Material Science

Option 1 — Primary vs Secondary Bonding

1. Fundamental differences (2 marks)

2. Category of ionic bonding and why (2 marks)

Ionic bonding belongs to the primary (chemical) bonding category. Reasons:

• It is formed by complete transfer of electrons from an electropositive atom to an electronegative atom, producing cations and anions — a genuine chemical change, not a mere physical attraction.
• The binding force is the strong electrostatic (Coulomb) attraction between the oppositely charged ions, giving bond energy of the order of several hundred kJ mol⁻¹ (NaCl ≈ 787 kJ mol⁻¹).
• Consequently ionic crystals are hard, brittle, high‑melting and poor conductors in the solid state but good conductors when molten — all typical of primary‑bonded solids.

3. Attractive and repulsive terms for an ionic crystal (2 marks)

For a pair of ions of charges $Z_1e$ and $Z_2e$ separated by a distance $r$:

Attractive (Coulomb) term: $E_A=-\frac{A}{r},\qquad A=\frac{Z_1Z_2e^{2}}{4\pi\varepsilon_0}$

For a whole crystal the Coulomb term is summed over all ion pairs, giving the Madelung constant $\alpha$: $E_A=-\frac{\alpha Z_1Z_2e^{2}}{4\pi\varepsilon_0 r}\qquad(\alpha_{\text{NaCl}}=1.748)$

Repulsive term (arising from overlap of electron clouds / Pauli exclusion): $E_R=\frac{B}{r^{\,n}}$

where $B$ is a constant and $n$ the Born exponent ($n\approx 9$ for NaCl, 5–12 in general).

Net potential energy: $E_N(r)=-\frac{A}{r}+\frac{B}{r^{\,n}}$

4. Equilibrium bonding energy $E_0$ (4 marks)

At equilibrium the net force is zero, i.e. $E_N$ is minimum:

$\left(\frac{dE_N}{dr}\right)_{r=r_0}=\frac{A}{r_0^{2}}-\frac{nB}{r_0^{\,n+1}}=0$

$\therefore\quad B=\frac{A\,r_0^{\,n-1}}{n}$

Substituting $B$ back into $E_N$:

$$E_0=-\frac{A}{r_0}+\frac{A\,r_0^{\,n-1}}{n\,r_0^{\,n}} =-\frac{A}{r_0}+\frac{A}{n\,r_0}$$

$\boxed{\,E_0=-\frac{A}{r_0}\left(1-\frac{1}{n}\right)\,}$

In terms of the Madelung constant (the Born–Landé equation):

$\boxed{\,E_0=-\frac{\alpha Z_1Z_2e^{2}}{4\pi\varepsilon_0 r_0}\left(1-\frac{1}{n}\right)\,}$

The negative sign shows that energy is released (the crystal is stabler than the separated ions); $|E_0|$ is the bonding or lattice energy per ion pair. For NaCl, $r_0\approx 0.28$ nm gives $E_0\approx -8$ eV per ion pair.

---

OR — Soft vs Hard Magnetic Materials (Hysteresis)

Comparative study (6 marks)

Applications based on hardness of magnetic properties (4 marks)

Soft magnetic materials — chosen where the material must be magnetised and demagnetised rapidly with minimum loss:

• Cores of transformers, generators, motors and inductors (silicon steel, ferrites)
• Electromagnet cores, relays, contactors, chokes
• Magnetic recording heads, memory cores, flux‑multiplying cores
• Ferrite cores in high‑frequency (radio/TV, SMPS) circuits

Hard magnetic materials — chosen where a strong, permanent magnetic field is required without any external supply:

• Permanent magnets in loudspeakers, microphones, earphones
• Field magnets of small DC motors and generators, magnetos, dynamos
• Magnetic chucks, separators, door catches, magnetic bearings
• MRI machines, particle accelerators, magnetrons (SmCo, NdFeB)
• Measuring instruments (galvanometers, ammeters, voltmeters) and toys

Option A: Relevancy of Spectroscopic Techniques in Material Science and Working Principle of Raman Spectroscopy

1. Relevancy of Spectroscopic Techniques in Material Science (3 marks)

Spectroscopic techniques study the interaction of electromagnetic radiation with matter (absorption, emission, scattering, or reflection) to reveal the structure, composition, and properties of materials. Their relevancy in material science includes:

1. Compositional analysis — Identify elements and chemical groups present in a material (e.g., XPS, EDX, IR).
2. Structural determination — Reveal crystal structure, bonding type, and molecular geometry (e.g., XRD, Raman, NMR).
3. Electronic structure study — Determine band gap, oxidation states, and density of states (e.g., UV-Vis, XPS).
4. Surface and interface analysis — Characterize thin films, coatings, and interfaces at nanometer depth (e.g., XPS, AES, SIMS).
5. Non-destructive testing — Examine valuable or fragile samples without damage (e.g., Raman, FTIR).
6. In-situ and real-time monitoring — Track phase transitions, chemical reactions, and growth processes during synthesis.
7. Quality control — Detect impurities, defects, dopants, and strain in semiconductors and nanomaterials.

Thus, spectroscopy is indispensable for designing, characterizing, and optimizing modern materials such as nanomaterials, semiconductors, polymers, and composites.

---

2. Working Principle of Raman Spectroscopy (7 marks)

Principle

When monochromatic light (usually a laser) strikes a sample, most photons are elastically scattered without any change in frequency — this is Rayleigh scattering. A very small fraction (~1 in 10⁶–10⁸ photons) is inelastically scattered with a change in frequency. This inelastic scattering is called Raman scattering, discovered by C. V. Raman in 1928.

The frequency shift arises because the incident photon exchanges energy with the vibrational or rotational modes of the molecules of the sample:

• Stokes lines: The molecule absorbs energy from the photon and moves to a higher vibrational level. The scattered photon has lower energy (lower frequency) than the incident photon.
  $\nu_{\text{scattered}} = \nu_{\text{incident}} - \nu_{\text{vib}}$
• Anti-Stokes lines: The molecule (already in an excited vibrational state) gives energy to the photon. The scattered photon has higher energy than the incident photon.
  $\nu_{\text{scattered}} = \nu_{\text{incident}} + \nu_{\text{vib}}$

The Raman shift is defined as: $\Delta \bar{\nu} = \frac{1}{\lambda_{\text{incident}}} - \frac{1}{\lambda_{\text{scattered}}} \quad (\text{in cm}^{-1})$

This shift is independent of the incident wavelength and is characteristic of the molecular bonds — hence it acts as a "fingerprint" of the material.

Selection Rule

A vibrational mode is Raman active only if it produces a change in the polarizability of the molecule during vibration. (In contrast, IR requires a change in dipole moment.) This is why Raman and IR are complementary techniques.

Instrumentation and Working

1. Laser source — Provides intense monochromatic light (e.g., He–Ne 632.8 nm, Ar-ion 514.5 nm, or NIR 785 nm).
2. Sample illumination — Light is focused on the sample; scattered light is collected at 90° or 180°.
3. Rayleigh filter (notch/edge filter) — Removes the strong Rayleigh line so that weak Raman signals can be detected.
4. Monochromator / Spectrograph — Disperses the scattered light according to wavelength.
5. Detector — CCD or photomultiplier tube records the intensity of scattered light at each wavelength.
6. Computer — Plots intensity versus Raman shift (cm⁻¹) to give the Raman spectrum.

Diagram

        Laser
          |
          v
   +-------------+
   |   Sample    |  <-- scattered light (Rayleigh + Raman)
   +-------------+
          |
          v
   [ Notch Filter ]  (blocks Rayleigh line)
          |
          v
   [ Monochromator / Grating ]
          |
          v
   [ Detector (CCD) ]
          |
          v
   [ Computer ] --> Raman Spectrum

Energy Level Diagram

Each process is an upward arrow (photon absorbed to a virtual state) followed by a downward arrow (photon scattered):

• Rayleigh: molecule goes up to a virtual state and falls back to the same level; photon energy is unchanged.
• Stokes: molecule starts in $v=0$, is raised to a virtual state, and falls to $v=1$. The molecule gains $h\nu_v$, so the scattered photon has energy $h(\nu_0-\nu_v)$.
• Anti-Stokes: molecule starts in $v=1$, is raised to a virtual state, and falls to $v=0$. The molecule loses $h\nu_v$, so the scattered photon has energy $h(\nu_0+\nu_v)$. It is weaker than Stokes because few molecules are in $v=1$ at room temperature.

Applications

• Identification of chemical bonds and functional groups.
• Study of carbon materials (diamond, graphite, graphene, CNTs).
• Detection of strain, defects, and phase changes in semiconductors.
• Non-destructive analysis of archaeological and biological samples.
• Pharmaceutical and forensic analysis.

---

Option B: Principle of Electron Microscopy and Comparison of SEM and TEM

1. Principle of Electron Microscopy (3 marks)

Electron microscopy uses a beam of accelerated electrons instead of light to image specimens. Since the de Broglie wavelength of electrons is extremely small (e.g., at 100 kV, λ ≈ 0.0037 nm), the resolution is far better than optical microscopes — down to atomic scale.

$\lambda = \frac{h}{\sqrt{2 m e V}}$

where (V) is the accelerating voltage. The electron beam is focused by electromagnetic lenses (condenser, objective, projector) and interacts with the sample to produce signals such as:

• Transmitted electrons (used in TEM),
• Secondary electrons (used in SEM for topography),
• Backscattered electrons (composition contrast),
• X-rays (elemental analysis, EDX).

The signal is collected by detectors and converted into an image on a fluorescent screen or digital display. Because electrons are scattered by air, the entire column operates under high vacuum.

---

2. Differences Between SEM and TEM (7 marks)

Summary: SEM images the surface of bulk samples with good 3D appearance, while TEM images the internal structure of ultra-thin samples at atomic resolution. They are complementary tools in material science.

Phase equilibrium is the state of a system in which two or more phases (solid, liquid, gas) of the same substance coexist at definite temperature and pressure without any net transfer of matter from one phase to another. At such a state the chemical potential (or Gibbs free energy per mole) of the substance is the same in all the coexisting phases, i.e. μ₁ = μ₂ = μ₃, and the system is in dynamic equilibrium — the rate of molecules leaving one phase equals the rate returning to it. The conditions of temperature, pressure and composition at which this happens are shown graphically by a phase diagram.

---

Nature of a one-component (unary) phase diagram

A one-component system contains only one chemical substance (e.g. water, CO₂, sulphur). Its phase diagram is a pressure–temperature (P–T) graph showing the regions of stability of solid, liquid and vapour, and the curves along which two phases coexist. Since the system has one component, by the phase rule

$F = C - P + 2 = 1 - P + 2 = 3 - P$

• P = 1 (one phase): F = 2 → both T and P can be varied independently (an area on the diagram).
• P = 2 (two phases): F = 1 → only one variable is free; the other is fixed (a curve).
• P = 3 (three phases): F = 0 → T and P are fixed (a point).

Main features of the water phase diagram:

1. Three regions (areas): solid (ice), liquid (water) and vapour (steam). Each area represents a single phase where T and P can be varied independently.
2. Three curves:
   • Fusion (melting) curve OA — separates solid and liquid; gives the melting point at each pressure. For water it slopes backwards (to the left) because ice is less dense than water (Clausius–Clapeyron: dP/dT = ΔH/TΔV, and ΔV is negative on melting).
   • Vaporisation curve OC — separates liquid and vapour; starts at the triple point and ends at the critical point (for water, T_c = 647 K, P_c = 218 atm). Beyond this point liquid and vapour become indistinguishable (supercritical fluid).
   • Sublimation curve OB — separates solid and vapour; gives the vapour pressure of the solid.
3. Triple point O — the unique point where all three phases coexist in equilibrium. For water it lies at 0.01 °C (273.16 K) and 4.58 mm Hg. Here F = 0, so the triple point is used to define the kelvin.
4. Metastable states: the vapour pressure curve of supercooled water can be extended below the triple point (dashed line) — a metastable condition.

Example — phase diagram of water: At 1 atm, moving horizontally from low to high temperature, ice melts at 0 °C (curve OA) and water boils at 100 °C (curve OC). At pressures below 4.58 mm Hg, ice sublimes directly to vapour without melting. Thus the diagram explains melting, boiling, sublimation, the effect of pressure on melting point, and the existence of the critical point — all from a single substance with one component.

Question: The optical phenomena that occur within solid materials involve interactions between electromagnetic radiation and atoms, ions, and/or electrons. Discuss the two most important of these interactions, with relevant examples. [4+4]

---

1. Electronic Polarization (Interaction with Electrons)

When electromagnetic radiation (light) is incident on a solid, its oscillating electric field component exerts a force on the bound electrons of the atoms. The electrons are displaced from their equilibrium positions relative to the nucleus, creating induced electric dipole moments. This phenomenon is called electronic polarization.

Key points:

• The electrons oscillate at the same frequency as the incident radiation.
• The extent of polarization depends on the ease with which the electron cloud can be distorted — this is measured by electronic polarizability (αₑ).
• This interaction is responsible for the refractive index of a material. According to the Lorentz–Lorentz relation:

$\frac{n^2 - 1}{n^2 + 2} = \frac{N \alpha_e}{3\varepsilon_0}$

where n is the refractive index, N is the number of atoms per unit volume, and αₑ is the electronic polarizability.

• When the frequency of the incident radiation approaches the natural frequency of electron oscillation, resonant absorption occurs — this is why materials absorb light at specific wavelengths (giving them colour).

Example:

• Glass (SiO₂): Electronic polarization of Si and O atoms gives glass its transparency in the visible range. The electrons are too tightly bound to absorb visible light, so light passes through with a characteristic refractive index (n ≈ 1.5).
• Diamond: Strong electronic polarization due to tightly bound carbon electrons gives diamond an exceptionally high refractive index (n = 2.42) and its brilliant sparkle.
• Coloured gemstones (e.g., ruby): The Cr³⁺ ions absorb specific wavelengths (green/yellow) via electronic transitions, transmitting red light — hence the red colour.

---

2. Interaction with Ions (Ionic Polarization / Lattice Absorption)

In ionic solids, the electromagnetic radiation interacts with the ions (cations and anions) of the crystal lattice. The oscillating electric field displaces the positive and negative ions in opposite directions, producing ionic polarization.

Key points:

• This interaction is significant in ionic crystals (e.g., NaCl, KCl, MgO).
• The ions oscillate about their lattice positions; the relative displacement of cation and anion creates a net dipole moment.
• Ionic polarization contributes to the dielectric constant and affects optical properties in the infrared region.
• When the radiation frequency matches the vibrational frequency of the lattice (optical phonon frequency), resonant absorption occurs — this is called lattice absorption or Reststrahlen absorption.

Example:

• NaCl crystal: The Na⁺ and Cl⁻ ions vibrate with a characteristic infrared frequency. When IR radiation of this frequency falls on NaCl, strong absorption occurs — NaCl is opaque in the IR region at these wavelengths.
• MgO: Ionic polarization in MgO leads to a high dielectric constant and characteristic IR absorption bands.
• Optical phonon interaction: In polar semiconductors like GaAs, the interaction of photons with longitudinal optical (LO) phonons (via the Fröhlich interaction) is a direct consequence of ionic polarization and is crucial in optoelectronic devices.

---

Summary Table

---

Conclusion: Electronic polarization dominates in the visible and UV regions and governs refractive index and colour, while ionic polarization dominates in the infrared region and governs lattice absorption and dielectric behaviour. Together, these two interactions explain most optical phenomena in solid materials.

Synthesis of Carbon Nanotubes from Graphene

Yes — in principle, carbon nanotubes (CNTs) can be synthesized from graphene. A single-walled carbon nanotube (SWCNT) is geometrically nothing but a rolled-up sheet of graphene. If a rectangular strip of graphene is curled so that its opposite edges join seamlessly, the hexagon network closes into a cylinder and a nanotube is formed. The chirality (the roll-up vector) decides whether the tube is metallic or semiconducting.

Practically this has been achieved by:

• Rolling/self-scrolling: depositing a graphene sheet on a substrate and using a solvent or surface tension to roll it into a tube.
• Nanotube unzipping in reverse: graphene nanoribbons obtained by unzipping CNTs can be re-rolled.
• CVD growth: graphene flakes act as the carbon source or as a template on which CNTs grow.

So CNTs and graphene are inter-convertible forms of the same sp² carbon allotrope.

Applications of Carbon Nanotubes as Advanced Materials

Present applications

1. Composites: CNTs are added to polymers, epoxy and concrete. Their very high tensile strength (~100 GPa) and Young's modulus (~1 TPa) give strong, light structural materials for sports goods, aircraft and automotive parts.
2. Electronics: Metallic CNTs carry current density up to ~10⁹ A/cm² without heating, so they are used as interconnects, field-emission displays and in transistors.
3. Energy storage: Extremely high surface area (~1300 m²/g) makes CNTs ideal electrode material for supercapacitors, lithium-ion batteries and fuel-cell catalyst supports.
4. Sensors: Semiconducting CNTs change conductance on adsorbing gases, so they detect NO₂, NH₃, and biomolecules with high sensitivity.
5. Conductive coatings and antistatic films: CNT films are transparent and conductive, used in touch screens and EMI shielding.
6. Biomedicine: Functionalized CNTs act as drug-delivery vehicles and as contrast agents in imaging; they can also be used in tissue scaffolds.

Future applications

1. Nanoelectronics: CNT transistors may replace silicon MOSFETs below 5 nm, extending Moore's law.
2. Quantum wires and interconnects: Ballistic conduction makes CNTs ideal for lossless wiring in future chips.
3. Hydrogen storage: CNTs can physisorb hydrogen, promising safe storage for fuel-cell vehicles.
4. Space elevators and ultra-strong cables: The theoretical strength-to-weight ratio of CNTs exceeds any known material.
5. Artificial muscles: CNT actuators respond to voltage and can mimic muscle movement.
6. Water purification and desalination: CNT membranes with controlled pore size can filter ions and pollutants.
7. Thermal management: CNTs have very high thermal conductivity (~3500 W/m·K), useful as heat sinks in electronics.

Limitations: high production cost, difficulty in separating metallic from semiconducting tubes, poor dispersion in matrices, and possible toxicity. These are the main barriers to large-scale commercial use.

1. Definition of solid solution (2 marks)

When foreign (impurity) atoms are added to a pure solid, they may either form a separate second phase or get dissolved into the parent lattice. If the impurity atoms enter the crystal lattice of the host solid and remain homogeneously distributed in it without forming any new phase or compound, the resulting single-phase crystalline solid is called a solid solution.

Thus a solid solution is a homogeneous, single-phase crystalline solid in which the lattice of the solvent (host) is retained, its composition can be varied continuously, and the solute (impurity) atoms occupy either:

• Substitutional sites — impurity atom replaces a host atom on a normal lattice point (e.g. Ni in Cu, Zn in Cu); or
• Interstitial sites — impurity atom occupies a void between host atoms (e.g. C, N in Fe).

Examples: Cu–Ni, Au–Ag, brass (Cu–Zn), steel (C dissolved in $\alpha$-Fe).

---

2. Features of solute and solvent atoms determining the degree of substitutional solid solution (6 marks)

The extent to which one atom can substitute for another in a lattice is governed by the Hume-Rothery rules. For extensive (complete) substitutional solubility, the following features must be satisfied:

(a) Atomic size factor (size difference < 15%) The difference in atomic radii of solute and solvent must be small. If $\frac{r_{\text{solute}}-r_{\text{solvent}}}{r_{\text{solvent}}}\times 100 < 15\%$ then extensive substitutional solubility occurs. If the difference is 15–30%, solubility is only partial (limited); above 30%, substitutional solubility is practically nil, because large size mismatch produces severe lattice distortion and high elastic strain energy. Example: Cu–Ni (radii differ by only ~2%) → complete solubility; Cu–Zn (~4%) → good solubility; Na in Cu → negligible.

(b) Crystal structure factor Both solute and solvent must have the same crystal structure (both FCC, both BCC, etc.) for complete solubility. If structures differ, solubility is restricted to a few percent. Example: Cu and Ni are both FCC → complete solid solution; Cu (FCC) and Zn (HCP) → limited solubility.

(c) Electronegativity factor The two atoms must have similar electronegativity. A large difference in electronegativity gives greater chemical affinity between them, so they tend to form an intermetallic compound rather than a solid solution. As a rough guide, if the electronegativity difference is less than about 0.4, solid solution formation is favoured; if greater, compound formation dominates. Example: Cu–Ni (small difference) → solid solution; Mg–Pb (large difference) → compound $\text{Mg}_2\text{Pb}$.

(d) Relative valency factor For a solvent of lower valency, a solute of higher valency dissolves more readily; for a solvent of higher valency, a solute of lower valency dissolves more readily. Example: divalent Zn dissolves more in monovalent Cu than in trivalent Al; conversely, in Ni (divalent) the solubility of lower-valency solutes is greater. A metal dissolves another metal of the same valency most easily.

(e) Temperature Solubility generally increases with temperature, since thermal energy helps to accommodate the distortion caused by the solute atom. Hence the solubility limit of a substitutional solid solution rises as temperature is raised (e.g. C in Fe).

(f) Other factors

• Chemical affinity / lattice distortion: strong affinity or large distortion reduces solubility.
• Electron-to-atom ratio: an increase in the electron concentration beyond a critical value destabilises the lattice and limits solubility.
• Ordering tendency: at low temperature, solute and solvent atoms may order into a superlattice, reducing random substitution.

Conclusion: Complete substitutional solid solution requires nearly equal atomic size (<15% difference), identical crystal structure, similar electronegativity, favourable relative valency, and sufficient temperature; failure of any one condition reduces the degree of substitution.

a) Current–voltage (I–V) characteristics of a p-n junction diode

The I–V curve is drawn with voltage (V) along the x-axis and current (I) along the y-axis. Forward bias is taken on the right, reverse bias on the left.

        I (forward)
          |
          |            /
          |           /   <- forward region
          |          /
          |         /
          |        /
  --------+-------+-------------------> V
   (reverse)  |   Vk (knee ~0.3 V Ge, 0.7 V Si)
              |
              |  <- reverse saturation current (small, ~µA)
              |

Key features to mark on the sketch:

• Forward bias: For small voltages the current is negligible. Beyond the knee (cut-in) voltage ((V_k \approx 0.3) V for Ge, (0.7) V for Si) the current rises exponentially and rapidly. The diode behaves as a closed switch.
• Reverse bias: A very small reverse saturation (leakage) current flows, due to minority carriers. It is almost independent of the applied voltage until breakdown.
• Breakdown: At a large reverse voltage (Zener or avalanche breakdown) the current suddenly increases sharply. Below breakdown the diode behaves as an open switch.

The diode equation governing the curve is $I = I_0\left(e^{eV/k_BT} - 1\right)$ where (I_0) is the reverse saturation current. For forward bias (V \gg k_BT/e), (I \approx I_0 e^{eV/k_BT}); for reverse bias (I \approx -I_0).

---

OR

Antiferromagnetism vs ferrimagnetism

Both have a critical (Néel) temperature above which the ordered arrangement breaks down and the material becomes paramagnetic.

---

b) Phonons and their role in thermal conductivity

Phonons: In a solid the atoms are not stationary but vibrate about their mean positions. These lattice vibrations are quantised, and each quantum of vibrational energy is called a phonon. Its energy is $E = \hbar\omega = h\nu$ where (\nu) is the vibration frequency. A phonon behaves like a particle of momentum (\hbar k) and is a boson (it obeys Bose–Einstein statistics). Phonons are not real particles but quasi-particles representing lattice vibration modes.

Role in thermal conductivity: In a solid, heat is carried from the hot end to the cold end by two agents — free electrons and lattice vibrations. In non-metals and insulators, which have few free electrons, heat conduction takes place almost entirely through phonons. The phonons travel through the lattice, collide with one another and with impurities and lattice defects, and transfer energy. The thermal conductivity due to phonons is $K = \frac{1}{3}C v l$ where (C) is the phonon heat capacity, (v) the phonon velocity and (l) the mean free path. At low temperature (K) rises as (C) increases; at high temperature phonon–phonon collisions shorten (l), so (K) falls. Thus phonons are the chief carriers of heat in insulators and a major contributor in semiconductors.

---

OR

Importance of recycling and disposability in material science

1. Environmental protection: Materials discarded after use pollute air, water and soil. Recycling reduces mining, landfill waste and toxic emissions, so material scientists must design materials that can be safely reused or degraded.
2. Conservation of resources: Many engineering materials (metals, rare-earth elements) are limited. Recycling recovers them and extends the life of natural reserves.
3. Energy saving: Producing a material from raw ore needs far more energy than reprocessing scrap — e.g. recycling aluminium saves about 95% of the energy needed for primary production.
4. Economic benefit: Recycled raw material is cheaper, lowers production cost and creates a recycling industry and employment.
5. Design requirement: Modern material selection is not based on properties alone. A material must be recyclable, biodegradable or easily disposable so that the product's full life cycle (cradle to grave) is eco-friendly. This is the basis of green material science and life-cycle assessment.
6. Legal and social pressure: Strict environmental laws and public awareness force industries to choose materials with low disposal hazards (e.g. lead-free solder, biodegradable polymers).

Hence recycling and disposability are treated as essential selection criteria along with mechanical, thermal and electrical properties.

(a) How lasers differ from ordinary light

The reason is that in a laser the photons are emitted by stimulated emission and are in phase, whereas in ordinary light they arise from spontaneous emission with random phases and directions.

---

(b) Fermi gas and its introduction on the basis of the free electron model

Fermi gas: A collection of a large number of non-interacting identical fermions (particles of half-integral spin, e.g. electrons) obeying the Pauli exclusion principle and the Fermi–Dirac statistics is called a Fermi gas. At absolute zero all energy states up to a maximum energy, the Fermi energy $E_F$, are filled and all states above it are empty.

On the basis of the free electron model:

• Valence electrons in a metal are treated as free particles moving in a uniform potential inside the metal (the "electron gas" or Fermi gas).
• Each electron occupies a quantum state defined by the wave vector $\vec{k}$, with energy $E = \frac{\hbar^2 k^2}{2m}$
• Since electrons are fermions, each state holds at most two electrons (spin up and spin down).
• At $T = 0$ K, states are filled from $k = 0$ up to the Fermi wave vector $k_F$, giving the Fermi energy $E_F = \frac{\hbar^2 k_F^2}{2m}, \qquad k_F = \left(3\pi^2 n\right)^{1/3}$ where $n$ is the electron number density.
• The density of states is $g(E) \propto E^{1/2}$, and the occupation probability is given by the Fermi–Dirac distribution $f(E) = \frac{1}{e^{(E-E_F)/k_BT} + 1}$
• At $T = 0$ K, $f(E) = 1$ for $E < E_F$ and $0$ for $E > E_F$; at $T > 0$ K the edge is smeared over about $k_BT$.
• The Fermi temperature $T_F = E_F/k_B$ is very high (~$10^4$–$10^5$ K), so even at room temperature the gas is highly degenerate.
• This model successfully explains the heat capacity (only electrons near $E_F$ contribute, giving $C \propto T$), electrical and thermal conductivity, and the Wiedemann–Franz law of metals.

---

(c) Conductivity of materials in terms of energy band structure

In a solid, the allowed electron energies form bands separated by forbidden gaps. Conductivity depends on how these bands are filled:

• Conductors (metals): The valence band is partially filled or overlaps the conduction band. There is no energy gap. A very small applied field excites electrons into nearby empty states, so current flows easily. Conductivity is high and decreases with rising temperature (due to increased lattice scattering).
• Insulators: The valence band is completely filled and the conduction band is completely empty, separated by a large forbidden gap ($E_g > 3$ eV, e.g. diamond ~5.5 eV). Thermal energy at room temperature cannot lift electrons across the gap, so essentially no free carriers exist and conductivity is extremely low.
• Semiconductors: The valence band is full and the conduction band is empty, but the gap is small ($E_g \approx 0.1$–$2$ eV, e.g. Si ~1.1 eV, Ge ~0.7 eV). At $T > 0$ K some electrons are thermally excited into the conduction band, leaving holes in the valence band. Both electrons and holes carry current, so conductivity is moderate and increases with temperature (negative temperature coefficient of resistance).
• Doping: Adding pentavalent (donor) or trivalent (acceptor) impurities creates n-type or p-type semiconductors, creating donor/acceptor levels just below the conduction band or just above the valence band, greatly increasing conductivity.

Thus the band gap and the degree of band filling decide whether a solid is a conductor, semiconductor or insulator.

---

(d) Historical examples of key synthetic discoveries

• Urea (1828) — Friedrich Wöhler synthesised urea from ammonium cyanate, the first organic compound made from inorganic starting materials, disproving the "vital force" theory.
• Aspirin / Acetylsalicylic acid (1897) — Felix Hoffmann at Bayer synthesised a stable, less irritating form of salicylic acid, the first widely used synthetic drug.
• Ammonia — Haber–Bosch process (1909–1913) — Fritz Haber and Carl Bosch synthesised ammonia from nitrogen and hydrogen, enabling synthetic fertilizers.
• Bakelite (1907) — Leo Baekeland produced the first fully synthetic plastic (thermosetting phenol-formaldehyde resin).
• Nylon (1935) — Wallace Carothers at DuPont synthesised the first fully synthetic fibre (polyamide).
• Penicillin (1928, isolated 1940s) — Alexander Fleming discovered it; Florey and Chain made it usable as the first antibiotic.
• Polio vaccine (1955) — Jonas Salk developed the first inactivated polio vaccine.
• CRISPR-Cas9 gene editing (2012) — Doudna and Charpentier developed a programmable tool for editing DNA.

For BCC iron the atoms touch along the body diagonal of the unit cell, so

$\sqrt{3}\,a = 4R \;\Rightarrow\; a=\frac{4R}{\sqrt3}=\frac{4(0.124)}{1.732}=0.2864\ \text{nm}$

The $[110]$ direction is a face diagonal. Its repeat length (from one corner atom to the next corner atom) is

$L=\sqrt{2}\,a = 1.414\times0.2864 = 0.405\ \text{nm}$

Along this line only the two end corner atoms lie on it. Each is shared with the neighbouring repeat length, so each contributes $\tfrac12$ atom, giving 1 atom per repeat length (the body-centre atom does not lie on a face diagonal).

$LD_{[110]}=\frac{\text{number of atoms centred on the direction vector}}{\text{length of the vector}}=\frac{1}{\sqrt2\,a}=\frac{1}{0.405\ \text{nm}}$

$\boxed{LD_{[110]}\approx 2.47\ \text{nm}^{-1}=2.47\times10^{9}\ \text{m}^{-1}}$

The stated composition (79.65\text{ wt% Fe} - 20.35\text{ wt% C}) is outside the Fe–Fe(_3)C phase diagram, since the maximum carbon content in cementite is (6.67\text{ wt% C}).
So this is almost certainly a typo for the standard problem:

[ 99.65\text{ wt% Fe} - 0.35\text{ wt% C} ]

i.e. (C_0 = 0.35\text{ wt% C}), just below the eutectoid temperature (727^\circ\text{C}).

Use:

[ C_\alpha = 0.022\text{ wt% C},\qquad C_{\text{Fe}_3\text{C}} = 6.67\text{ wt% C},\qquad C_e = 0.76\text{ wt% C} ]

where (C_e) is the eutectoid composition.

---

a. Fraction of total ferrite and cementite

For total ferrite:

[ W_\alpha=\frac{C_{\text{Fe}3\text{C}}-C_0}{C{\text{Fe}3\text{C}}-C\alpha} ]

[ W_\alpha=\frac{6.67-0.35}{6.67-0.022} =\frac{6.32}{6.648} =0.9507 ]

For total cementite:

[ W_{\text{Fe}3\text{C}}=\frac{C_0-C\alpha}{C_{\text{Fe}3\text{C}}-C\alpha} ]

[ W_{\text{Fe}_3\text{C}}=\frac{0.35-0.022}{6.67-0.022} =\frac{0.328}{6.648} =0.0493 ]

[ \boxed{W_\alpha \approx 95.07%} ]

[ \boxed{W_{\text{Fe}_3\text{C}} \approx 4.93%} ]

---

b. Fractions of proeutectoid ferrite and pearlite

For proeutectoid ferrite:

[ W_{\alpha,\text{pro}}=\frac{C_e-C_0}{C_e-C_\alpha} ]

[ W_{\alpha,\text{pro}}=\frac{0.76-0.35}{0.76-0.022} =\frac{0.41}{0.738} =0.5556 ]

For pearlite:

[ W_{\text{pearlite}}=\frac{C_0-C_\alpha}{C_e-C_\alpha} ]

[ W_{\text{pearlite}}=\frac{0.35-0.022}{0.76-0.022} =\frac{0.328}{0.738} =0.4444 ]

[ \boxed{W_{\alpha,\text{pro}} \approx 55.56%} ]

[ \boxed{W_{\text{pearlite}} \approx 44.44%} ]

---

Final answers

[ \boxed{\text{Total ferrite} = 95.07%,\quad \text{Total cementite} = 4.93%} ]

[ \boxed{\text{Proeutectoid ferrite} = 55.56%,\quad \text{Pearlite} = 44.44%} ]

If the composition is literally (20.35\text{ wt% C}), then the problem is not valid for the Fe–C system.

(a) Arsenic (group V) has five valence electrons, one more than silicon needs for bonding. It acts as a donor, so electrons are the majority carriers and the material is n-type.

(b) $N_D = 10^{23}\ \text{m}^{-3}=10^{17}\ \text{cm}^{-3}$. This is far above the intrinsic concentration of Si at both temperatures ($n_i\approx10^{16}\ \text{m}^{-3}$ at 300 K and about $10^{19}\ \text{m}^{-3}$ at 400 K), and the donors are fully ionised, so

$n \approx N_D = 10^{23}\ \text{m}^{-3}$

Conductivity due to impurity: $\sigma = n e\mu_n$.

The electron mobility must be taken at this doping level. Impurity scattering lowers it well below the 0.135 m$^2$/V$\cdot$s of nearly pure Si. From the mobility versus doping curves for Si:

• at 300 K: $\mu_n \approx 0.078\ \text{m}^2/\text{V}\cdot\text{s}$
• at 400 K: $\mu_n \approx 0.06\ \text{m}^2/\text{V}\cdot\text{s}$ (lattice scattering increases with temperature, so mobility falls)

$\sigma_{300}=10^{23}\times1.6\times10^{-19}\times0.078\approx1.25\times10^{3}\ \text{S/m}$

$\sigma_{400}=10^{23}\times1.6\times10^{-19}\times0.06\approx0.96\times10^{3}\ \text{S/m}$

The carrier concentration is fixed by doping, so the conductivity falls slightly with temperature only because mobility decreases.

(a) When the ends of the rod are rigidly fixed, free thermal expansion is prevented. Therefore a compressive stress develops.

Free thermal strain:

[ \epsilon_{th}=\alpha \Delta T ]

[ \Delta T=85-15=70^\circ\text{C} ]

[ \epsilon_{th}=(20.0\times 10^{-6})(70)=1.4\times 10^{-3} ]

Since the ends are rigid, the mechanical strain is equal and opposite:

[ \epsilon_m=-1.4\times 10^{-3} ]

Taking Young’s modulus for brass as (E=100\text{ GPa}=100\times 10^9\text{ Pa}),

[ \sigma=E\epsilon_m ]

[ \sigma=(100\times 10^9)(-1.4\times 10^{-3}) ]

[ \sigma=-140\times 10^6\text{ Pa} ]

[ \sigma=-140\text{ MPa} ]

The negative sign indicates compressive stress.

[ \boxed{\text{Type: compressive; magnitude } = 140\text{ MPa}} ]

(b) For a fully constrained rod, thermal stress is independent of the length of the rod. The same temperature change and same material properties give the same stress.

[ \boxed{\text{For a 1 m long rod also: compressive stress } = 140\text{ MPa}} ]

So changing the length from (0.35\text{ m}) to (1\text{ m}) does not change the magnitude of thermal stress.
(If (E) for brass is taken as (105\text{ GPa}), the stress would be (147\text{ MPa}).)

Given:

• Radius of curvature at crack tip, $\rho = 1.9 \times 10^{-4}\ \text{mm}$
• Crack length (total internal crack length) $= 3.8 \times 10^{-2}\ \text{mm}$ So half-crack length, $a = \dfrac{3.8 \times 10^{-2}}{2} = 1.9 \times 10^{-2}\ \text{mm}$
• Applied tensile stress, $\sigma_0 = 140\ \text{MPa}$

Formula:

The maximum stress at the crack tip is given by

$\sigma_m = 2\sigma_0\sqrt{\frac{a}{\rho}}$

Substitution:

$\sigma_m = 2 \times 140 \times \sqrt{\frac{1.9 \times 10^{-2}}{1.9 \times 10^{-4}}}$

$\sigma_m = 280 \times \sqrt{100}$

$\sigma_m = 280 \times 10 = 2800\ \text{MPa}$

Result:

$\boxed{\sigma_m = 2800\ \text{MPa} = 2.8\ \text{GPa}}$

The stress concentration factor is $K = \dfrac{\sigma_m}{\sigma_0} = \dfrac{2800}{140} = 20$, so the stress at the crack tip is 20 times the applied stress.

For a cuboctahedral fcc nanoparticle with $n$ complete shells (layers):

$N_{\text{total}}=\frac{10n^3+15n^2+11n+3}{3},\qquad N_{\text{surface}}=10n^2+2$

The surface fraction is $f_s = N_{\text{surface}}/N_{\text{total}}$.

Discussion

• Total atoms grow as $n^3$ (volume) but surface atoms only as $n^2$ (area), so for large $n$, $f_s\approx 30n^2/(10n^3)=3/n$.
• For very small particles ($n=2$) about three quarters of the atoms are on the surface; by $n=100$ (a few tens of nm) only about 3 % are.
• Surface atoms have fewer neighbours (lower coordination), so they are more reactive, have higher surface energy, and give a lower melting point and a higher catalytic activity. These effects are strong for small particles and fade as the particle grows towards bulk behaviour.