Quantum Physics And Applications Notes Unit 2 – B.Tech 1st Year AKTU

Course: AKTU B.Tech First Year | Semester I & II

Course Codes: AAS101A–AAS101I | AAS201A–AAS201I

Streams: CSE, EE, ECE, CE, ME, BT, CH, TX & AG

Unit II Overview

Connect electronic states with conduction, material structure with mechanical response, fluid motion with pressure, and heat transfer with soil energy. The optics routes explain fibres, lasers, interference, diffraction and polarisation. The branch guide identifies each stream’s prescribed Unit II.

Topics Covered

Learning Outcomes

• Explain bands, Fermi statistics, carrier concentrations and conductivity.
• Connect atomic packing, elastic response, wetting and viscoelasticity.
• Analyse pressure, viscosity, pipe flow, capillarity and fluidization.
• Apply heat-transfer laws and interpret soil energy and evaporation.
• Explain fibre guidance, laser action and the main wave-optics experiments.

Included in These Notes

50 solved problems with methods and checks; 35 labelled diagrams and plots; complete derivations; a formula sheet with conditions; unit conversions and answered revision checks.

Study Approach

Choose your route. Use the branch guide to find your branch, paired semester codes and assigned sections.

Follow the mechanism. Read the axes, ray paths, energy flows and labels before memorising a relation.

Check the assumptions. Confirm equilibrium, flow regime, heat-transfer signs and optical conventions.

Practise and revise. Solve your route’s examples, then review the derivations and unit conversions.

Coverage note: Unit II differs across the streams. This is one combined study volume with separate syllabus routes, not one identical syllabus for every branch.

Unit II: Branch and Topic Guide

Select your stream below. Semester I and II codes are paired within each row; the assigned sections cover that stream’s prescribed Unit II topics.

Materials route: ME reads atoms, bonding, BCC/FCC/HCP, lattice parameters, stress, strain and Young/bulk/shear moduli (11–18). BT reads those foundations plus amorphous structure, viscoelasticity, surface energy, wetting, contact angle, adhesion and cohesion (19–20).

Solids route: CSE, EE and ECE read free electrons, Fermi-Dirac statistics, density of states, bands, gap, effective mass, intrinsic/extrinsic carriers, conductivity and Fermi level (1–10).

Other routes: CH reads fluid properties through fluidization and slurry/polymer rheology (21–26); AG reads thermodynamic laws through soil energy and evaporation (27–30); CE reads fibres and lasers (31–35); TX reads interference, diffraction and polarisation including Nicol and optical activity (36–41).

Important distinction: Shared concepts can appear in another stream at a different unit number. This guide identifies Unit II only. CH uses suffix G from the course scheme; its detailed subject heading contains a conflicting suffix in the supplied document.

1. A. Quantum Theory of Solids: From Atoms to Current

Prescribed for CSE, EE and ECE. A solid contains many interacting atoms. Electrons occupy quantum states constrained by the crystal potential and the exclusion principle. Electrical behaviour depends on the available states, their occupation and how carriers respond to forces. A large number of electrons alone does not guarantee easy conduction.

Begin with Three Separate Questions

Which energies are allowed, which allowed states are occupied, and how does occupation change when a field is applied?

Density of states answers the first counting question; Fermi-Dirac statistics answers the second; velocity, scattering and effective mass help answer the third.

The classical Drude model treats mobile electrons as a gas undergoing collisions. Between collisions a field changes momentum. With mean relaxation time τ, the steady mean drift is opposite the electric field for electrons. Conventional current remains along the field. The random microscopic speed is much larger than the net drift speed.

The simplest free-electron model neglects the detailed periodic potential. It is useful for basic metallic estimates but does not explain band gaps by itself. Quantum statistics improves state filling; adding the periodic potential explains allowed bands and forbidden gaps. These are successive model refinements, not interchangeable assumptions.

2. Drude Conductivity and the Quantum Improvement

For an electron, m dv/dt = -eE – mv/τ in the relaxation-time picture. At steady drift, vd = -eτE/m. Current density is J = -nevd, so J = ne²τE/m and σ = ne²τ/m. Mobility is μ = |vd|/E = eτ/m; resistivity is ρ = 1/σ.

The mean free path estimate is ℓ = vτ. State which speed v is used. In a degenerate metal the relevant electronic speed is normally of the order of the Fermi speed, not the tiny drift speed. A specimen resistance is R = ρL/A, so it also depends on geometry.

The classical model gives useful transport intuition but fails quantitatively for electronic heat capacity and some temperature dependences. Sommerfeld retains an approximately free gas but fills states using Pauli exclusion and Fermi-Dirac statistics. Only a thin energy region near the Fermi level responds strongly to small thermal perturbations in a metal.

At ordinary temperatures, lattice vibrations and defects scatter carriers. A relaxation time is an effective description, not a statement that every electron collides after exactly τ. Mobility and carrier concentration can both vary with temperature; predicting conductivity requires considering both.

Worked Problem 2.1: Metal Conductivity

Problem: A metal has n = 8.50 × 10^28 m^-3 and τ = 2.50 × 10^-14 s. Use the free-electron mass to find conductivity and resistivity.

Method: σ = ne²τ/m; ρ = 1/σ.

Calculation: σ = (8.50 × 10^28)e²(2.50 × 10^-14)/m; take the reciprocal for ρ.

Answer: σ = 5.988 × 10^7 S/m; ρ = 1.67 × 10^-8 ohm m.

Check: Carrier density alone is not a conductivity; scattering enters through τ.

Worked Problem 2.2: Drift and Current

Problem: A metal has n = 8.0 × 10^28 m^-3, area 1.0 mm² and current 2.0 A. Find electron drift-speed magnitude.

Method: I = neA|vd|.

Calculation: A = 1.0 × 10^-6 m²; |vd| = 2/[8.0 × 10^28 e × 10^-6].

Answer: |vd| = 0.000156 m/s.

Check: Electron drift is opposite conventional current. This is not the random microscopic speed.

3. Counting Free-Electron States and Fermi Energy

Use a large cubic specimen with periodic boundary conditions to count states in k-space. Each allowed k-state occupies volume (2π)³/V in k-space, and spin supplies a factor of two. Counting the states inside a sphere gives N = VkF³/(3π²) at zero temperature.

With n = N/V, kF = (3π²n)^(1/3). The free-electron dispersion is E = ħ²k²/(2m), giving EF = ħ²(3π²n)^(2/3)/(2m). The zero of energy is the chosen band minimum. A semiconductor parabolic-band estimate replaces m by an appropriate effective mass, with valley degeneracy accounted for separately.

The Fermi speed is vF = ħkF/m and the Fermi temperature is TF = EF/kB. TF is an energy expressed as a temperature scale; it does not mean the specimen is physically at TF. When T is much smaller than TF, the electron gas is degenerate.

Integrating energy over the occupied free-electron sphere gives mean energy 3EF/5 at zero temperature. This differs from the classical 3kBT/2. EF is the highest occupied energy at T = 0, not the energy of every electron.

Worked Problem 2.3: Fermi Scales

Problem: Estimate kF, EF, vF and TF for n = 8.50 × 10^28 m^-3 in a three-dimensional free-electron gas at T = 0.

Method: kF = (3π²n)^(1/3); EF = ħ²kF²/(2m).

Calculation: Use vF = ħkF/m and TF = EF/kB after evaluating kF.

Answer: kF = 1.36 × 10^10 m^-1; EF = 7.049 eV; vF = 1.575 × 10^6 m/s; TF = 8.18 × 10^4 K.

Check: Room temperature is far below this TF, consistent with degeneracy.

Worked Problem 2.4: Density Scaling

Problem: A free-electron density increases by a factor eight with mass fixed. Find the factors for kF, EF and vF.

Method: kF ∝ n^(1/3), EF ∝ n^(2/3), vF ∝ n^(1/3).

Calculation: 8^(1/3) = 2 and 8^(2/3) = 4.

Answer: kF doubles; EF quadruples; vF doubles.

Check: The Fermi energy does not scale linearly with carrier density.

4. Fermi-Dirac Distribution and Thermal Smearing

The occupation probability of an available one-electron state is f(E) = 1/[1 + exp((E – EF)/(kBT))]. At finite temperature EF denotes the equilibrium chemical potential in this notation. The probability is dimensionless and lies between zero and one.

At T = 0, states below EF are occupied and those above EF are empty. At finite T, the sharp edge becomes smooth over a few kBT. At E = EF the probability is one half. For equally spaced energies about EF, f(EF + Δ) + f(EF – Δ) = 1.

When E – EF is several kBT above zero, f(E) is approximately exp[-(E – EF)/(kBT)]. This Maxwell-Boltzmann tail is useful in a non-degenerate semiconductor. It is not valid near EF or inside a heavily occupied band.

Thermal excitation creates vacant states below the former edge and occupied states above it while conserving the total electron number in an isolated specimen. Do not infer a carrier concentration from f alone: the number of available states must also be included.

Worked Problem 2.5: Occupation Above EF

Problem: At 300 K find f(E) for E – EF = 0.100 eV.

Method: f = 1/[1 + exp(ΔE/kBT)].

Calculation: kBT = 0.02585 eV; exponent = 0.100/(kBT).

Answer: f = 0.02047; the state 0.100 eV below EF has occupation 0.97953.

Check: The sum for symmetric offsets is one; this is not a carrier-density calculation.

Worked Problem 2.6: Temperature for Specified Occupation

Problem: At E – EF = 0.150 eV the occupation is 0.100. Assume EF is fixed for this exercise. Find T.

Method: exp(ΔE/kBT) = 1/f – 1 = 9.

Calculation: T = 0.150/[kB(in eV/K) ln9].

Answer: T = 792.2 K.

Check: Taking a logarithm, rather than multiplying by the exponential, isolates T.

5. Density of States: Availability Is Not Occupation

For a three-dimensional isotropic parabolic conduction band, the density of available states per unit volume per unit energy is gC(E) = (1/2π²)(2m*/ħ²)^(3/2)√(E – EC), for E ≥ EC. This expression includes spin degeneracy two and one valley; extra equivalent valleys need their degeneracy factor.

For a free electron gas take EC = 0 and m* = m. Integrating g(E) from zero to E gives the number of states per volume up to E. Multiplying g(E) by f(E) gives the occupied-electron density per energy interval. Therefore n = ∫ gC(E)f(E)dE, integrated over the conduction band.

For holes use the valence-band states weighted by 1 – f(E). The density of states is not the same as electron density and need not be populated. A forbidden gap has no ideal bulk band states even if the Fermi function is mathematically defined there.

If energy is in joules, g has units m^-3 J^-1. To express a numerical density per eV, multiply the per-joule value by the joule-per-eV conversion. Effective density NC has units m^-3, unlike g(E).

Worked Problem 2.7: Density-of-States Ratio

Problem: For the same isotropic parabolic band, compare g at E – EC = 0.36 eV and 0.09 eV.

Method: g ∝ √(E – EC).

Calculation: Ratio = √(0.36/0.09).

Answer: The density of available states is twice as large at the higher energy.

Check: The ratio of occupied densities also depends on f(E); it need not be two.

6. Formation of Bands and Classification of Solids

When N identical atoms are brought together, interactions split an isolated atomic level into many closely spaced levels. In a macroscopic crystal these form an allowed energy band. The periodic potential creates ranges of forbidden energy between bands; this cannot be obtained from a completely potential-free gas alone.

The valence band is the highest relevant band normally filled at low temperature; the conduction band supplies higher accessible mobile-electron states. Eg = EC – EV is the band gap. A partially filled band or overlapping relevant bands allows metallic conduction without a large thermal activation across a gap.

In an ideal intrinsic semiconductor, thermal excitation promotes electrons across a moderate gap, leaving holes. An insulator has a larger gap relative to ordinary excitation energies. The distinction is quantitative and temperature dependent; do not use one fixed gap threshold as a universal definition.

A completely filled ideal band carries no net ordinary current under a small field because symmetry-related velocities cancel and nearby states are unavailable. Empty bands have no electrons to carry current. Holes represent missing electrons near the valence-band top and behave as positive mobile carriers.

7. Effective Mass from Band Curvature

An electron wave packet in a band has group velocity v = (1/ħ)dE/dk. Under a weak external force, ħ dk/dt = F. Differentiating the velocity gives a = (1/ħ²)(d²E/dk²)F, so in one dimension m* = ħ²/(d²E/dk²).

Near a conduction minimum, E ≈ EC + ħ²(k – k0)²/(2m*), and the curvature is positive. A flatter band means a larger effective mass. Near a valence maximum the electron curvature is negative; the hole description uses a positive hole mass and positive charge.

Effective mass is the crystal response parameter of a carrier, not a literal change of its vacuum rest mass. In general it is a tensor. The mass used for density of states need not equal a directional conductivity mass; introductory isotropic problems must specify the intended approximation.

When E is given in eV and k in m^-1, convert the energy curvature to joule units before using ħ in SI. For E = Ak² with A in J m², d²E/dk² = 2A and m* = ħ²/(2A).

Worked Problem 2.8: Effective Mass from a Parabola

Problem: A band near its minimum is E = EC + A k² with A = 3.05213 × 10^-38 J m². Find m* relative to the free-electron mass.

Method: d²E/dk² = 2A; m* = ħ²/(2A).

Calculation: Differentiate twice, substitute the given SI coefficient and divide by m.

Answer: m* = 1.822 × 10^-31 kg = 0.200 m.

Check: A smaller curvature would give a larger effective mass, not a smaller one.

8. Intrinsic Semiconductors and the Fermi Level

A pure intrinsic semiconductor in thermal equilibrium has n = p = ni. Electron-hole pairs are thermally generated and recombine. The material remains neutral; a hole is not an extra ion or a proton moving through the lattice.

In the non-degenerate parabolic-band approximation, n = NC exp[-(EC – EF)/(kBT)] and p = NV exp[-(EF – EV)/(kBT)]. The effective densities NC = 2(2πmnkBT/h²)^(3/2) and NV = 2(2πmpkBT/h²)^(3/2) use suitable density-of-states masses and any required degeneracy conventions.

Multiplying gives np = NCNV exp[-Eg/(kBT)] = ni². Thus ni = √(NCNV)exp[-Eg/(2kBT)]. The factor of two in the gap exponent is essential. As temperature increases, pair generation rises strongly, although the band gap and mobility also change.

Equating n and p gives Ei = (EC + EV)/2 + (kBT/2)ln(NV/NC). The intrinsic Fermi level is exactly midgap only when NC = NV in this model. A Fermi level inside the gap is an occupation parameter, not an allowed stationary electron energy there.

Worked Problem 2.9: Intrinsic Concentration

Problem: At 300 K, use NC = 2.80 × 10^25 m^-3, NV = 1.04 × 10^25 m^-3 and Eg = 1.12 eV. Find ni.

Method: ni = √(NCNV) exp[-Eg/(2kBT)].

Calculation: Use kBT = 0.025852 eV; exponent = -1.12/(2 × 0.025852).

Answer: ni = 6.676 × 10^15 m^-3, for these supplied model parameters.

Check: Do not replace the factor 2kBT by kBT; material parameters are temperature dependent.

Worked Problem 2.10: Intrinsic Level Offset

Problem: For the preceding NC and NV at 300 K, locate Ei relative to midgap.

Method: Ei – Emid = (kBT/2)ln(NV/NC).

Calculation: NV/NC = 1.04/2.80; its logarithm is negative.

Answer: Ei – Emid = -0.01280 eV; the level lies slightly below midgap.

Check: Intrinsic does not require exactly midgap when the band densities differ.

9. Extrinsic Semiconductors, Charge Neutrality and Regimes

A donor such as phosphorus in silicon contributes an electron through a shallow donor state. An acceptor such as boron accepts an electron and supplies a hole. Donor states lie near EC and acceptor states near EV in the simple shallow-impurity picture.

For n-type material with complete donor ionisation and negligible acceptors, n ≈ ND when ND is much greater than ni; p = ni²/n. For p-type material, p ≈ NA and n = ni²/p. Ionised donors are positive fixed centres; ionised acceptors are negative fixed centres. The bulk specimen can still be electrically neutral.

Charge neutrality is p + ND+ = n + NA-. With fully ionised compensation, define D = ND – NA. Combining n – p = D and np = ni² gives n = [D + √(D² + 4ni²)]/2 and p = ni²/n. Do not subtract dopants and then discard minority carriers when the difference is comparable to ni.

Temperature regimes are freeze-out, extrinsic and intrinsic. At low temperature incomplete ionisation invalidates n ≈ ND. At high temperature intrinsic generation dominates. Heavy doping can invalidate the Boltzmann approximation, requiring full Fermi-Dirac statistics.

Worked Problem 2.11: Donors and Minority Holes

Problem: Use ni = 1.0 × 10^16 m^-3 and fully ionised ND = 1.0 × 10^22 m^-3, with no acceptors. Find n and p in the extrinsic approximation.

Method: n ≈ ND; np = ni².

Calculation: p = 10^32 / 10^22.

Answer: n ≈ 1.0 × 10^22 m^-3; p ≈ 1.0 × 10^10 m^-3.

Check: n greatly exceeds ni, supporting the stated extrinsic approximation.

Worked Problem 2.12: Compensated Sample Near Intrinsic

Problem: At equilibrium ni = 1.0 × 10^16 m^-3, ND = 3.0 × 10^16 and NA = 1.0 × 10^16 m^-3, all ionised. Find both carrier densities.

Method: n – p = ND – NA; np = ni².

Calculation: D = 2.0 × 10^16; n = [D + √(D² + 4ni²)]/2; p = ni²/n.

Answer: n = 2.414 × 10^16 m^-3; p = 4.142 × 10^15 m^-3.

Check: Here simply setting n = ND – NA loses a significant minority-carrier contribution.

10. Semiconductor Conductivity and a Complete Solution Method

The drift conductivity is σ = e(nμn + pμp). Both terms are positive because conventional electron current and hole current are along the field. Electron velocity itself is opposite the field. With field magnitude E, |J| = σE and total current I = JA for a uniform specimen.

Intrinsic conductivity is σi = eni(μn + μp). In an n-type extrinsic sample the electron term often dominates, but verify the relative products nμn and pμp. Resistivity is the reciprocal of conductivity, and specimen resistance additionally requires L/A.

The equilibrium Fermi-level displacement follows EF – Ei = kBT ln(n/ni) = -kBT ln(p/ni), assuming the same temperature and non-degenerate statistics. Mobility may be quoted in cm²/(V s): multiply by 10^-4 to obtain m²/(V s). Concentration in cm^-3 must be multiplied by 10^6 to obtain m^-3.

Working sequence: Identify intrinsic or doped conditions; state ionisation and equilibrium assumptions; enforce neutrality; calculate minority carriers; convert all units; evaluate σ; then calculate any geometry-dependent resistance. Check that the predicted Fermi level is consistent with the assumed non-degenerate regime.

Common errors: A larger carrier density does not automatically mean a proportionally larger conductivity when mobility changes. The Fermi level shifts toward the majority-carrier band; it is not itself a band edge.

Worked Problem 2.13: Intrinsic Conductivity

Problem: Use ni = 1.0 × 10^16 m^-3, μn = 0.135 and μp = 0.048 m²/(V s). Find σi and ρi.

Method: σi = eni(μn + μp).

Calculation: Add the mobilities, multiply by e ni, then invert.

Answer: σi = 0.0002932 S/m; ρi = 3410.7 ohm m.

Check: Both carrier contributions are positive in conventional current.

Worked Problem 2.14: Mobility Units and Specimen Resistance

Problem: An n-type specimen has n = 10^16 cm^-3, μn = 1200 cm²/(V s), L = 2 mm and A = 1 mm². Neglect holes. Find R.

Method: σ ≈ enμn; R = L/(σA).

Calculation: n = 10^22 m^-3; μn = 0.120 m²/(V s); L/A = 2000 m^-1.

Answer: σ = 192.3 S/m; R = 10.4 ohm.

Check: Converting only mobility while leaving concentration in cm^-3 gives a million-fold error.

Worked Problem 2.15: Doping and Fermi-Level Shift

Problem: At 300 K an equilibrium non-degenerate sample has n/ni = 10^4. Find EF – Ei.

Method: EF – Ei = kBT ln(n/ni).

Calculation: Use natural logarithm ln(10^4), with kBT in eV.

Answer: EF – Ei = 0.23811 eV.

Check: The positive shift is toward the conduction band. Check non-degeneracy against the actual band edges.

11. B. Atomic Structure and Bonding – ME and BT

For ME and BT, Unit II begins with atomic structure and the structure-property connection. An atom contains a small positive nucleus and electrons occupying quantum states. Atomic number gives proton count; mass number gives protons plus neutrons. Isotopes share atomic number but differ in neutron count.

Valence electrons strongly influence bonding and hence electrical, mechanical and thermal behaviour. A material property cannot be inferred from atomic mass alone. Bond type, atomic arrangement, defects, temperature and microstructure all contribute.

A crystal has a repeating long-range arrangement described using a lattice and basis. An amorphous solid lacks that long-range periodic order, although it can retain local atomic organisation. A grain is one crystal region in a polycrystalline material; grain boundaries separate differently oriented regions.

Structure is described at several scales: electronic bonding, atomic packing, defects/grains and the macroscopic component. This unit introduces the ideal structures and elastic measures before more specialised process-material behaviour.

Worked Problem 2.16: Atomic Number and Isotope Count

Problem: A neutral atom is written with atomic number 26 and mass number 56. Find proton, electron and neutron counts.

Method: Atomic number fixes protons; neutrality fixes electrons.

Calculation: Protons = 26; electrons = 26.

Calculation: Neutrons = 56 – 26.

Answer: 26 protons, 26 electrons and 30 neutrons.

Check: An isotope changes neutron count while retaining atomic number. Ionisation changes electron count, not the number of protons in the element.

12. Ionic, Covalent and Metallic Bonding

Ionic bonding: Electron transfer produces oppositely charged ions whose electrostatic attraction helps bind the solid. Charge neutrality and ion size influence the structure. Ionic solids are often stiff and brittle; an insulating solid can conduct when ions become mobile under suitable conditions.

Covalent bonding: Atoms share electrons in directional bonds. The directionality constrains local geometry and can produce strong networks, such as diamond or silicon. Properties vary widely: bonding alone does not imply every covalent material has identical conductivity or hardness.

Metallic bonding: Valence electrons are delocalised through a lattice of positive ion cores. This supports electrical/thermal conduction and nondirectional bonding. Atomic planes can rearrange through deformation mechanisms, explaining why many metals are ductile rather than brittle.

Bond energy and equilibrium separation help explain elastic response. Stretching slightly from equilibrium produces a restoring force; temperature and composition can modify the response. Actual materials can exhibit mixed bonding rather than fitting one perfectly pure category.

13. Unit Cells and Lattice Parameters

A unit cell repeats by translation to build a crystal. Lattice parameters are the edge lengths a, b, c and inter-edge angles. For a cubic cell, a = b = c and all angles are 90°. HCP uses a hexagonal description with basal parameter a and axial parameter c.

An atom at a conventional cubic corner is shared by eight cells and contributes 1/8 to one cell. A face-centred atom is shared by two cells and contributes 1/2. A body-centred atom belongs completely to one cell. Count shared fractions before assigning atoms per cell.

Coordination number counts nearest neighbours, not atoms per unit cell. Atomic packing factor (APF) is ideal hard-sphere volume inside a chosen cell divided by that cell’s volume. It is not the mass density and must not include spheres extending outside the cell without the proper sharing fractions.

The lattice parameter is not always the nearest-neighbour distance. Derive the contact direction for each structure: a cube edge, body diagonal or face diagonal. A memorised formula without its geometry can be applied to the wrong structure.

14. BCC, FCC and HCP Packing

BCC: Eight corners plus one body centre give Z = 8/8 + 1 = 2 atoms per conventional cubic cell. Contact along the body diagonal gives √3 a = 4r. Its nearest-neighbour coordination is 8 and ideal APF is π√3/8 ≈ 0.680.

FCC: Eight corners and six face centres give Z = 8/8 + 6/2 = 4. Contact along a face diagonal gives √2 a = 4r. Coordination is 12 and ideal APF is π/(3√2) ≈ 0.740. FCC is a close-packed arrangement with ABCABC layer sequence.

HCP: Ideal close-packed layers follow ABAB rather than FCC’s ABCABC sequence. Coordination is 12, ideal c/a = √(8/3) ≈ 1.633 and APF is approximately 0.740. Actual HCP materials can depart from the ideal axial ratio.

For the conventional full hexagonal prism, the atom count is 6 and the volume is (3√3/2)a²c. A smaller primitive hexagonal cell can use 2 atoms and one-third that volume. Either convention gives the same density when atom count and volume are paired consistently.

Worked Problem 2.17: BCC Radius from Lattice Parameter

Problem: An ideal BCC material has cubic lattice parameter 0.2866 nm. Find atomic hard-sphere radius and atoms per cell.

Method: Atoms touch along the body diagonal; count shared corner sites.

Calculation: √3a = 4r, so r = √3a/4.

Calculation: Z = 8 × (1/8) + 1 = 2.

Answer: r ≈ 0.1241 nm; Z = 2.

Check: Coordination number is 8, not 2. Atoms per cell and nearest-neighbour count describe different properties.

Worked Problem 2.18: FCC Lattice Parameter

Problem: An ideal FCC hard-sphere radius is 0.128 nm. Find a and coordination number.

Method: Contact follows a face diagonal.

Calculation: √2a = 4r, giving a = 2√2r.

Calculation: FCC has 12 nearest neighbours.

Answer: a ≈ 0.362 nm; coordination number 12.

Check: Using the BCC body-diagonal formula would give a different structure. The specified contact geometry determines the relation.

Worked Problem 2.19: Ideal HCP Axial Ratio

Problem: An ideal HCP structure has basal parameter a = 0.320 nm. Find c and the hard-sphere radius.

Method: Use ideal c/a = √(8/3) and a = 2r.

Calculation: c = 1.632993 × a.

Calculation: r = a/2.

Answer: c ≈ 0.5226 nm; r = 0.160 nm.

Check: Actual HCP materials need not have the ideal axial ratio. Use a supplied measured ratio instead when the problem provides one.

Worked Problem 2.20: Derive BCC Packing Factor

Problem: Show the ideal BCC packing factor using Z = 2 and a = 4r/√3.

Method: Divide the sphere volume assigned to the cell by a³.

Calculation: Sphere volume = 2 × (4πr³/3).

Calculation: Cell volume = (4r/√3)³; take their ratio.

Answer: APF = π√3/8 ≈ 0.6802.

Check: APF is dimensionless. It does not equal the mass density and does not count coordination neighbours.

15. Crystal Density and Structure Calculations

Mass density is cell mass divided by cell volume. If Z atoms of molar mass M belong to a cell, mass per cell is ZM/NA. For a cubic cell, ρ = ZM/(NAa³). Use M in kg/mol and a in metres to obtain kg/m³.

When M is in g/mol and a in centimetres, the same formula gives g/cm³. Mixing a in nm with a volume assumed to be cm³ creates errors of many powers of ten. Record the unit conversion before cubing the lattice parameter.

For a specimen volume V, the ideal number of cells is V/a³ and atom count is ZV/a³ for a cubic perfect-crystal model. Surface truncation, vacancies, porosity and multiple phases are not included in this simple estimate.

Density alone does not always identify a unique crystal structure because composition, defects and temperature also matter. Structure calculations should specify the supplied model and check whether the resulting scale is plausible. One g/cm³ equals 1000 kg/m³.

Worked Problem 2.21: Density of a Supplied BCC Model

Problem: Use BCC Z = 2, a = 0.2866 nm and molar mass M = 55.845 g/mol. Find ideal density.

Method: ρ = ZM/(NA a³), with consistent SI units.

Calculation: M = 0.055845 kg/mol; a = 2.866 × 10^-10 m.

Calculation: Find cell mass 2M/NA and divide by a³.

Answer: ρ ≈ 7878 kg/m³ = 7.878 g/cm³.

Check: The estimate assumes an ideal fully dense structure and the supplied lattice parameter. Porosity and defects can alter measured bulk density.

16. Stress, Strain and Elastic Response

Engineering normal stress is force divided by original cross-sectional area: σ = F/A0. Tensile stress stretches a specimen; compressive stress shortens it under the chosen sign convention. Stress has units N/m² = Pa. One MPa equals one N/mm².

Engineering longitudinal strain is ε = ΔL/L0. It is dimensionless and may be expressed as a fraction, percent or microstrain. A value 0.001 is 0.1%, not 1%. Keep original and deformed dimensions distinct when a problem asks for engineering rather than true measures.

Shear stress is tangential force divided by the relevant area. For a small simple shear, strain γ ≈ lateral displacement/height ≈ shear angle in radians. Volumetric strain is ΔV/V; its sign depends on whether compression is represented as negative or by a positive magnitude.

Elastic deformation is recovered on unloading within the relevant regime. Linear elasticity is the more restrictive region where stress is proportional to strain. Elastic does not automatically mean exactly linear up to every load.

Worked Problem 2.22: Engineering Tensile Stress

Problem: A tensile force 5000 N acts on original cross-sectional area 25.0 mm². Find engineering stress.

Method: σ = F/A0; 1 N/mm² = 1 MPa.

Calculation: σ = 5000/25.0 = 200 N/mm².

Calculation: Equivalently A0 = 25.0 × 10^-6 m² and σ = 2.00 × 10^8 Pa.

Answer: σ = 200 MPa.

Check: The area is the original area for engineering stress. Use current area only when the requested definition is true stress.

Worked Problem 2.23: Strain, Percent and Microstrain

Problem: A 2.00 m gauge length extends by 0.500 mm. Find longitudinal strain, percent strain and microstrain.

Method: Convert extension to the same length unit as the gauge length.

Calculation: ε = 0.000500/2.00 = 0.000250.

Calculation: Percent = 100ε; microstrain = 10^6 ε.

Answer: ε = 2.50 × 10^-4 = 0.0250% = 250 microstrain.

Check: Strain is dimensionless. A numerical fraction and the percentage written from it differ by a factor of 100.

17. Young, Bulk and Shear Moduli

Young modulus Y is normal stress divided by longitudinal strain in a uniaxial linear elastic test: Y = σ/ε = FL0/(A0ΔL). It measures stiffness of the material, while a component’s extension also depends on its dimensions.

Bulk modulus K measures resistance to volume change under hydrostatic pressure: K = -Δp/(ΔV/V) for a sufficiently small change. With compression magnitudes, K = Δp/|ΔV/V|. The minus sign makes K positive for ordinary stable materials.

Shear modulus G is shear stress divided by shear strain in the linear regime: G = τ/γ. It measures resistance to shape change rather than direct hydrostatic volume compression. All three moduli have pressure units.

Poisson ratio νP is minus transverse strain divided by axial strain in a uniaxial test. For a homogeneous isotropic linear elastic material, Y = 2G(1 + νP) = 3K(1 – 2νP) and Y = 9KG/(3K + G). Do not impose these isotropic formulas blindly on direction-dependent single-crystal properties.

Worked Problem 2.24: Extension of a Linear Elastic Wire

Problem: A wire has L0 = 2.00 m, A0 = 10.0 mm² and Y = 200 GPa. A 1000 N axial force stays in the linear elastic regime. Find extension.

Method: Rearrange Y = FL0/(A0 ΔL).

Calculation: A0 = 10.0 × 10^-6 m²; Y = 2.00 × 10^11 Pa.

Calculation: ΔL = FL0/(A0Y).

Answer: ΔL = 1 mm.

Check: The relation assumes a uniform uniaxial specimen and small linear elastic strain. A load beyond the stated regime requires a different constitutive model.

18. Stress-Strain Curve and Unit Revision

The initial slope of a uniaxial stress-strain curve gives Young modulus in its linear elastic region. The proportional limit, elastic limit, yielding, maximum engineering stress and fracture are different features. Their exact appearance depends on the material and test conditions.

A ductile specimen can undergo substantial permanent strain before fracture. A brittle specimen fractures with comparatively little plastic strain. High stiffness, high strength and high toughness are different properties; one should not be substituted for another in an explanation.

For uniaxial linear loading, elastic strain-energy density is u = σε/2 = σ²/(2Y). It is the area under the linear stress-strain segment and has units J/m³. The full area to fracture describes a different energy measure and cannot be found from the initial slope alone.

Revision path: Atomic/valence structure → bonding → crystal geometry → sharing/counting → density → stress/strain → modulus and elastic response. For every calculation state the material model, unit-cell convention or loading mode before substituting values.

Worked Problem 2.25: Check Isotropic Modulus Relations

Problem: For a homogeneous isotropic linear elastic material, Y = 200 GPa and Poisson ratio νP = 0.250. Find G and K.

Method: Use the isotropic relations only under the supplied assumptions.

Calculation: G = Y/[2(1 + νP)] = 200/2.5 = 80.0 GPa.

Calculation: K = Y/[3(1 – 2νP)] = 200/1.5 GPa.

Answer: G = 80.0 GPa; K ≈ 133.3 GPa.

Check: Check Y = 9KG/(3K + G). A direction-dependent single crystal cannot be reduced to these isotropic relations without an appropriate model.

19. Amorphous Solids and Viscoelastic Response – BT

BT extension. Crystalline solids possess long-range periodic order; amorphous solids have short-range order without a repeating lattice over large distances. Glass and many polymers illustrate amorphous structure. A material can be partly crystalline, so the labels are ideal descriptions rather than a requirement that all specimens belong to one extreme.

Elastic strain responds reversibly to stress. Viscous flow accumulates deformation with time. A viscoelastic material combines both, so its response depends on duration and loading rate. Biological tissues, polymer implants and gels often show this behaviour. A single Young modulus cannot describe every timescale.

Under constant stress, creep means strain increases with time. Under fixed strain, stress relaxation means stress decreases. Hysteresis during cyclic loading represents dissipated energy per volume, equal to the area inside the stress-strain loop. Elastic does not automatically mean linear.

A Maxwell spring-dashpot series model has total strain equal to elastic plus viscous strain. At fixed total strain its stress decays as σ(t) = σ0 exp(-t/τ), with τ = η/E. A Kelvin-Voigt parallel model gives creep ε(t) = (σ0/E)[1 – exp(-tE/η)]. These simple models illustrate different responses; neither describes all biomaterials.

Worked Problem 2.26: Relaxation Time

Problem: A Maxwell material has E = 2.0 MPa and η = 6.0 × 10^6 Pa s. Find τ and the remaining fraction of stress after 6.0 s at fixed strain.

Method: τ = η/E; σ/σ0 = exp(-t/τ).

Calculation: τ = 6.0 × 10^6/(2.0 × 10^6) = 3.0 s; t/τ = 2.

Answer: τ = 3.0 s; σ/σ0 = 0.1353.

Check: Stress relaxation holds total strain fixed; constant-stress creep is a different test.

20. Surface Energy, Contact Angle, Adhesion and Cohesion – BT

Surface energy is the reversible work required to create unit area of surface, in J/m². Liquid surface tension γ has units N/m; these are dimensionally equivalent. Molecules at an interface have a different environment from molecules in the bulk, so increasing interfacial area generally requires work.

The equilibrium contact angle θ is measured through the liquid at the solid-liquid-gas contact line. On an ideal smooth, chemically homogeneous surface, Young’s relation is γSV – γSL = γLV cosθ. A smaller θ indicates better wetting of that particular surface by that liquid. Roughness, contamination and contact-angle hysteresis can change real observations.

Cohesion refers to attraction within one material; adhesion refers to attraction between different materials. They help determine spreading and attachment but are not synonyms for surface tension. The ideal reversible work of adhesion is WA = γSV + γLV – γSL = γLV(1 + cosθ) under Young-Dupré assumptions.

In biotechnology, wetting influences coating, cell-substrate interaction, fluid access to scaffolds and device cleaning. Contact angle alone does not fully predict biological compatibility: chemistry, roughness, protein adsorption and mechanical response also matter.

Worked Problem 2.27: Wetting and Adhesion

Problem: A liquid has γLV = 0.072 N/m and contact angle 60° on an ideal surface. Find γSV – γSL and ideal WA.

Method: Young: difference = γLV cosθ; WA = γLV(1 + cosθ).

Calculation: cos60° = 0.5; multiply by 0.072.

Answer: γSV – γSL = 0.036 J/m²; WA = 0.108 J/m².

Check: These are ideal equilibrium relations; they do not by themselves measure biological compatibility.

21. C. Fluid Properties, Pressure and Buoyancy – CH

Prescribed for CH. A fluid cannot sustain a static shear stress without continuing deformation. Density is ρ = m/V, specific volume is 1/ρ and specific gravity is a density ratio. Compressibility measures volume response to pressure. Liquids are often approximately incompressible over modest pressure ranges; gases generally need an equation of state.

Pressure is normal compressive force per area. In a static liquid, dp/dz = -ρg when z increases upward. For constant density, pressure rises by ρgh at depth h. Absolute pressure equals gauge pressure plus atmospheric pressure. A negative gauge pressure does not automatically imply a negative absolute pressure.

Pascal’s principle describes the transmission of an applied pressure increment through a confined static fluid. In a hydraulic press, F1/A1 = F2/A2 for equal elevations and negligible losses. Force amplification comes with a corresponding displacement change; it does not create energy.

Archimedes’ principle gives upward buoyancy equal to the weight of displaced fluid, FB = ρfluid g Vdisplaced. A floating object displaces its own weight. For a uniform object floating in one liquid, the submerged fraction is ρobject/ρfluid. Use displaced volume, which need not be the entire object volume.

Worked Problem 2.28: Pressure and Floating Fraction

Problem: Find gauge pressure at 2.0 m depth in water and the submerged fraction of a uniform block of density 600 kg/m³. Use ρwater = 1000 kg/m³.

Method: p = ρgh; Vsub/V = ρblock/ρwater.

Calculation: p = 1000 × 9.81 × 2; fraction = 600/1000.

Answer: p = 19.62 kPa gauge; submerged fraction = 0.600.

Check: Atmospheric pressure is added only if absolute pressure is requested.

22. Viscosity, Newtonian Fluids and Rheology – CH

Viscosity describes resistance to relative motion between neighbouring fluid layers. For a Newtonian fluid in simple shear, τs = μ du/dy, with μ the dynamic viscosity in Pa s. Kinematic viscosity is ν = μ/ρ, in m²/s. Shear stress and pressure have the same units but different directions and physical roles.

A Newtonian fluid has a linear shear-stress versus shear-rate curve through the origin at fixed state. Shear-thinning fluids show falling apparent viscosity with increasing shear rate; shear-thickening fluids show rising apparent viscosity. A yield-stress material needs a finite stress before sustained flow in the ideal model.

The power-law model is τs = K γdot^n for positive simple shear rate. n < 1 represents shear thinning, n = 1 Newtonian behaviour, and n > 1 shear thickening. K has units Pa s^n. Apparent viscosity is τs/γdot = Kγdot^(n-1); it is not generally constant.

A Bingham model after yielding is τs = τ0 + μp γdot. Thixotropy and rheopexy describe time-dependent changes under shear, distinct from shear-rate dependence alone. Slurries and polymers may combine yield stress, viscoelasticity and history dependence; report temperature, concentration and measurement protocol.

Worked Problem 2.29: Shear-Thinning Apparent Viscosity

Problem: For τs = 2.0 γdot^0.5 in SI units, find stress and apparent viscosity at shear rates 25 and 100 s^-1.

Method: μapp = τs/γdot; K = 2.0 Pa s^0.5.

Calculation: At 25: τs = 2√25 = 10 Pa. At 100: τs = 20 Pa.

Answer: μapp = 0.400 Pa s at 25 s^-1 and 0.200 Pa s at 100 s^-1.

Check: Stress rises but apparent viscosity falls; this is shear thinning, not lower stress at higher shear.

23. Measuring Viscosity and Interpreting Slurry Data – CH

A capillary viscometer relates laminar flow rate to pressure drop in a known tube. For a Newtonian incompressible liquid with fully developed laminar flow and no slip, Q = πr^4 Δp/(8μL). Measure radius carefully: a small radius error strongly affects the inferred viscosity.

A falling-sphere method uses a sphere reaching terminal speed. In the creeping-flow limit, viscous drag is 6πμrv. Balancing sphere weight, buoyancy and drag gives μ = 2r²(ρs – ρf)g/(9vt). This requires a small particle Reynolds number and negligible wall/end effects. A sphere still accelerating does not satisfy the terminal balance.

A rotational rheometer imposes a controlled torque or motion and measures the conjugate response. Geometry converts torque and angular motion into shear stress and shear rate. Run a sweep of shear rate rather than forcing one apparent viscosity to describe a non-Newtonian suspension.

For slurries, increasing solids loading often raises resistance and may introduce a yield stress. Polymer solutions can exhibit shear thinning as structures orient and rearrange. Numerical model parameters apply only over the stated range; extrapolating a power law to zero or infinite shear can be unphysical.

Worked Problem 2.30: Capillary Viscosity Measurement

Problem: A Newtonian liquid flows at 1.00 mL/s through a capillary of radius 1.00 mm and length 0.200 m with Δp = 2.00 kPa. Find μ, assuming laminar developed flow.

Method: μ = πr^4 Δp/(8LQ).

Calculation: Q = 1.00 × 10^-6 m³/s; substitute SI dimensions.

Answer: μ = 0.00393 Pa s.

Check: With density 1000 kg/m³ the mean speed is 0.3183 m/s and Re ≈ 81, consistent with laminar flow.

24. Surface Tension and Capillarity – CH

Surface tension pulls tangentially along a liquid interface. The net pressure jump across a curved interface follows the Young-Laplace relation Δp = γ(1/R1 + 1/R2). A spherical liquid drop with one interface has inside excess pressure 2γ/r. A thin soap bubble has two interfaces and approximately 4γ/r.

In a circular capillary of radius r, vertical surface-tension force is 2πrγ cosθ. Balancing the weight ρgπr²h gives h = 2γ cosθ/(ρgr), neglecting meniscus volume. θ is measured through the liquid. A wetting liquid with θ < 90° rises; a non-wetting one can be depressed.

Do not confuse capillary radius with diameter. If diameter d is specified, h = 4γ cosθ/(ρgd). The liquid density and contact angle must match the system; using cosθ = 1 silently assumes complete wetting.

Capillarity matters in pores, filtration and impregnation. For a real porous network, tortuosity and pore-size distribution complicate flow. The simple rise law predicts an equilibrium height in an ideal tube, not the time needed for a complex slurry or polymer liquid to reach that height.

Worked Problem 2.31: Capillary Rise

Problem: Water fully wets a tube of radius 0.500 mm. Use γ = 0.0720 N/m and ρ = 1000 kg/m³. Find the equilibrium rise.

Method: h = 2γ cosθ/(ρgr), θ = 0.

Calculation: Use r = 5.00 × 10^-4 m and cosθ = 1.

Answer: h = 0.02936 m = 29.358 mm.

Check: If 0.500 mm were the diameter instead, the predicted rise would double.

25. Continuity and Bernoulli Energy Balance – CH

Steady mass conservation gives ρ1A1v1 = ρ2A2v2 for one inlet and one outlet without leakage or accumulation. For an incompressible fluid, Q = Av is constant, so narrowing the area increases mean speed. A velocity field can vary even when the material density is constant.

For steady, incompressible, inviscid flow along a streamline, Bernoulli’s equation is p + ρv²/2 + ρgz = constant. It expresses mechanical energy per volume; dividing by ρg gives pressure, velocity and elevation heads. For rotational flow the constant can differ between streamlines.

For engineering pipe systems include pumps, turbines and head loss: p1/(ρg) + v1²/(2g) + z1 + hp = p2/(ρg) + v2²/(2g) + z2 + ht + hL. Irreversible loss is nonnegative along the flow. Real viscous pipes do not obey loss-free Bernoulli over an arbitrary length.

A Venturi uses continuity and a pressure difference to infer flow under a specified discharge coefficient. A Pitot tube relates stagnation and static pressure to local speed when the assumptions apply. Check elevations and whether the question asks static pressure, stagnation pressure or head.

Worked Problem 2.32: Horizontal Constriction

Problem: Water flows through a horizontal ideal pipe. Area falls from 4.0 to 1.0 cm² and initial speed is 1.0 m/s. Find v2 and p1 – p2, ignoring losses.

Method: A1v1 = A2v2; p1 – p2 = ρ(v2² – v1²)/2.

Calculation: v2 = 4.0 m/s; Δp = 1000(16 – 1)/2.

Answer: v2 = 4.0 m/s; p1 – p2 = 7.50 kPa.

Check: With real losses, additional pressure decrease is needed for the same flow and elevations.

26. Pipe Flow, Pressure Drop and Fluidization – CH

The Reynolds number Re = ρvD/μ compares inertia with viscous effects. For smooth circular-pipe flow, Re below about 2300 is commonly laminar; transition depends on disturbances, and high values are usually turbulent. This threshold is not a universal criterion for every geometry.

For fully developed laminar Newtonian flow, v(r) is parabolic, maximum speed is twice the mean, and Δp = 32μLv/D². The Darcy-Weisbach equation is Δp = fD(L/D)ρv²/2. In laminar circular flow fD = 64/Re. The Fanning factor is one quarter of the Darcy factor: identify the convention.

Turbulent friction depends on Reynolds number and relative roughness. Minor losses use Kρv²/2 and may be important even in a short pipe. A slurry can need a non-Newtonian correlation and a minimum transport velocity to avoid settling; a Newtonian formula is not automatically valid.

In a packed particle bed, upward flow creates a pressure drop. At incipient fluidization the fluid drag approximately balances the effective bed weight: Δp ≈ (ρs – ρf)(1 – ε)Lg, neglecting wall support. Increasing velocity expands and mixes the bed. Entrainment, channeling and particle-size effects limit the simple model.

Worked Problem 2.33: Pipe Loss and Bed Support

Problem: Water of μ = 0.001 Pa s flows at 0.10 m/s through a 5.0 m pipe of diameter 10 mm. Also find incipient bed Δp for ρs = 2500, ρf = 1000 kg/m³, ε = 0.40 and Lbed = 0.20 m.

Method: Re = ρvD/μ; Δppipe = 32μLv/D²; Δpbed = (ρs – ρf)(1 – ε)Lbed g.

Calculation: Re = 1000, so use laminar pipe law. Pipe Δp = 32(0.001)(5)(0.1)/0.01².

Answer: Pipe loss = 160 Pa; bed support Δp = 1765.8 Pa.

Check: The bed and pipe are separate stated examples; do not add them unless they belong to one system.

27. D. Thermal Physics in Agriculture: Laws and State Variables

Prescribed for AG. Temperature describes thermal equilibrium; internal energy is a state function; heat and work are transfers across a system boundary. The zeroth law supports temperature measurement. A thermometer should reach equilibrium without significantly disturbing the sample.

Using work done by the system as positive, the first law is ΔU = Q – W. For a quasistatic expansion W = ∫p dV. Constant-volume heating has no boundary expansion work, whereas an isothermal ideal-gas expansion can absorb heat while maintaining constant internal energy. State the sign convention before calculating.

The second law sets the direction of spontaneous processes and limits heat-engine efficiency. For a reversible transfer dS = δQrev/T; total entropy generation is nonnegative. A heat engine between reservoirs TH and TC cannot exceed Carnot efficiency 1 – TC/TH, with temperatures in kelvin.

The third law concerns the limiting entropy of a perfect crystal as temperature approaches absolute zero; absolute zero cannot be reached by a finite sequence of ordinary cooling operations. In crop and soil systems, energy conservation still applies even though real processes are irreversible and exchange matter with the surroundings.

Worked Problem 2.34: First Law with Work Output

Problem: A closed system receives 600 J of heat and does 250 J of work. Find ΔU. Then state the maximum engine efficiency between 600 K and 300 K.

Method: ΔU = Q – W; ηCarnot = 1 – TC/TH.

Calculation: ΔU = 600 – 250; ηmax = 1 – 300/600.

Answer: ΔU = +350 J; maximum efficiency = 0.500 = 50.0%.

Check: The efficiency is an upper reversible bound; the two calculations describe separate systems.

28. Conduction, Convection and Thermal Radiation – AG

Fourier’s law for one-dimensional conduction is qz = -k dT/dz, where qz is heat flux in W/m² and k is thermal conductivity in W/(m K). Heat flows down the temperature gradient. For a uniform slab at steady state, heat rate is kA(Th – Tc)/L and thermal resistance is L/(kA).

Convection transfers heat between a surface and a moving fluid. Newton’s law is Qdot = hA(Ts – Tfluid). The coefficient h depends on flow and geometry and is not the same as material conductivity k. Wind can increase convective exchange but does not necessarily cool a surface below the air temperature without another energy sink.

Thermal radiation does not require a material medium. For a small grey surface exchanging with large isothermal surroundings, Qdot = εσSB A(Ts^4 – Tsur^4). All temperatures must be absolute. Emissivity is between zero and one under the stated spectral/total convention.

Conduction, convection and radiation usually act together around plants and protected cultivation. Mulch changes radiation absorption, contact resistance and evaporation as well as conduction. A formula for one mechanism is not automatically the complete field energy balance.

Worked Problem 2.35: Conduction and Convection in Series

Problem: A 0.10 m slab has k = 0.50 W/(m K), area 2.0 m² and one exposed film with h = 10 W/(m² K). Between the inner face and outside air is a 15 K difference. Find heat rate.

Method: Rtotal = L/(kA) + 1/(hA); Qdot = ΔT/Rtotal.

Calculation: Rslab = 0.100 K/W; Rfilm = 0.050 K/W.

Answer: Rtotal = 0.150 K/W; Qdot = 100 W.

Check: Use the temperatures at the ends of the full series path, not the surface difference across only one element.

Worked Problem 2.36: Net Thermal Radiation

Problem: A grey surface at 300 K exchanges with large surroundings at 280 K. Use ε = 0.90. Find net outward radiant flux.

Method: q = εσSB(Ts^4 – Tsur^4).

Calculation: Use absolute temperatures and σSB = 5.670374419 × 10^-8 W/(m² K^4).

Answer: q = 99.6906 W/m² outward.

Check: Using the fourth power of a Celsius temperature or of the temperature difference is incorrect.

29. Soil Heat Capacity, Diffusivity and Temperature Profiles – AG

Soil is a composite of solids, water, air and organic matter. Volumetric heat capacity Cvol is approximately a volume-weighted sum of constituent density times specific heat. Adding water usually raises heat storage markedly. Thermal conductivity also depends on moisture, packing and contact between particles.

Thermal diffusivity α = k/Cvol measures how rapidly a temperature disturbance spreads. For one-dimensional conduction in a uniform medium without internal sources, ∂T/∂t = α ∂²T/∂z². A larger α gives faster equilibration over a fixed length scale, approximately t ~ L²/α.

For a sinusoidal surface temperature of angular frequency ω, a semi-infinite homogeneous-soil solution has amplitude proportional to exp(-z/δ) and phase lag z/δ, where δ = √(2α/ω). Thus daily temperature oscillations weaken and occur later at greater depth. This is an ideal conductive model, not a complete model of water movement and roots.

Soil temperature influences germination, root activity and microbial processes. Compare readings at the same depth and time; a surface reading is not a substitute for the root-zone temperature. Ground heat flux is positive downward in the convention used in the next section.

Worked Problem 2.37: Daily Soil Temperature Damping

Problem: Uniform soil has α = 5.0 × 10^-7 m²/s and a daily surface-temperature amplitude of 10 K. Estimate amplitude and lag at 0.10 m depth.

Method: δ = √(2α/ω); A(z) = A0 exp(-z/δ); lag time = (z/δ)/ω.

Calculation: ω = 2π/86400 s^-1; use the semi-infinite sinusoidal conduction model.

Answer: δ = 0.1173 m; amplitude = 4.2623 K; lag = 3.257 h.

Check: The deeper oscillation is weaker and later; the model neglects water transport and heterogeneous layers.

30. Agricultural Energy Balance and Evaporation – AG

For a simplified surface with negligible storage, net radiation satisfies Rn = H + LE + G. Rn is incoming minus outgoing radiative energy, H is upward sensible heat flux, LE is upward latent heat flux, and G is downward soil heat flux. All fluxes use W/m². If storage or other terms matter, add them explicitly.

Evaporation consumes latent energy while liquid becomes vapour. If latent heat per mass is Lv, the mass flux is mdot/A = LE/Lv. Dividing by water density converts it to depth per time. One kg/m² of water corresponds to one millimetre depth; one day is 86400 seconds.

Evaporation depends on available energy, vapour-pressure difference, air movement and water supply. Potential atmospheric demand is not the same as actual evaporation from a dry soil. Transpiration adds plant control through stomata; evapotranspiration combines evaporation and transpiration.

A greenhouse modifies radiative and convective exchanges, ventilation and humidity. Covering a surface can reduce evaporation while changing sensible and ground fluxes. When comparing irrigation strategies, close the energy balance before assigning the entire net radiation to latent heat.

Worked Problem 2.38: Energy Balance to Evaporation

Problem: Daily-average fluxes are Rn = 400, H = 100 and downward G = 50 W/m², with negligible storage. Find LE and water loss per day using Lv = 2.45 MJ/kg.

Method: LE = Rn – H – G; mass per area = LE × 86400/Lv.

Calculation: LE = 250 W/m². One kg/m² corresponds to one mm water depth.

Answer: LE = 250 W/m²; evaporation equivalent = 8.8163 mm/day.

Check: This assumes the residual latent flux is supplied by available water and the declared balance terms.

31. E. Photonics: Optical Fibre and Total Internal Reflection – CE

Prescribed for CE Unit II. An optical fibre has a transparent core with refractive index n1 surrounded by lower-index cladding n2. A protective coating is mechanically useful but is not the optical cladding. Light is guided by the core-cladding structure; the elementary ray picture uses repeated total internal reflection.

At an interface Snell’s law is n1 sin i = n2 sin r. Total internal reflection requires incidence from higher to lower refractive index and incidence angle larger than ic = sin^-1(n2/n1), measured from the normal. At ic the refracted ray is tangent to the boundary.

A step-index fibre has an abrupt refractive-index change; a graded-index fibre changes index across the core. A single-mode fibre supports one spatial mode over its operating range; multimode fibres support several. The ray description is most useful when geometric optics is appropriate, while true guidance is a wave-mode problem.

Fibre links are useful for communication, inspection and structural monitoring. A fibre sensor can encode strain or temperature in an optical change, but a bare communication fibre does not automatically measure every structural quantity. Coupling, calibration and packaging determine a practical sensing system.

Worked Problem 2.39: Critical Angle

Problem: For core n1 = 1.48 and cladding n2 = 1.46 find the internal critical angle.

Method: ic = sin^-1(n2/n1), measured from the normal.

Calculation: ic = sin^-1(1.46/1.48).

Answer: ic = 80.5700 degrees.

Check: This is an internal interface angle, not the fibre entrance acceptance angle.

32. Acceptance Cone, Numerical Aperture and Attenuation – CE

Let an external medium have index n0 and let θa be the largest meridional entrance angle to the fibre axis. At the limiting guided ray, the internal incidence equals ic. Combining Snell’s law at the end face with the critical-angle geometry gives NA = n0 sinθa = √(n1² – n2²) for an ideal step-index fibre.

The acceptance cone has half-angle θa and full apex angle 2θa. In air n0 ≈ 1. In another medium, do not set sinθa equal to NA without dividing by n0. Relative index difference Δ = (n1 – n2)/n1 gives NA ≈ n1√(2Δ) when Δ is small.

Attenuation reduces optical power along the link through absorption, scattering, bending and other losses. With a power coefficient αp in inverse length, P(L) = P0 exp(-αp L). A decibel loss is 10 log10(Pin/Pout). The coefficient in dB per length is 4.343αp, using the same length unit.

Connector and splice losses add in decibels to distributed fibre loss. Power attenuation and pulse broadening are different limitations: dispersion can broaden pulses even if the power loss is small. Specify whether an attenuation coefficient applies to power or field amplitude.

Worked Problem 2.40: Numerical Aperture and Cone

Problem: For the same indices, find NA and the acceptance half-angle in air and in water of index 1.33.

Method: NA = √(n1² – n2²); θa = sin^-1(NA/n0).

Calculation: Evaluate the square-root difference before dividing by the external index.

Answer: NA = 0.24249; air half-angle = 14.0334°; water half-angle = 10.5050°.

Check: The full cone angle is twice the half-angle. NA is dimensionless.

Worked Problem 2.41: Fibre Link Budget

Problem: A 3 km link has 0.40 dB/km fibre loss, two 0.50 dB connectors and four 0.10 dB splices. Input power is 2.0 mW. Find output power.

Method: Total loss is a sum in dB; Pout = Pin × 10^(-loss/10).

Calculation: Loss = 1.2 + 1.0 + 0.4 = 2.6 dB.

Answer: Pout = 1.09908 mW.

Check: Add dB losses before converting to a power ratio; do not add linear transmission fractions.

33. Radiation-Matter Interaction and Einstein Coefficients – CE

Two levels separated by E2 – E1 = hν can exchange radiation at frequency ν. Absorption promotes an atom from level 1 to level 2 by taking energy from the field. Spontaneous emission occurs without an inducing photon and has random emission timing and phase across independent atoms.

Stimulated emission occurs when incident radiation induces an excited atom to emit into the stimulating optical mode. It supplies the coherent amplification mechanism of a laser. A laser still needs a pumped gain medium and a resonator or feedback arrangement; a fluorescent source alone is not a laser.

If uν is spectral energy density per unit frequency, transition rates are N1B12uν for absorption, N2B21uν for stimulated emission and N2A21 for spontaneous emission. Thermal equilibrium with Planck radiation gives g1B12 = g2B21 and A21/B21 = 8πhν³/c³ for the vacuum spectral-frequency convention.

For equal degeneracies B12 = B21. When degeneracies differ, inversion for net gain requires N2/g2 > N1/g1. At thermal equilibrium this is not maintained; pumping creates a non-equilibrium population. A long-lived metastable level helps accumulate excited population.

Worked Problem 2.42: Einstein Spontaneous-to-Stimulated Ratio

Problem: For a 632.8 nm transition, evaluate A21/B21 with spectral energy density per Hz in vacuum.

Method: A21/B21 = 8πhν³/c³ = 8πh/λ³.

Calculation: Use λ = 632.8 × 10^-9 m in the last expression.

Answer: A21/B21 = 6.572 × 10^-14 J s/m³.

Check: Its units match spectral energy density per Hz. This ratio is not a dimensionless rate ratio without specifying uν.

34. Laser Action, Pumping and Resonator Threshold – CE

LASER means light amplification by stimulated emission of radiation. The essential components are an active medium, an energy pump and optical feedback. A partly transmitting mirror permits output while another reflects more strongly. Light repeatedly crosses the gain region so suitable modes build up.

Population inversion produces gain only when stimulated emission exceeds competing absorption in the relevant transition. A simple two-level system cannot sustain ordinary inversion through resonant pumping alone because absorption and stimulated emission compete. Three- and four-level arrangements provide additional relaxation paths.

For a linear cavity of length L with intensity gain g and distributed intensity loss α per length, threshold requires R1R2 exp[2(g – α)L] = 1. Thus gth = α + ln[1/(R1R2)]/(2L). Here R1 and R2 are intensity reflectivities; avoid mixing field and intensity coefficients.

Laser beams can have narrow spectral width, strong directionality, high intensity and coherence, but the degree depends on the system. Coherence is not the same as brightness. Applications include surveying, alignment, ranging, machining, optical communication and spectroscopy; select wavelength and power for the task.

Worked Problem 2.43: Laser Cavity Threshold

Problem: A 0.30 m cavity has R1 = 0.99, R2 = 0.95 and distributed intensity loss α = 0.10 m^-1. Find threshold intensity gain per length.

Method: gth = α + ln[1/(R1R2)]/(2L).

Calculation: Use the intensity reflectivity product 0.9405 and the round-trip distance 0.60 m.

Answer: gth = 0.20224 m^-1.

Check: A field-amplitude gain coefficient would use a different convention; do not mix them.

35. Ruby Laser and He-Ne Laser: Construction and Comparison – CE

A ruby laser uses Cr3+ ions in aluminium oxide, optically pumped by a flash lamp. Pumped ions relax rapidly to a metastable level, then emit near 694.3 nm to the ground manifold. It is a three-level laser: the lower laser level is associated with the ground state, so substantial pumping is required. Introductory ruby systems normally operate in pulses.

A He-Ne laser uses an electrical discharge in a helium-neon gas mixture. Electron collisions excite helium metastable states; near-resonant collisions transfer energy to neon. A suitable neon transition produces the familiar 632.8 nm red output. The helium is an energy-transfer partner, not the species emitting that red laser line.

Population depletion of the lower laser level and continued discharge pumping allow the common He-Ne arrangement to operate continuously. Its level scheme is effectively a multi-level system, often represented as four levels. Mirror coatings select useful optical transitions among the available neon transitions.

For an exam comparison, give medium, pumping, energy transfer, laser transition, operating mode and one application. A labelled diagram should distinguish pump arrows, rapid non-radiative decay and stimulated emission. Do not draw the ruby pump transition as the 694.3 nm output or label helium as the red-emitting atom.

Worked Problem 2.44: Laser Photon Energy and Flux

Problem: A 632.8 nm He-Ne beam has power 2.0 mW. Find photon energy in eV and photon rate.

Method: Eph = hc/λ; photon rate = P/Eph.

Calculation: Convert power to 0.0020 W and λ to metres.

Answer: Eph = 1.95930 eV; rate = 6.3712 × 10^15 photons/s.

Check: The photon energy depends on wavelength; increasing beam power raises photon rate at fixed wavelength.

36. F. Coherence and Interference in Thin Films – TX

Prescribed for TX Unit II. Interference is redistribution of intensity when coherent optical amplitudes superpose. Coherent sources maintain a stable relative phase over the observation time. For two beams, I = I1 + I2 + 2√(I1I2)cosδ. For equal intensities, perfect overlap and coherence, maxima are four times one-beam intensity and minima can be zero.

In a film of refractive index μ, thickness t and refraction angle r, the geometrical optical path difference between the two main reflected rays is 2μt cos r. Reflection from lower to higher refractive index adds a phase reversal π; reflection from higher to lower does not. Count the reversals before selecting a condition.

For the common reflected-light arrangement with one net phase reversal, dark fringes satisfy 2μt cos r = mλ and bright fringes satisfy 2μt cos r = (m + 1/2)λ, where λ is the vacuum wavelength and m is a nonnegative integer. Two or zero net reversals interchange the reflected bright and dark conditions.

A uniform film can produce fringes of equal inclination; a wedge of small angle produces nearly parallel equal-thickness fringes. In a near-normal air wedge, adjacent reflected dark fringes have spacing β = λ/(2αw), where αw is the small wedge angle in radians. Textile coatings can show thin-film colours, but broadband illumination and thickness variation affect the appearance.

Worked Problem 2.45: Thin Film and Wedge

Problem: A reflected air wedge with one net reflection phase reversal has fringe spacing 0.50 mm at λ = 600 nm. Find its small angle and the minimum nonzero dark-film thickness for μ = 1.50 at normal incidence.

Method: β = λ/(2αw); reflected dark: 2μt = mλ.

Calculation: αw = 600 × 10^-9/(2 × 0.50 × 10^-3); use m = 1 for the nonzero thickness.

Answer: Wedge angle = 6.00 × 10^-4 rad; t = 200 nm.

Check: The zero-thickness dark condition is m = 0; the question explicitly asks for the next dark condition.

37. Newton Rings: Geometry, Diameter Method and Liquids – TX

A plano-convex lens on a flat glass plate encloses a thin air film. At near-normal incidence its reflected pattern contains concentric rings because thickness depends on radius. With lens curvature radius R and small gap, t ≈ r²/(2R). For the usual one-reversal reflected system, dark rings satisfy r_m² = mλR.

The dark-ring diameter is D_m² = 4mλR, so λ = [D_(m+p)² – D_m²]/(4pR). Using a difference removes sensitivity to an uncertain order origin and reduces some measurement errors. Measure ring diameters on both sides of the centre to avoid treating a radius as a diameter.

With a liquid of index μ between lens and plate, D_m² = 4mλR/μ under the same geometry and phase convention. Therefore μ equals the air diameter-squared difference divided by the liquid difference for the same order separation and wavelength. The lens curvature radius is unchanged.

The centre is dark only for ideal contact and the expected reflection phases. Dust or a finite central separation shifts the pattern. Rings become closer together at larger order because r grows as √m. Newton rings are equal-thickness fringes, not a grating spectrum.

Worked Problem 2.46: Newton-Ring Wavelength and Liquid

Problem: Dark rings separated by five orders have diameter-squared difference 12.0 mm² in air with R = 1.00 m. The same difference is 9.00 mm² in a liquid. Find λ and μ.

Method: λ = ΔD²/(4pR); μ = ΔDair²/ΔDliquid².

Calculation: Convert 12.0 mm² to 12.0 × 10^-6 m² and use p = 5.

Answer: λ = 600 nm; μ = 1.3333.

Check: These are squared diameters, not squared radii. The same lens, wavelength and order separation are required.

38. Single-Slit Fraunhofer Diffraction – TX

Diffraction is wave spreading and interference across an aperture. For a uniformly illuminated slit of width a, the far-field intensity is I(θ) = I0(sinβ/β)², where β = πa sinθ/λ. The central limit at β = 0 is one, not an undefined intensity.

The minima occur at a sinθ = mλ for nonzero integers m. The central maximum extends between the first minima on either side. Its angular width is approximately 2λ/a for small angles. On a screen distance D away, its linear width is approximately 2Dλ/a.

A narrower slit produces a wider central diffraction maximum but transmits less total light for fixed incident illumination. The secondary maxima are weaker than the central one and are not located exactly halfway between all minima. Keep the slit width a separate from a grating period d.

Fraunhofer conditions require effectively plane-wave illumination and far-field observation, often achieved with lenses. In near-field Fresnel diffraction the pattern depends differently on geometry. When using a screen approximation, check that sinθ ≈ tanθ ≈ y/D is justified.

Worked Problem 2.47: Single-Slit Central Width

Problem: A slit of width 0.20 mm is illuminated at 500 nm. Find the central maximum width on a screen 2.0 m away.

Method: w ≈ 2Dλ/a for small angles.

Calculation: w = 2(2.0)(500 × 10^-9)/(0.20 × 10^-3).

Answer: w ≈ 0.0100 m = 10.0 mm.

Check: This is the full width between first minima; the half-width is 5.0 mm.

39. Diffraction Grating, Dispersion and Resolving Power – TX

A plane transmission grating has period d = a + b, the distance between corresponding points of neighbouring slits. For normal incidence, principal maxima satisfy d sinθ = mλ. If the grating has Nlines per unit length, d is the reciprocal of that line density after converting units.

Angular dispersive power is dθ/dλ = m/(d cosθ). It describes angular separation per wavelength difference. Resolving power is R = λ/Δλ = mN under the Rayleigh criterion for N coherently illuminated identical slits in the ideal grating. Dispersion and resolution are different quantities.

A larger illuminated width increases N and hence resolution at a fixed order even if angular dispersion is unchanged. Higher order increases ideal resolution but is limited by |mλ/d| ≤ 1 and possible overlap of spectra. A grating with many lines printed on it only uses those actually illuminated.

The finite width of each slit supplies a diffraction envelope. A principal grating order can be missing when it coincides with a single-slit minimum. In a numerical solution, determine d, verify order feasibility, calculate angle, and only then evaluate dispersion or resolution as requested.

Worked Problem 2.48: Grating Angle and Resolution

Problem: A grating has 600 lines/mm and illuminated width 20 mm. For second-order 500 nm light find angle, angular dispersion and minimum resolvable wavelength difference.

Method: d sinθ = mλ; dθ/dλ = m/(d cosθ); λ/Δλ = mN.

Calculation: d = 1/(600 × 10^3) m; N = 600 × 20 = 12000; mN = 24000.

Answer: θ = 36.8699°; dispersion = 0.001500 rad/nm; Δλ = 0.02083 nm.

Check: Use the illuminated line count, and verify sinθ = 0.600 is physically possible.

40. Polarisation, Double Refraction and Huygens Construction – TX

Polarisation specifies the transverse electric-field behaviour. Unpolarised light has a fluctuating direction with no preferred transverse orientation over the measurement interval. Linear polarisation maintains one fixed vibration direction. An ideal linear polariser transmits half the intensity of unpolarised incident light.

For linearly polarised light through an ideal analyser at angle θ, Malus’ law gives I = I0 cos²θ, where I0 is the intensity incident on that analyser. With an initially unpolarised beam and two polarisers, include the initial factor one half. Crossed polarisers extinguish an ideal linearly polarised beam.

In an anisotropic uniaxial crystal such as calcite, a suitable incident ray generally splits into ordinary and extraordinary rays with different polarisation directions. Along the optic axis there is no ordinary-extraordinary speed splitting. The ordinary refractive index is direction independent, while the extraordinary response depends on direction.

Huygens’ explanation uses a spherical secondary wave surface for the ordinary wave and an ellipsoidal wave surface for the extraordinary wave in a uniaxial medium. Their intersection along the optic-axis direction expresses the common speed there. Wave normals and energy-ray directions for the extraordinary wave need not coincide.

Worked Problem 2.49: Polariser and Analyser

Problem: Unpolarised light of intensity 100 W/m² passes through an ideal polariser and then an analyser at 60° to it. Find final intensity.

Method: First polariser: I1 = Iin/2. Analyser: I2 = I1 cos²θ.

Calculation: I1 = 50 W/m²; cos²60° = 1/4.

Answer: I2 = 12.5 W/m².

Check: Applying Malus directly to the original unpolarised beam misses the initial half-intensity factor.

41. Nicol Prism and Optical Activity – TX

A Nicol prism is made from a suitably cut calcite crystal whose parts are cemented with Canada balsam. In the usual visible-light description the indices satisfy no(calcite) > nbalsam > ne(calcite). For the chosen prism geometry the ordinary ray meets the cement layer above its critical angle and is totally internally reflected to an absorbing side.

The extraordinary ray is transmitted and emerges plane polarised. The construction therefore separates orthogonal polarisations through birefringence and selective total internal reflection. It is not simply a pair of ordinary glass prisms; the anisotropy and index ordering are essential.

An optically active medium rotates the plane of linear polarisation. For a solution, observed rotation θrot = [α] l c under a specified wavelength, temperature and concentration convention. When l is in dm and c in g/mL, the specific rotation has corresponding units degree mL/(g dm). Different conventions change numerical values.

A polarimeter uses a polariser, sample tube and analyser. Determine the reference position with the solvent or empty setup and measure rotation with the sample. Optical rotation is different from dichroic absorption and from the double image caused by calcite. A racemic mixture can have zero net rotation despite containing optically active molecules.

Worked Problem 2.50: Specific Optical Rotation

Problem: A solution rotates the polarisation by 13.0°. Its concentration is 0.100 g/mL and tube length is 2.00 dm. Find specific rotation in this convention.

Method: [α] = θrot/(lc).

Calculation: [α] = 13.0/(2.00 × 0.100).

Answer: [α] = 65.0 degree mL/(g dm).

Check: Quote wavelength and temperature when reporting a measured material value; these are held fixed in this exercise.

Complete Derivations and Exam Explanations

A. Derive the Intrinsic Concentration and Intrinsic Fermi Level

Assume thermal equilibrium, non-degenerate carrier statistics and parabolic conduction and valence bands. Integrating the density of available states against the Boltzmann occupation tail gives n = NC exp[-(EC – EF)/(kBT)] and p = NV exp[-(EF – EV)/(kBT)]. The band-edge separation is Eg = EC – EV. These expressions refer to concentrations per volume, not the total number in a finite specimen.

Multiply the two equations. The Fermi-level terms cancel because -(EC – EF) -(EF – EV) = -(EC – EV). Therefore np = NCNV exp[-Eg/(kBT)]. For intrinsic material, neutrality and pair generation give n = p = ni, so ni² equals that product. Taking the positive square root gives ni = √(NCNV) exp[-Eg/(2kBT)]. This cancellation explains why the equilibrium mass-action product is independent of doping under the model assumptions.

Next set the original n and p expressions equal, take natural logarithms and collect EF terms. The result is Ei = (EC + EV)/2 + (kBT/2)ln(NV/NC). If the effective densities are equal the logarithm vanishes. If NV is smaller, Ei lies below midgap. The derivation requires neither an allowed state exactly at Ei nor an electron whose energy equals Ei.

For a doped sample combine the same mass-action relation with charge neutrality. Complete ionisation gives n – p = ND – NA; incomplete ionisation needs actual charged-dopant populations. If the calculated EF lies too close to or inside a band, the original Boltzmann-tail assumption must be revisited rather than extending the approximation silently.

B. Explain Atomic Packing and Elastic Response as Separate Scales

Count a cubic conventional cell by assigning shared sites fractionally. Eight corner atoms contribute eight times one eighth, or one atom. A body-centred atom contributes one more, giving Z = 2 for BCC. Six face-centred sites contribute six times one half, which together with corners gives Z = 4 for FCC. These counts describe the chosen repeating cell; coordination number instead counts the nearest neighbours of one atom.

In BCC, touching hard spheres lie along the body diagonal. Its length is √3 a and it spans four radii, giving 4r = √3 a. In FCC the analogous contact is along a face diagonal, so 4r = √2 a. Insert the appropriate radius relation into the occupied-sphere volume Z(4πr³/3) divided by a³ to obtain packing fractions about 0.680 and 0.7405 respectively. HCP has the same ideal close-packing fraction as FCC but a different stacking sequence.

Atomic bonding and structure influence stiffness, but cell geometry alone does not numerically determine Young modulus without an interatomic response model. At the continuum scale, impose the correct loading: tensile stress and longitudinal strain define Y, hydrostatic pressure and relative volume contraction define K, and shear stress and angular distortion define G. A uniaxial test also gives lateral contraction through Poisson ratio.

For a homogeneous isotropic linear solid, Y = 2G(1 + νP) and Y = 3K(1 – 2νP). Crystals can be anisotropic, and viscoelastic polymers depend on timescale, so these simple constants and identities have a restricted domain. A professional answer connects the atomic picture with the engineering measurement while keeping the assumptions of each scale explicit.

C. Obtain the Laminar Pipe-Flow Relation and Test Its Use

Consider a horizontal circular tube of radius R and length L carrying a steady incompressible Newtonian liquid. Assume fully developed, axisymmetric flow, no slip at the wall and negligible entrance effects. Let Δp be the positive pressure decrease along the flow. For a cylindrical fluid core of radius r, pressure force Δp πr² balances viscous traction 2πrLτs, giving τs(r) = Δp r/(2L).

The axial velocity falls toward the wall, so its derivative in the outward radial direction is negative. The relevant magnitude relation is -μ dv/dr = Δp r/(2L). Integrating and using v(R) = 0 gives v(r) = Δp(R² – r²)/(4μL). The centreline speed is ΔpR²/(4μL); the average is half that value.

Integrate flow through annular rings: Q = ∫0 to R v(r)2πr dr = πR^4 Δp/(8μL). With mean speed vbar = Q/(πR²) and D = 2R, rearrange to Δp = 32μLvbar/D². Combining with Darcy-Weisbach gives fD = 64/Re, provided Re = ρvbarD/μ and fD denotes the Darcy friction factor.

Before applying this result check flow regime, rheology, geometry and development length. A power-law slurry does not have a constant Newtonian μ; a turbulent flow does not have this parabolic profile. State those limitations in a solution. Pressure energy is dissipated into internal energy, so loss-free Bernoulli cannot describe the same long viscous tube without a loss term.

D. Translate a Soil-Energy Balance into a Water-Loss Estimate

Define a control surface near the soil or crop canopy and a consistent observation period. Take Rn as net radiative input, H as upward sensible transfer, LE as upward latent transfer and G as downward conduction into the soil. If storage and additional terms are negligible, conservation gives Rn – H – LE – G = 0. The signs are declarations, not universal conventions used identically by every instrument.

Solve for the latent residual: LE = Rn – H – G. If each term is an average flux in W/m² over duration Δt, energy per area is LEΔt in J/m². Divide by the latent heat Lv in J/kg to obtain evaporated mass per area in kg/m². For water of density near 1000 kg/m³, one kg spread over one square metre has depth 0.001 m, or one millimetre.

A negative latent residual under the model indicates condensation or an inconsistency in neglected terms or signs; it should not be presented as ordinary positive evaporation without interpretation. A dry surface may be water limited even under strong radiation. At night, radiation can be a loss and heat can move upward from warmer subsurface layers.

Temperature profiles supply the conductive component through Fourier’s law, G = -k dT/dz for downward-positive z under the declared sign. The moisture dependence of k and volumetric heat capacity affects both storage and transport. Measurements should therefore specify depth, time, moisture condition and averaging interval when comparing agricultural treatments.

E. Derive Numerical Aperture and the Laser Threshold Condition

For a meridional ray entering an ideal straight step-index fibre, let its internal angle to the axis be θ1. At the cylindrical core-cladding boundary the incidence to the normal is 90° – θ1. At the limiting guided condition, sin ic = n2/n1, so sinθ1,max = √[1 – (n2/n1)²]. Snell’s law at the perpendicular entrance face gives n0 sinθa = n1 sinθ1,max = √(n1² – n2²).

This derivation shows why the external entrance angle and internal critical angle are not interchangeable. It also shows why immersing the same fibre end in a higher-index medium narrows the entrance half-angle. Bending, imperfect coupling, finite core size and wave-mode effects can modify a practical system; the derived expression is the introductory ideal numerical aperture.

For a laser, follow one round trip rather than one pass. A beam crossing a length L twice receives the intensity factor exp[2(g – α)L]. Two reflections multiply intensity by R1R2. At threshold the round-trip multiplier equals one: R1R2 exp[2(g – α)L] = 1. Taking logarithms yields gth = α + ln[1/(R1R2)]/(2L).

Below threshold the circulating intensity decays from pass to pass. Above threshold gain saturation and other dynamics determine steady output; the unsaturated exponential model cannot predict unlimited power. The mirrors select optical feedback, while pumping maintains the population distribution responsible for gain. Mirror reflectivity alone does not create inversion.

F. Derive Newton Rings and Distinguish Dispersion from Resolution

Near the contact between a plano-convex lens and a plane plate, circle geometry gives r² = 2Rt – t². For a thin film t is much smaller than R, so neglect t² and use t = r²/(2R). In the standard reflected air-film arrangement there is one net reflection phase reversal, so a dark ring requires 2t = mλ. Substituting gives r_m² = mλR and hence D_m² = 4mλR.

Subtract the expression for order m from that for m + p. The unknown starting order cancels, leaving ΔD² = 4pλR. If a uniform liquid fills the same gap, the optical thickness becomes μt and the difference is divided by μ. Compare squared-diameter differences in air and liquid to obtain the liquid refractive index. Keep both the geometry approximation and phase convention in the explanation.

For a normally illuminated grating, d sinθ = mλ. Differentiating at fixed order and period gives d cosθ dθ = m dλ, so angular dispersion is m/(d cosθ). This is an angular separation rate; it does not specify the width of each spectral peak.

With N coherent illuminated slits, the principal peak narrows with N. The Rayleigh criterion leads to resolving power λ/Δλ = mN in the ideal grating. Increasing illuminated width increases resolution by increasing N even at unchanged angular dispersion. The maximum feasible order still satisfies |mλ| ≤ d, and overlapping spectra or missing orders can restrict a real measurement.

Formulas: Select the Correct Model

Unit Conversions and Symbol Checks

Final calculation check: Identify your branch route, declare the model, convert all quantities to one unit system, keep guard digits until the last line and interpret the sign or scale of the answer.

Answered Revision Checks

Why Does Density of States Differ from Occupation?

The first counts available states per interval; the second gives the probability that each state is occupied. Their product determines occupied density.

Can the Fermi Level Lie in a Forbidden Gap?

Yes. It is an equilibrium chemical potential used in occupation statistics; it need not itself be an allowed electron state.

Does a Hole Require a Mobile Ion?

No. It describes a missing valence electron whose transport can be represented as positive-carrier motion.

Does Larger Mass Mean Larger Band Curvature?

No. Effective mass is inversely proportional to the local second derivative of E(k) in the stated one-dimensional model.

How Do FCC and HCP Differ?

Both have ideal close-packing fraction about 0.7405 and coordination twelve, but FCC stacks ABC and HCP stacks AB.

What Distinguishes Creep from Stress Relaxation?

Creep measures increasing strain at fixed stress; relaxation measures decreasing stress at fixed strain.

Is Apparent Viscosity a Constant in a Power-Law Liquid?

Only for exponent n = 1. Otherwise it changes with shear rate over the range where the fitted law applies.

When Is Loss-Free Bernoulli Unsafe?

Over appreciably viscous or pumped pipe systems. Include head loss and any pump or turbine work with a consistent sign convention.

What Supports an Incipient Fluidized Bed?

The upward fluid drag, represented by the bed pressure drop, balances the effective weight of the solids in the simple model.

Why Do Root-Zone Temperatures Lag the Surface?

Heat requires time to diffuse downward; the periodic conductive solution predicts both decreasing amplitude and increasing phase delay.

Does Net Radiation Equal Latent Flux?

Only if sensible heat, soil flux, storage and other terms are negligible under the specified balance. Usually they are not all zero.

Why Distinguish Acceptance and Critical Angles?

One is measured to the fibre axis outside the entrance; the other is measured to the normal at the internal core-cladding boundary.

Can Thermal Equilibrium Sustain Laser Inversion?

Not for the usual positive-temperature level populations. Pumping maintains the required non-equilibrium distribution.

Why Is Reflected Zero-Thickness Air Film Dark?

In the usual arrangement one of the reflected rays acquires a π reversal. The phase convention and physical interfaces must match.

Does More Dispersion Guarantee More Resolution?

No. Dispersion separates wavelengths angularly; resolution also depends on peak width and the number of illuminated grating slits.

Why Does a Nicol Reject the Ordinary Ray?

The chosen calcite-cement geometry produces total internal reflection of that ray, while the extraordinary ray is transmitted.

Related B.Tech First Semester Study Materials

CSE 1st Year All Subject PYQs → CSE 1st Year All Subject Notes → CSE 1st Year All Subject Syllabus →