Q: Fermi–Dirac statistics at the Fermi level For an electron energy level exactly equal to the Fermi level EF, what is the Fermi–Dirac occupation probability f(EF) at thermal equilibrium?

Introduction / Context: Fermi–Dirac statistics govern the occupancy of electron energy states in metals and semiconductors. The Fermi level EF is a key reference energy at which the probability of occupation yields an elegant and temperature-independent result. This concept underpins carrier distributions, density-of-states calculations, and device physics. Given Data / Assumptions: Thermal equilibrium and non-degenerate statistics are not assumed; the exact Fermi–Dirac form is used. We evaluate the probability precisely at E = EF. Temperature T can be any non-negative absolute temperature. Concept / Approach: The Fermi–Dirac distribution is f(E) = 1 / (1 + exp[(E − EF) / (kT)]). At E = EF, the exponential term becomes exp(0) = 1, hence f(EF) = 1 / (1 + 1) = 1/2. This holds at all temperatures, including T = 0 K and T > 0 K, making 0.5 a universal result at the Fermi level in equilibrium. Step-by-Step Solution: Write Fermi–Dirac distribution: f(E) = 1 / (1 + exp[(E − EF)/kT]).Substitute E = EF → exponent = 0.Compute f(EF): 1 / (1 + 1) = 0.5. Verification / Alternative check: At T = 0 K, all states below EF are filled and above EF are empty; the limiting value at EF is 0.5 by convention. At finite T, thermal smearing occurs, but the symmetry about EF keeps f(EF) = 0.5. Why Other Options Are Wrong: 0 or 1 correspond to energies far from EF at T = 0 K; 0.25 is not supported by the Fermi–Dirac formula at E = EF; “Depends on temperature” is incorrect because the value remains 0.5 for any T in equilibrium. Common Pitfalls: Confusing f(EF) with the average occupancy near EF where temperature does influence the slope but not the exact mid-point probability at E = EF. Final Answer: 0.5

Question 1

Curie–Weiss temperature dependence of magnetic susceptibility According to the Curie–Weiss law, how does the magnetic susceptibility χ of a material vary with absolute temperature T (for T sufficiently above any ordering temperature)?

Accepted Answer

Introduction / Context: The Curie–Weiss law extends Curie’s law by including mean-field interactions via the Weiss temperature θ. It is widely used to analyze susceptibility data of paramagnets and ferromagnets above the Curie temperature. Given Data / Assumptions: Linear response regime and temperatures well above ordering temperature. Local-moment picture with mean-field coupling. Concept / Approach: For interacting moments, the effective field is H_eff = H + λM, which shifts the temperature scale. This yields χ = C / (T − θ), so χ varies inversely with (T − θ). The sign and magnitude of θ reveal the nature and strength of magnetic interactions. Step-by-Step Solution: Start from Curie law: χ = C / T for non-interacting spins.Introduce mean-field interaction: replace T by (T − θ).Conclude: χ ∝ 1 / (T − θ). Verification / Alternative check: Plotting 1/χ versus T gives a straight line with intercept θ. This is a standard diagnostic in magnetism labs. Why Other Options Are Wrong: Proportionalities to T, T^2, or independence of T contradict data for local-moment systems. T^-2 is not the Curie–Weiss form. Common Pitfalls: Applying Curie–Weiss law below Tc or in itinerant magnets where deviations occur. Misinterpreting negative θ as antiferromagnetic correlations. Final Answer: χ ∝ 1 / (T − θ)

Question 2

Effect of magnetic field on superconducting transition temperature Tc Consider a superconductor cooled near its transition temperature. What is the general effect of applying an external magnetic field on the observed transition temperature?

Accepted Answer

Introduction / Context: Superconductivity is suppressed by magnetic fields. Engineers must understand how fields reduce the temperature margin for superconducting magnets, MRI systems, and power devices. Given Data / Assumptions: Type I or type II superconductor in the vicinity of its transition temperature. External magnetic field is applied steadily. We consider the general trend, not special cases like strong pinning or mixed states far below Tc. Concept / Approach: In type I superconductors, the thermodynamic critical field Hc decreases with temperature, often approximated by Hc(T) ≈ Hc(0) * [1 − (T/Tc(0))^2]. Equivalently, at fixed H, the transition occurs at a lower temperature than Tc(0). In type II materials, Hc2(T) shows a similar suppression trend: higher fields push the transition to lower temperatures. Step-by-Step Solution: Recognize that superconductivity is a delicate state destabilized by magnetic fields.At fixed H, solve the critical condition H = Hc(T) (type I) or H = Hc2(T) (type II).Because Hc(T) and Hc2(T) decrease with T, a nonzero H forces the loss of superconductivity at a temperature below the zero-field Tc. Verification / Alternative check: Experimental phase diagrams T–H for materials like Pb (type I) and NbTi or Nb3Sn (type II) show Tc(H) < Tc(0) for H > 0. Why Other Options Are Wrong: Increase or no effect contradicts standard phase boundaries. Unpredictable behavior is incorrect; the trend is systematic. Room-temperature superconductivity is not achieved by applying a magnetic field. Common Pitfalls: Confusing critical current or critical field with Tc; all three are interrelated but distinct. Ignoring mixed states in type II; despite vortices, Tc still decreases with applied field. Final Answer: It decreases the transition temperature

Question 3

Superconductivity basics for electrical engineering When a material is in the superconducting state (below its critical temperature, magnetic field, and current limits), what is the electrical resistance of the superconductor under steady direct cur…

Accepted Answer

Introduction / Context: Superconductivity is a quantum state of certain materials in which they lose all electrical resistance and expel magnetic fields (Meissner effect) below specific critical limits. Understanding the correct value of resistance in this state is fundamental for power applications, magnetic resonance, cryogenic electronics, and high-field magnets. Given Data / Assumptions: The material is already in the superconducting phase (temperature below Tc). Applied magnetic field and transport current are below their respective critical values. We are considering steady direct current (DC) conduction, not alternating current losses due to flux motion. Concept / Approach: In a superconductor carrying DC below its critical limits, the electrical resistivity is exactly zero. This is not merely a very small value; it is identically zero within the DC model. Persistent currents can flow indefinitely without a drop in terminal voltage because no Joule heating occurs. This is distinct from ordinary conductors where resistivity increases with temperature due to phonon scattering. Step-by-Step Solution: Identify regime: T < Tc, H < Hc (or Hc2), I < Ic.Recall defining property: ρ_superconductor = 0 for DC transport.Conclude R = 0 for a finite-length specimen (ignoring contacts and leads). Verification / Alternative check: Persistent currents in superconducting rings have been observed to persist for years without measurable decay, which is only possible if resistance is zero. AC conditions can introduce losses due to flux motion in type-II superconductors, but this does not contradict the DC zero-resistance property. Why Other Options Are Wrong: “Low”, “very low”, or “very very low” imply a finite resistance and hence Joule heating; that contradicts the defining feature of superconductivity. “Negligible but not zero” is also incorrect because zero is exact in the DC superconducting state. Common Pitfalls: Confusing DC zero resistance with AC losses; mixing up cryogenic high-purity metals (which have low but non-zero resistance) with the superconducting state (exactly zero DC resistance). Final Answer: zero

Question 4

Fermi–Dirac statistics at the Fermi level For an electron energy level exactly equal to the Fermi level EF, what is the Fermi–Dirac occupation probability f(EF) at thermal equilibrium?

Accepted Answer

Introduction / Context: Fermi–Dirac statistics govern the occupancy of electron energy states in metals and semiconductors. The Fermi level EF is a key reference energy at which the probability of occupation yields an elegant and temperature-independent result. This concept underpins carrier distributions, density-of-states calculations, and device physics. Given Data / Assumptions: Thermal equilibrium and non-degenerate statistics are not assumed; the exact Fermi–Dirac form is used. We evaluate the probability precisely at E = EF. Temperature T can be any non-negative absolute temperature. Concept / Approach: The Fermi–Dirac distribution is f(E) = 1 / (1 + exp[(E − EF) / (kT)]). At E = EF, the exponential term becomes exp(0) = 1, hence f(EF) = 1 / (1 + 1) = 1/2. This holds at all temperatures, including T = 0 K and T > 0 K, making 0.5 a universal result at the Fermi level in equilibrium. Step-by-Step Solution: Write Fermi–Dirac distribution: f(E) = 1 / (1 + exp[(E − EF)/kT]).Substitute E = EF → exponent = 0.Compute f(EF): 1 / (1 + 1) = 0.5. Verification / Alternative check: At T = 0 K, all states below EF are filled and above EF are empty; the limiting value at EF is 0.5 by convention. At finite T, thermal smearing occurs, but the symmetry about EF keeps f(EF) = 0.5. Why Other Options Are Wrong: 0 or 1 correspond to energies far from EF at T = 0 K; 0.25 is not supported by the Fermi–Dirac formula at E = EF; “Depends on temperature” is incorrect because the value remains 0.5 for any T in equilibrium. Common Pitfalls: Confusing f(EF) with the average occupancy near EF where temperature does influence the slope but not the exact mid-point probability at E = EF. Final Answer: 0.5

Question 5

Thermal class limits for electrical insulation systems According to standard insulation classes used in electrical machines and equipment, what is the maximum permissible operating temperature for Class B insulation?

Accepted Answer

Introduction / Context: Insulation classes (A, B, F, H, etc.) define the thermal endurance and maximum operating temperatures for insulating materials used in transformers, motors, and generators. Selecting the appropriate class is essential to ensure reliability and life expectancy under thermal stresses. Given Data / Assumptions: Standard thermal classes are considered (e.g., A, B, F, H). We seek the maximum continuous operating temperature (hot-spot limit) for Class B. Ambient and temperature rise components combine to stay within this limit in design calculations. Concept / Approach: Typical standardized limits: Class A = 105°C, Class B = 130°C, Class F = 155°C, Class H = 180°C. These values guide machine sizing, cooling requirements, and permissible temperature rises for the windings and core. Step-by-Step Solution: Recall standardized class limits.Identify Class B value from the list.Report maximum permissible operating temperature for Class B as 130°C. Verification / Alternative check: Manufacturers’ datasheets and standards consistently list 130°C for Class B systems. Derivations in thermal life models (Arrhenius-type aging) also reference these nominal limits. Why Other Options Are Wrong: 90°C and 105°C correspond to lower classes or specific components; 120°C is not a standard class limit; 155°C corresponds to Class F, not Class B. Common Pitfalls: Confusing insulation class with permissible temperature rise; the class is the absolute hot-spot temperature limit, not merely the rise above ambient. Final Answer: 130°C

Question 6

Matthiessen-type decomposition of metallic resistivity Evaluate the statement: “The resistivity of metals consists of two parts, one approximately temperature independent (residual) and the other temperature dependent (e.g., phonon scattering).” Is …

Accepted Answer

Introduction / Context: Engineers often model metallic resistivity using Matthiessen’s rule, which separates contributions from different scattering mechanisms. Recognizing a residual (impurity/defect) term plus a temperature-dependent term (phonon scattering) is crucial for designing precision resistors, interconnects, and cryogenic instrumentation. Given Data / Assumptions: Ordinary metals in their normal (non-superconducting) state. Temperature range not too close to phase transitions. Impurity and phonon scattering are the dominant mechanisms. Concept / Approach: Matthiessen’s rule states that total resistivity ρ(T) ≈ ρ_residual + ρ_lattice(T). The residual portion arises from static defects (impurities, dislocations, grain boundaries) and is weakly temperature dependent, while the lattice term grows with temperature due to increased electron–phonon scattering. Although exact additivity can break down under some conditions, the decomposition is widely useful and accurate for engineering calculations. Step-by-Step Solution: Identify two contributors: impurity/defect scattering and phonon scattering.Write engineering model: ρ(T) = ρ0 + ρ_ph(T).Use this model to interpret trends: as T increases, ρ_ph(T) increases; as T decreases, ρ(T) approaches ρ0. Verification / Alternative check: Low-temperature data show ρ(T) saturating to a finite residual value; high-purity metals exhibit very small ρ0 and large residual-resistivity ratios (RRR), which validates the decomposition. Why Other Options Are Wrong: Saying “False” ignores well-established scattering physics; limiting it to 0 K or alloys only is unnecessarily restrictive; mentioning superconductors is irrelevant since their DC resistivity is zero below Tc. Common Pitfalls: Assuming perfect additivity at all temperatures and fields; in reality, deviations can occur, but the engineering statement remains correct. Final Answer: True

Question 7

Atomic arrangement, X-ray diffraction, and material properties Assertion (A): The regular or irregular stacking of atoms has an important effect on the properties of materials. Reason (R): The arrangement of atoms in a given material can be studied …

Accepted Answer

Introduction / Context: Crystal structure, defects, and atomic stacking directly influence mechanical, electrical, magnetic, and optical properties. At the same time, X-ray diffraction (XRD) is a primary tool for studying atomic arrangements. This assertion–reason item asks whether the reason explains the assertion, not merely whether both are true. Given Data / Assumptions: A: Atomic stacking order/disorder affects properties. R: X-rays can probe atomic arrangements (lattice parameters, phases, texture). We seek the logical explanatory link between R and A. Concept / Approach: Assertion A is correct: defects, grain size, stacking faults, and crystallographic texture alter yield strength, conductivity, coercivity, and more. Reason R is also correct: XRD measures interplanar spacings and symmetry to characterize such order. However, R does not explain why stacking affects properties; it merely states a method to study the structure. A proper explanation would reference how bonding and electron/phonon scattering depend on structure. Step-by-Step Solution: Evaluate A for truth: well-established in materials science.Evaluate R for truth: XRD is a standard structural probe.Check explanatory link: R is not the underlying reason for property changes; thus, R does not explain A. Verification / Alternative check: Examples include strengthening by precipitation and work hardening (dislocation interactions), and anisotropic properties from crystallographic texture—none are “caused” by the existence of X-ray methods; XRD only reveals them. Why Other Options Are Wrong: (a) incorrectly claims R explains A; (c) and (d) falsely negate either A or R; (e) contradicts established science. Common Pitfalls: Confusing diagnostic tools (XRD) with causal mechanisms of properties (bonding, defects). Final Answer: Both A and R are true but R is not correct explanation of A

Question 8

Ferromagnetism and crystal structure: sodium vs. iron (ferrous) Assertion (A): Both sodium and ferrous (iron) are ferromagnetic. Reason (R): Both sodium and ferrous have a body-centred cubic (BCC) structure.

Accepted Answer

Introduction / Context: Magnetic ordering (ferromagnetic, paramagnetic, etc.) depends on electronic configuration and exchange interactions, while crystal structure can influence but does not alone determine ferromagnetism. This assertion–reason item contrasts sodium and iron at room temperature. Given Data / Assumptions: Sodium (Na) at room temperature: metallic, weakly paramagnetic. Iron (Fe) at room temperature: ferromagnetic (α-Fe, BCC). Crystal structures: Na is BCC; α-Fe is also BCC at room temperature. Concept / Approach: The assertion claims both are ferromagnetic; this is false because sodium is not ferromagnetic. The reason states both have BCC structure; that is true for Na and α-Fe. However, having BCC does not guarantee ferromagnetism (e.g., many BCC metals are not ferromagnetic). Thus the correct logical combination is A false, R true. Step-by-Step Solution: Check magnetism: Na → paramagnetic; Fe → ferromagnetic.Check structure: Na → BCC; α-Fe → BCC.Conclude: Assertion false; Reason true; R does not make A true. Verification / Alternative check: Empirical data show Curie temperature of iron is well above room temperature (∼1043 K), ensuring ferromagnetism. Sodium shows Pauli paramagnetism with no long-range ordering. Why Other Options Are Wrong: (a) and (b) wrongly treat assertion as true; (c) makes R false; (e) denies the true structural statement. Common Pitfalls: Assuming crystal structure alone dictates magnetic order; electronic structure and exchange dominate the outcome. Final Answer: A is false but R is true

Question 9

Permittivity units and constitutive relation Assertion (A): The permittivity of free space ε0 has the dimensions of farad per metre (F/m). Reason (R): The electric flux density satisfies D = ε0 εr E, where εr is dimensionless relative permittivity a…

Accepted Answer

Introduction / Context: Permittivity connects electric field E to electric flux density D. Correct unit analysis is vital in electromagnetics, dielectric design, and capacitor calculations. This item tests whether the constitutive equation explains the dimensionality of ε0. Given Data / Assumptions: D = ε0 εr E in linear, isotropic media. εr is dimensionless; ε0 carries the physical units. Standard SI base units are used for dimensional analysis. Concept / Approach: From D = ε E in vacuum (ε = ε0), the units must satisfy [D] = [ε0][E]. Using SI, [D] = C/m^2 and [E] = V/m. Therefore [ε0] = (C/m^2) / (V/m) = C/(V·m) = F/m, because 1 F = C/V. Hence the constitutive relation directly explains the unit of ε0. Step-by-Step Solution: Write D = ε0 εr E; set εr = 1 for free space.Units: [D] = C/m^2, [E] = V/m.Compute [ε0] = (C/m^2) / (V/m) = C/(V·m) = F/m. Verification / Alternative check: Capacitance per unit length of a parallel-plate capacitor in vacuum, C = ε0 A/d, also yields [ε0] = F/m when solving for units. Why Other Options Are Wrong: Any option denying the equation or the dimensional derivation conflicts with standard SI definitions and Maxwell’s equations. Common Pitfalls: Confusing ε0 (F/m) with εr (dimensionless), or mixing Gaussian and SI unit systems. Final Answer: Both A and R are true and R is correct explanation of A

Question 10

Drift velocity and carrier mobility Assertion (A): The drift velocity of electrons in a conductor or semiconductor is proportional to the applied electric field E. Reason (R): The ratio of drift velocity to electric field is called the mobility of t…

Accepted Answer

Introduction / Context: Carrier transport under an applied electric field is a cornerstone of electronic device physics. The concepts of drift velocity and mobility appear in conductivity, Ohm’s law at the microscopic level, and current–voltage characteristics of semiconductors and metals. Given Data / Assumptions: Low-field (ohmic) regime where velocity is proportional to field. Scattering time and effective mass define mobility in simple Drude-like models. Temperature and impurity concentration influence mobility but not the definition. Concept / Approach: In the low-field regime, v_d = μ E, where v_d is drift velocity and μ is mobility. Thus v_d ∝ E, and μ ≡ v_d / E. This linearity breaks down only at very high fields (velocity saturation, hot-carrier effects), but for ordinary operation the proportionality is valid and mobility provides the constant of proportionality. Step-by-Step Solution: State relation: v_d = μ E.Conclude proportionality: increase E → v_d increases linearly.Define mobility: μ = v_d / E, consistent with the reason. Verification / Alternative check: Conductivity σ = q n μ in semiconductors (or σ = q p μ_p for holes) connects mobility to measurable current density J = σ E, reinforcing the proportional relationship. Why Other Options Are Wrong: Claiming the statements are false contradicts basic transport models; denying the explanatory role of mobility ignores the definition μ = v_d / E. Common Pitfalls: Extending linearity into high-field regimes without accounting for velocity saturation; confusing thermal velocity with drift velocity. Final Answer: Both A and R are true and R is correct explanation of A

Question 11

Curie temperature of iron (ferromagnetic to paramagnetic transition) Approximate the Curie temperature for elemental iron, above which it loses ferromagnetism and becomes paramagnetic. Choose the closest value.

Accepted Answer

Introduction / Context: The Curie temperature Tc marks the phase transition from ferromagnetic to paramagnetic behavior. Knowing Tc is essential in materials selection for magnets, transformers, and sensors where operating temperatures must remain well below Tc to maintain magnetic performance. Given Data / Assumptions: Element: iron (Fe). We seek an approximate numerical value. Room temperature is far below iron’s Tc. Concept / Approach: Iron’s Curie temperature is approximately 1043 K (about 770°C). Above this temperature, thermal agitation disrupts long-range spin alignment, and the material behaves as a paramagnet. Therefore, “about 1000 K” is the closest rounded option commonly used in quick design estimates. Step-by-Step Solution: Recall Tc(Fe) ≈ 1043 K ≈ 770°C.Compare with options: 1000 K is closest to 1043 K.Select “about 1000 K”. Verification / Alternative check: Reference data for α-Fe report Tc between about 1040–1045 K depending on purity and measurement, validating the rounded choice. Why Other Options Are Wrong: 20 K and -20°C are far too low; 100°C is still ferromagnetic; 300 K (room temperature) is below Tc, so iron remains ferromagnetic there. Common Pitfalls: Confusing Celsius and Kelvin scales or mixing iron with other ferromagnets that have different Curie temperatures (e.g., nickel ≈ 627 K). Final Answer: about 1000 K

Question 12

In a crystalline solid with metallic (valence) bonding, how are the outer (valence) electrons shared among atoms? Provide the best description that reflects the collective behavior of valence electrons in a typical metallic crystal.

Accepted Answer

Introduction / Context: Understanding how valence electrons behave inside a solid distinguishes metallic, covalent, and ionic bonding. In metals (often called valence crystals in older texts), outer electrons are not confined to a single atom or a single bond. Instead, they are delocalized over the entire lattice and form an 'electron sea' that enables high electrical and thermal conductivity, ductility, and characteristic metallic luster. Given Data / Assumptions: We are discussing a crystalline solid exhibiting metallic bonding. Atoms contribute valence electrons to a common pool. No external fields or extreme conditions are assumed. Concept / Approach: Metallic bonding arises when metal atoms donate their valence electrons to a shared conduction band. Positive ion cores (atomic nuclei plus inner electrons) are arranged in a regular lattice, while the valence electrons are free to move throughout the crystal. This delocalization explains metallic properties such as conductivity and malleability. Step-by-Step Solution: Identify bonding type: metallic (valence) bonding in a crystalline solid.Assess electron behavior: valence electrons are not localized between specific neighbouring atoms.Select the statement that reflects delocalization across the lattice.Therefore, the correct choice is that electrons are shared by all atoms. Verification / Alternative check: Band theory shows that overlapping atomic orbitals form conduction bands that extend over the crystal; electrons occupy these bands and are mobile. This is incompatible with electrons being confined to specific pairs of atoms as in purely covalent bonds. Why Other Options Are Wrong: Are not shared: contradicts metallic bonding. Shared only between neighbours: describes localized covalent bonds rather than metallic delocalization. Any of the above depending on temperature: temperature affects occupancy and scattering, not the fundamental bonding model. Tightly bound to ions only: inconsistent with conductivity of metals. Common Pitfalls: Confusing metallic delocalization with covalent sharing or assuming electrons remain localized at low temperature; even at low temperatures, metallic electrons remain delocalized across the lattice. Final Answer: Are shared by all atoms (delocalized electron sea)

Question 13

In a semiconductor, the probability of electron–hole recombination is proportional to which carrier measure? Choose the best expression for how recombination rate depends on carrier concentrations.

Accepted Answer

Introduction / Context: Carrier recombination is central to semiconductor devices (diodes, LEDs, solar cells). The rate at which electrons and holes annihilate each other determines lifetime, minority-carrier dynamics, and device efficiency. A basic qualitative rule is that recombination requires both species to be present: electrons and holes must meet to recombine. Given Data / Assumptions: Uniform semiconductor material at steady conditions. Nonradiative and radiative recombination mechanisms are possible. We focus on the basic concentration dependence. Concept / Approach: In its simplest form, the recombination rate R is proportional to the probability of an electron meeting a hole, which scales with the product of their concentrations, n and p. Many models express this as R ∝ n p − ni^2 under nonequilibrium conditions, where ni is the intrinsic concentration. Thus, higher density of either species alone is not sufficient; both must be present for recombination events to occur. Step-by-Step Solution: Recognize recombination requires an electron and a hole simultaneously.Meeting probability ∝ n * p under well-mixed conditions.Therefore, select the dependence on the product n * p. Verification / Alternative check: Device equations (e.g., low-level injection in p–n junctions) commonly use recombination terms proportional to n p − ni^2 with appropriate coefficients for SRH, radiative, or Auger processes, reinforcing the product dependence. Why Other Options Are Wrong: Electrons only or holes only: ignores the need for the other carrier. Sum (n + p): does not represent pairwise interaction probabilities. None of the above: incorrect because n * p dependence is well-established. Common Pitfalls: Assuming linear dependence on a single carrier or forgetting the equilibrium correction term n p ≈ ni^2 in intrinsic conditions. Final Answer: Product of electron and hole densities (n * p)

Question 14

Identify the material class: A photoconductor is best described as a(n) ________. (Choose the most accurate fundamental classification.)

Accepted Answer

Introduction / Context: Photoconductors are materials whose electrical conductivity increases when illuminated. They are essential in light sensors, imaging arrays, and some photovoltaic components. Recognizing their fundamental classification reveals why light controls their conductivity and how devices using them operate. Given Data / Assumptions: Illumination generates additional free carriers via photon absorption. Energy of incident photons is sufficient to excite electrons across the band gap or liberate carriers from traps. Steady temperature and bias conditions. Concept / Approach: In semiconductors, the band gap is small enough that photons of visible/near-IR energy can promote electrons from the valence band to the conduction band, creating electron–hole pairs and increasing conductivity. In pure conductors (metals), abundant carriers already exist and light does not significantly change conductivity. In insulators, very large band gaps prevent significant photogeneration at ordinary photon energies. Step-by-Step Solution: Define photoconduction: conductivity increases under illumination.Relate to band-gap physics: photon absorption generates carriers in semiconductors.Conclude that photoconductors are semiconductors by nature. Verification / Alternative check: Classic photoconductive materials include CdS, CdSe, PbS, and a-Si:H—each a semiconductor whose conductivity rises with illumination intensity. Why Other Options Are Wrong: Conductor: already high carrier density; little photoconductive effect. Insulator: band gap too large for effective photogeneration. Superconductor: zero DC resistance below critical temperature; unrelated mechanism. Dielectric-only: ignores the semiconducting band structure required. Common Pitfalls: Confusing photoconductors with photodiodes or photovoltaic devices; while related, photoconductors rely primarily on increased conductivity rather than junction-based current generation. Final Answer: Semiconductor

Question 15

Under alternating fields, is the dielectric constant (relative permittivity) a complex quantity that captures both energy storage and dielectric loss?

Accepted Answer

Introduction / Context: In AC fields, dielectrics exhibit not only energy storage but also energy dissipation (loss). To represent both effects, engineers use a complex permittivity or complex dielectric constant, whose real part captures stored energy and whose imaginary part captures losses due to polarization lag and conduction-like behavior at the frequency of interest. Given Data / Assumptions: Alternating electric field of angular frequency ω. Linear, time-invariant dielectric response (within small-signal regime). Material may have multiple polarization mechanisms with relaxation times. Concept / Approach: The complex dielectric constant is written as εr(ω) = ε′(ω) − j ε″(ω). The real part ε′ represents energy storage (capacitive behavior), while ε″ represents dielectric loss (out-of-phase response), related to dissipation factor and loss tangent tan δ = ε″/ε′. Frequency-dependent relaxation (Debye, Cole–Cole, etc.) governs how these components vary with ω. Step-by-Step Solution: Identify need to model both storage and loss in AC fields.Represent permittivity as complex quantity: ε = ε′ − j ε″.Interpretation: ε′ → stored energy; ε″ → loss per cycle. Verification / Alternative check: Measurement techniques (LCR meters, impedance spectroscopy) directly report ε′ and ε″ (or tan δ), confirming the complex nature over frequency. Why Other Options Are Wrong: False or restricted cases: Complex permittivity is general across frequencies and material classes, not limited to optics or liquids. Common Pitfalls: Assuming ε is purely real; ignoring dielectric relaxation and conductivity contributions that introduce phase lag and losses. Final Answer: True

Question 16

Name the time parameter: The interval between the generation of a free electron–hole pair and its recombination in a semiconductor is called the ________.

Accepted Answer

Introduction / Context: Carrier lifetime is a key figure of merit in semiconductors. It governs diffusion lengths, photoconductive gain, quantum efficiency, and transient response in devices such as photodiodes, LEDs, and bipolar transistors. Correct terminology helps connect physical understanding to equations used in design and analysis. Given Data / Assumptions: A free electron–hole pair is created (e.g., by light absorption or injection). Recombination eventually annihilates the pair. We refer to the average time interval between creation and recombination. Concept / Approach: The average time a nonequilibrium carrier exists before recombining is called carrier lifetime (electron lifetime or hole lifetime; often minority-carrier lifetime). It is distinct from relaxation time (momentum-scattering time) and collision time (average time between scattering events), which impact mobility but not directly the recombination process duration. Step-by-Step Solution: Clarify the event sequence: generation → existence → recombination.Identify the metric for this interval: lifetime.Reject alternatives that refer to momentum scattering (relaxation/collision time). Verification / Alternative check: Device equations (Shockley–Read–Hall recombination) use lifetime τ as the characteristic time constant for carrier decay back to equilibrium after a perturbation. Why Other Options Are Wrong: Relaxation time / collision time: relate to momentum randomization, not recombination. Lorentz time: not a standard semiconductor parameter. Transit time: time for carriers to traverse a device region, not recombination time. Common Pitfalls: Confusing mobility-related times with recombination lifetime; they influence different aspects of device behavior. Final Answer: Lifetime

Question 17

Statement check: In ferromagnetic materials (below Curie temperature), neighbouring permanent magnetic dipoles tend to align parallel within a domain. Is this statement correct?

Accepted Answer

Introduction / Context: Ferromagnetism arises from exchange interactions that energetically favor parallel alignment of neighboring atomic moments. This leads to regions (domains) with spontaneous magnetization even without an external field. Recognizing domain-level alignment clarifies why ferromagnets can be strongly magnetized and exhibit hysteresis. Given Data / Assumptions: Temperature below Curie temperature TC. Crystalline ferromagnetic material with domain structure. No external field is necessary for spontaneous ordering inside a domain. Concept / Approach: Within a magnetic domain, exchange coupling minimizes energy when neighboring spins align parallel, producing a net magnetization. Different domains can orient differently to reduce magnetostatic energy, so a bulk specimen may have near-zero net magnetization until a field reorients domain walls. Step-by-Step Solution: Identify temperature regime: below TC → ferromagnetic order possible.Within each domain, spins align parallel due to exchange interactions.Conclude the statement is true for domain interiors. Verification / Alternative check: Magnetic imaging (e.g., Kerr microscopy) shows domain patterns with uniform magnetization directions separated by domain walls, validating parallel alignment within domains. Why Other Options Are Wrong: False options ignore exchange interaction or conflate with paramagnetism, where alignment is weak and field-induced. Common Pitfalls: Confusing domain-level order (parallel inside domains) with macroscopic randomness (different domain orientations) that can yield a small net magnetization without an external field. Final Answer: True

Question 18

Units and dimensions: The relative permittivity (dielectric constant) εr of a material has which unit?

Accepted Answer

Introduction / Context: Relative permittivity εr, commonly called the dielectric constant, scales a material’s absolute permittivity relative to vacuum. Correctly identifying its units helps in avoiding dimensional mistakes when using C = ε0 εr A / d or analyzing wave propagation and capacitive effects in materials. Given Data / Assumptions: Absolute permittivity ε = ε0 εr. Vacuum permittivity ε0 has units F/m. εr is defined as a ratio. Concept / Approach: εr = ε / ε0. Since both ε and ε0 have the same units (F/m), their ratio is unitless. Thus εr is a pure number, often frequency dependent and sometimes complex under AC conditions, but dimensionless nonetheless. Step-by-Step Solution: Write εr = ε / ε0.Units cancel: (F/m) / (F/m) → dimensionless.Hence, εr has no units. Verification / Alternative check: Datasheets list εr as a numeric value (e.g., εr ≈ 2.2 for PTFE, ≈ 80 for water at low frequency) without units. Why Other Options Are Wrong: F/m or F: these are units for ε or capacitance, not εr. N/C and V/m: electric field-related units, not permittivity ratios. Common Pitfalls: Confusing absolute permittivity (ε, with units) with relative permittivity (εr, no units). Final Answer: Dimensionless (no units)

Question 19

Assertion–Reason (Semiconductors) Assertion (A): For any semiconductor in thermal equilibrium, the carrier concentrations satisfy n p = n_i^2. Reason (R): A p-type semiconductor is obtained by adding a trivalent impurity to intrinsic material.

Accepted Answer

Introduction / Context: The mass–action law n p = ni^2 is a cornerstone of equilibrium semiconductor physics, relating electron and hole concentrations at a given temperature. Doping determines whether the material is p-type or n-type but does not invalidate the equilibrium relation itself. This question checks clarity on these two separate ideas. Given Data / Assumptions: Thermal equilibrium (no external excitation or strong injection). n and p denote electron and hole concentrations; ni is intrinsic concentration. p-type doping uses acceptor (trivalent) impurities in group-IV semiconductors like Si/Ge. Concept / Approach: Mass–action law: n p = ni^2 arises from detailed balance between generation and recombination processes and the position of Fermi level. Doping shifts individual values of n and p but their product remains fixed at equilibrium. Meanwhile, p-type doping via trivalent atoms (e.g., B in Si) creates acceptor states leading to hole majority carriers. Step-by-Step Solution: Evaluate A: n p = ni^2 → true in thermal equilibrium.Evaluate R: p-type via trivalent dopants → true statement.Causal link: R does not explain A; doping type does not derive mass–action law.Therefore, the correct choice is “both true, but R not the explanation”. Verification / Alternative check: Textbook derivations from Fermi statistics show n = NC exp [−(EC − EF)/kT], p = NV exp [−(EF − EV)/kT]; multiplying yields n p = NCNV exp [−(EC − EV)/kT] = ni^2. Why Other Options Are Wrong: A true, R false: R is factually correct. R explains A: incorrect linkage. A false: contradicts equilibrium semiconductor theory. Common Pitfalls: Believing that heavy doping breaks the mass–action law; at equilibrium it still holds (with bandgap narrowing caveats at extreme levels, but the principle remains). Final Answer: Both A and R are true but R is not the correct explanation of A

Materials and Components Questions