice in angstrom units)?

Accepted Answer

Introduction / Context: The visible spectrum corresponds to the wavelengths of electromagnetic radiation that stimulate the human visual system, spanning violet to red. Familiarity with this range is important in optics, photometry, and spectroscopy, and helps relate units like nanometres and angstroms. Given Data / Assumptions: 1 angstrom (Å) = 10^-10 m; 1 nm = 10 Å. Typical visible range in nm ≈ 390–760 nm (often quoted as ~400–700 nm). We must choose the angstrom interval matching that range. Concept / Approach: Converting nm to Å: multiply by 10. Thus 390–760 nm becomes 3900–7600 Å. Options suggesting 1300–3000 Å are in the ultraviolet; values near 8500–9800 Å are near the infrared. The narrow 7800–8000 Å band is far too small to represent the entire visible spectrum. Step-by-Step Solution: Recall visible nm range ≈ 390–760 nm.Convert: λ_visible ≈ 3900–7600 Å.Match with options: select 3900–7600 angstrom.Eliminate UV and IR ranges. Verification / Alternative check: Check colour anchors: violet ~400 nm (4000 Å), red ~700–760 nm (~7000–7600 Å). The chosen option brackets these well, confirming correctness. Why Other Options Are Wrong: 8500–9800 Å: Infrared region beyond red. 7800–8000 Å: Too narrow; also infrared boundary. 1300–3000 Å: Ultraviolet (far below visible). Common Pitfalls: Forgetting unit conversion between nm and Å. A quick tip: multiply nm by 10 to get Å; divide Å by 10 to get nm. Final Answer: 3900 - 7600 angstrom

Question 12

Nuclear Physics – Binding in the Nucleus Inside an atomic nucleus, protons and neutrons are held together predominantly by which interaction?

Accepted Answer

Introduction / Context: Nuclear stability arises from a strong, short-range attraction that overcomes the electrostatic repulsion between protons. This question asks you to identify the nature of that binding interaction in standard textbook language. Given Data / Assumptions: Nucleons = protons + neutrons. Protons repel via Coulomb force, yet nuclei are bound. The binding force must be much stronger at femtometre scales but short-ranged. Concept / Approach: The modern view describes the strong nuclear force (residual strong interaction) between nucleons, often modelled historically as an “exchange force” (Yukawa meson exchange). In introductory treatments, “exchange forces” captures this short-range, saturating attraction. Gravitational and magnetic forces are far too weak at nuclear scales, and Coulomb forces are repulsive among protons, not binding. Step-by-Step Solution: Identify requirement: strong, attractive, short-range binding.Map to concept: residual strong interaction ≈ exchange forces (Yukawa).Eliminate alternatives: Coulomb is repulsive (p–p); gravity/magnetism are negligible here.Therefore, select “exchange forces.” Verification / Alternative check: Potential form is short-ranged (falls rapidly beyond ~1–2 fm). This explains why distant nucleons do not bind strongly and why nuclear forces saturate, consistent with measured binding energies per nucleon. Why Other Options Are Wrong: gravitational forces: Incredibly weak at subatomic masses. coulombic forces: Repel protons; cannot be the net binder. magnetic forces: Too weak and not the primary binding mechanism. Common Pitfalls: Answering “Coulombic” because it is a familiar term. Remember: Coulomb explains repulsion, not nuclear cohesion; the strong (exchange) interaction does the binding. Final Answer: exchange forces

Question 13

Nuclear Energy – Fissionable Uranium Which isotope of uranium is capable of sustaining a nuclear chain reaction with thermal (slow) neutrons?

Accepted Answer

Introduction / Context: In reactor physics, only certain nuclides are “fissile,” meaning they readily undergo fission upon absorbing slow (thermal) neutrons and can sustain a chain reaction. Knowing which uranium isotope is fissile is essential for understanding nuclear power and weapons basics. Given Data / Assumptions: We focus on uranium isotopes used in reactors. Thermal neutrons have low kinetic energy (~0.025 eV at room temperature). A self-sustaining chain reaction requires a fissile isotope. Concept / Approach: U-235 is fissile with thermal neutrons and is the key component in natural and enriched uranium fuels. U-238 is fertile (captures a neutron to eventually form Pu-239) but is not fissile with thermal neutrons. U-239 is a short-lived intermediate in plutonium breeding, and U-245 is not a standard naturally occurring or reactor-relevant isotope in basic curricula. Step-by-Step Solution: Identify fissile uranium: U-235.Differentiate U-238: fertile, not thermally fissile.Exclude U-239 (transient in breeding) and U-245 (not standard).Thus, select U-235. Verification / Alternative check: Commercial light-water reactors use fuel enriched in U-235 (typically ~3–5%) precisely because U-235 sustains a thermal-neutron chain reaction. This operational fact corroborates the choice. Why Other Options Are Wrong: U-238: Not fissile with thermal neutrons; requires fast neutrons or breeding to Pu-239. U-239: Short-lived beta emitter, not a fuel. U-245: Not a standard fissile isotope in basic reactor discussions. Common Pitfalls: Confusing “fissile” (readily fissions with slow neutrons) with “fertile” (can be converted into a fissile nuclide). U-238 is fertile, not fissile. Final Answer: U-235

Question 14

Astrophysics – Solar Spectrum (Fraunhofer Lines) The dark lines observed in the solar spectrum are primarily due to which phenomenon?

Accepted Answer

Introduction / Context: The solar spectrum contains numerous dark absorption lines known as Fraunhofer lines. Understanding why they occur links atomic spectroscopy with stellar atmospheres and is a classic application of absorption/emission concepts. Given Data / Assumptions: Light originates in the hot, dense solar interior (continuum source). Light passes through cooler, rarified outer layers (photosphere/chromosphere). Atoms/ions in these cooler layers absorb specific wavelengths. Concept / Approach: Hot, dense sources produce continuous spectra. Cooler gases in front absorb radiation at discrete wavelengths corresponding to electronic transitions, producing dark lines against the continuum. The lines identify elements present in the solar atmosphere. Interference or prism absorption do not explain the precise, element-specific pattern seen across instruments and methods. Step-by-Step Solution: Model: continuum emission from the interior + line absorption in cooler outer layers.Result: missing intensity at specific λ → dark lines in recorded spectrum.Therefore choose: absorption by outer solar layers.Reject alternatives that do not account for element-specific line positions. Verification / Alternative check: Laboratory spectra of elements match solar Fraunhofer lines (e.g., sodium D lines), proving that absorption by outer layers is the cause, independent of instruments like prisms or gratings. Why Other Options Are Wrong: destructive interference: Would depend on path differences and setup, not produce a fixed elemental pattern. absorption by the prism: Instrumental effect cannot mimic the rich, reproducible solar line spectrum. absence from the core: The core emits a broad continuum; absence is not the driver—selective absorption is. Common Pitfalls: Attributing dark lines to the instrument. Spectrographs with different dispersing elements still record the same lines, confirming an astrophysical origin. Final Answer: absorption of corresponding wavelengths by the outer layers of the sun

Question 15

Atomic Structure – Composition of Atoms Which statement best describes the basic composition of atoms in modern atomic theory?

Accepted Answer

Introduction / Context: Atoms consist of a tiny, massive nucleus surrounded by a cloud of electrons. Recognising this structure is fundamental to chemistry, spectroscopy, and nuclear physics. This question asks you to pick the most accurate concise description. Given Data / Assumptions: Nucleus contains protons (positive charge) and neutrons (neutral), together most of the mass. Electrons carry negative charge and occupy orbitals around the nucleus. We need a minimal but correct description. Concept / Approach: “Electrons and nuclei” accurately captures the two major components without listing substructure. Saying “electrons and protons” omits neutrons and misrepresents the nucleus. “Electrons only” or “protons only” are obviously incomplete. Thus, the best answer is the pair that partitions matter into its nuclear core and electronic cloud. Step-by-Step Solution: Identify nucleus + electrons as the two-part model.Note that nucleus = protons + neutrons; electrons are separate.Choose “electrons and nuclei.”Eliminate incomplete or misleading alternatives. Verification / Alternative check: Rutherford scattering and later experiments show a compact, heavy nucleus; spectral lines and quantum mechanics confirm bound electrons. These observations collectively support the nucleus–electron model. Why Other Options Are Wrong: electrons and protons: Excludes neutrons; not the standard phrasing. electrons only / protons only: Discarded by basic evidence. Common Pitfalls: Forgetting neutrons because they carry no charge. Despite neutrality, they are essential to nuclear mass and stability, which is captured by the term “nucleus.” Final Answer: electrons and nuclei

Question 16

History of Atomic Models – Nuclear Atom Who proposed, based on scattering experiments, that most of an atom’s mass is concentrated in a small, dense nucleus?

Accepted Answer

Introduction / Context: Early twentieth-century experiments on alpha-particle scattering transformed our view of atomic structure. This question asks you to identify the scientist who concluded that atoms have a tiny, massive nucleus containing most of their mass. Given Data / Assumptions: Gold-foil experiment with alpha particles. Large-angle scattering events inconsistent with a diffuse “plum pudding” model. We must name the proposer of the nuclear atom. Concept / Approach: Ernest Rutherford interpreted Geiger–Marsden results and proposed the nuclear model: a compact, positively charged nucleus with electrons orbiting (later refined by Bohr and quantum mechanics). Thomson (not “Thompson”) proposed the earlier plum-pudding model; Bohr introduced quantized orbits around Rutherford’s nucleus; Einstein’s work was not on atomic structure models of this type. Step-by-Step Solution: Recall experiment: thin gold foil, alpha particles, scintillation detection.Observation: rare but significant large deflections → concentrated positive charge.Inference: most mass and all positive charge reside in a tiny nucleus.Therefore, attribute to Rutherford. Verification / Alternative check: The calculated nuclear radius (~10^-15 m) compared with atomic radius (~10^-10 m) explains why most alpha particles pass through, while a few scatter strongly—matching experimental statistics. Why Other Options Are Wrong: Thompson (Thomson): Proposed the diffuse model contradicted by scattering results. Bohr: Built on Rutherford by quantizing electron orbits; not the discoverer of the nucleus. Einstein: Not associated with proposing the nuclear atom model. Common Pitfalls: Spelling confusion (Thomson vs. Thompson) and crediting Bohr with the nucleus discovery. Bohr refined energy levels; Rutherford identified the nucleus. Final Answer: Rutherford

Question 17

General Science — Nuclear Radiation and Materials Which of the following has the least penetrating power?

Accepted Answer

Introduction / Context: Radiation safety questions often test your understanding of how different nuclear radiations behave in matter. Penetrating power indicates how deeply a radiation type can pass through materials such as air, paper, aluminum, or lead. Knowing the relative order helps in choosing proper shielding and assessing biological hazards. Given Data / Assumptions: Candidates: alpha particles, beta particles, gamma rays, and the distractor claim that all have the same penetration. Standard conditions: comparison through simple barriers like paper, thin metal sheets, or lead. No special high-energy edge cases are being considered; we use typical energies from common nuclear decays. Concept / Approach: Penetrating power depends on mass, charge, and interaction cross-section. Heavily charged, massive particles interact more strongly with matter and lose energy quickly. Neutral, massless (or effectively massless in the case of photons) radiations interact less frequently and thus penetrate more. The classic ordering is: alpha (least) < beta < gamma (most). Step-by-Step Solution: Alpha particles are helium nuclei (2 protons + 2 neutrons), carrying +2 charge and relatively large mass.Because of high ionization density, alphas are stopped by a few centimeters of air or even a single sheet of paper.Beta particles (electrons/positrons) have far less mass and a single charge, so they penetrate more than alphas; thin aluminum can usually stop them.Gamma rays are high-energy photons with no charge or rest mass; they interact via photoelectric effect, Compton scattering, and pair production and thus penetrate the most, requiring thick lead or concrete for significant attenuation. Verification / Alternative check: A quick rule used in health physics is: alpha blocked by paper/skin, beta blocked by a few mm of aluminum, gamma needs lead/concrete. This mnemonic confirms alpha as the least penetrating. Why Other Options Are Wrong: All have same penetrating power: False; their interactions differ drastically. Beta particles: More penetrating than alpha but less than gamma. Gamma rays: These are the most penetrating, not the least. Common Pitfalls: Confusing ionizing power with penetrating power. Alpha has the highest ionizing power but the lowest penetration; gamma is the opposite. Final Answer: Alpha particles

Question 18

Geochronology — Dating the Earth’s Oldest Rocks The age of the most ancient geological formations is estimated by which method?

Accepted Answer

Introduction / Context: Geologists use radiometric dating to determine the absolute ages of rocks. Different isotopic systems work over different timescales. The question targets the method suitable for dating the most ancient (billions of years old) formations, such as crustal zircons and ancient igneous bodies. Given Data / Assumptions: We must choose among several radiometric methods. Target materials include very old minerals (e.g., zircon) and rocks. We need a method that remains reliable over billions of years and resists alteration. Concept / Approach: Uranium–lead (U–Pb) dating uses the decay of U-238 to Pb-206 and U-235 to Pb-207. Zircon (ZrSiO4) incorporates uranium but rejects lead during crystallization. Any lead present later is radiogenic, making U–Pb in zircon one of the most robust geochronometers for the Archean and Hadean eons. By contrast, radiocarbon (C-14) is limited to recent organic remains (< 50,000 years). Potassium–argon is useful for volcanic rocks but is generally less precise than U–Pb for the very oldest terrestrial materials. Step-by-Step Solution: Identify the age window: “most ancient” implies billions of years.Match isotopic system: U–Pb works from ~1 million years up to Earth’s age and excels with zircon.Eliminate methods with limited ranges: C-14 too young; K–Ar acceptable but typically secondary to U–Pb for the oldest zircons.Select Uranium–lead method. Verification / Alternative check: Oldest known zircons (~4.3–4.4 billion years) are dated by U–Pb concordia methods, confirming its suitability for the most ancient materials. Why Other Options Are Wrong: Ra–Si method: Not a standard geochronometer. Potassium–argon method: Useful but not the premier choice for the very oldest zircons. C14 method: Time window is only tens of thousands of years. Common Pitfalls: Assuming any radiometric method works for all ages. Always match the half-life and mineral system to the age range and rock type. Final Answer: Uranium - lead method

Question 19

Atomic Energy — Source of Energy in Nuclear Explosions In an atomic explosion, the enormous energy released is mainly due to:

Accepted Answer

Introduction / Context: Nuclear explosions (fission or fusion) release energy far beyond chemical reactions. Understanding why requires Einstein’s relation E = m * c^2, which links mass loss to enormous energy output. This principle underlies both nuclear power and nuclear weapons. Given Data / Assumptions: Nuclear processes involve changes in binding energy of nuclei. Small differences in mass (mass defect) occur between reactants and products. Speed of light, c, is very large, so even tiny mass loss produces huge energy. Concept / Approach: During fission, a heavy nucleus splits into smaller fragments; during fusion, light nuclei combine. In both, the total mass of products is slightly less than that of reactants. The “missing” mass converts to energy as dictated by E = m * c^2. This is neither a chemical process nor a transformation of mechanical energy; it is a nuclear-level mass-to-energy conversion. Step-by-Step Solution: Recognize that nuclear binding energy differences create a mass defect.Apply E = m * c^2: small m multiplied by c^2 (a huge number) yields enormous E.Conclude the dominant source is mass-to-energy conversion.Reject options describing chemical or mechanical changes. Verification / Alternative check: Energy per nucleon change in fission/fusion is orders of magnitude higher than in chemical bonds, consistent only with nuclear binding energy differences tied to mass defect. Why Other Options Are Wrong: conversion of chemical energy into heat energy: Chemical reactions cannot account for nuclear yields. conversion of mechanical energy into nuclear energy: Mechanically impossible description. conversion of neutrons into protons: Beta processes may occur but are not the main energy source in an explosion. Common Pitfalls: Confusing the mechanism (mass defect) with byproducts like heat and radiation. Heat is a result, not the fundamental source. Final Answer: conversion of mass into energy

Question 20

Light and Optics — Visible Spectrum How many distinct colours are traditionally recognized in the sunlight (visible) spectrum?

Accepted Answer

Introduction / Context: The visible spectrum is the portion of electromagnetic radiation detectable by the human eye. When white sunlight is dispersed by a prism or water droplets, it separates into constituent wavelengths perceived as colours. Educational convention often enumerates these as seven distinct colours, remembered by the mnemonic VIBGYOR (Violet, Indigo, Blue, Green, Yellow, Orange, Red). Given Data / Assumptions: Traditional school-level classification is used. The question seeks the standard count, not a discussion of continuous spectra. Human perception groups continuous wavelengths into categories. Concept / Approach: Although the spectrum is continuous, cultural and pedagogical tradition recognizes seven colours for simplicity. This aligns with common curricula and basic optics problems. The count arises from historical choices (e.g., Newton’s division aligning with musical notes), not from discrete physical boundaries. Step-by-Step Solution: Recall the mnemonic: Violet, Indigo, Blue, Green, Yellow, Orange, Red.Count these categories: total = 7.Select “Seven.”Acknowledge that the underlying spectrum is continuous, but the answer follows standard teaching. Verification / Alternative check: Observing a rainbow or a prism experiment reproduces the ordered sequence; textbooks list seven named bands. Why Other Options Are Wrong: Three, Four, Five: Undercount the commonly taught categories. Common Pitfalls: Overthinking the continuous nature of spectra. Exams typically expect the conventional “seven colours” answer unless otherwise specified. Final Answer: Seven

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