Physics भौतिकशास्त्र

Quick summary

In a crystal, atoms are arranged in a regular, repeating 3-D pattern. This topic explains lattice, basis and unit cell, the seven crystal systems and fourteen Bravais lattices, the simple cubic, BCC and FCC structures, Miller indices and the spacing between lattice planes.

Key points

• Crystalline solids have long-range order and sharp melting points; amorphous solids (glass) do not.
• Lattice: a regular array of points in space, each with identical surroundings.
• Basis: the atom or group of atoms attached to each lattice point. Crystal structure = lattice + basis.
• Unit cell: the smallest repeating block; described by edges a, b, c and angles α, β, γ.
• Seven crystal systems: cubic, tetragonal, orthorhombic, monoclinic, triclinic, trigonal (rhombohedral), hexagonal.
• Bravais (1848) showed there are only 14 distinct space lattices.
• Cubic lattices: simple cubic (SC), body-centred cubic (BCC) and face-centred cubic (FCC).
• Coordination number: number of nearest neighbours of an atom.
• Packing fraction: fraction of unit-cell volume filled by atoms.
• Miller indices (h k l) describe a plane: take intercepts in units of a, b, c, take reciprocals, clear fractions to smallest integers.
• Planes parallel to an axis have intercept ∞ on it, so index 0.
• Examples: Polonium is SC; iron (α), sodium are BCC; copper, aluminium, gold are FCC; NaCl is FCC lattice with a two-ion basis.

Important terms

• Lattice parameters – a, b, c and α, β, γ.
• Primitive cell – unit cell with lattice points only at corners (one lattice point per cell).
• Symmetry elements – centre, planes and axes of symmetry.
• Crystal (स्फटिक) – a solid with regular repeating arrangement of atoms.

Formulas

• Interplanar spacing (cubic): d = a/√(h² + k² + l²)
• Density: ρ = nM/(N_A a³) (n = atoms per cell, M = molar mass in kg/mol)

5 important questions

Q1. What is a Bravais lattice? How many are there?

A Bravais lattice is an infinite array of points in which every point has exactly the same surroundings. In three dimensions there are 14 Bravais lattices grouped in 7 crystal systems.

Q2. Find the Miller indices of a plane with intercepts 2a, 3b and ∞ (parallel to the z-axis).

Intercepts in units: 2, 3, ∞. Reciprocals: 1/2, 1/3, 0. Multiply by LCM 6: 3, 2, 0. Miller indices = (3 2 0).

Q3. Calculate the packing fraction of a BCC lattice.

1. Atoms per cell = 8 × (1/8) + 1 = 2.
2. Atoms touch along the body diagonal: √3 a = 4r, so r = √3a/4.
3. Volume of atoms = 2 × (4/3)πr³ = (8/3)π(√3a/4)³ = (8/3)π × 3√3 a³/64 = √3πa³/8.
4. Packing fraction = (√3πa³/8)/a³ = √3π/8 ≈ 0.68, i.e. 68%.

Q4. Copper is FCC with a = 3.61 Å and M = 63.5 g/mol. Find its density (N_A = 6.022 × 10²³).

1. n = 4, M = 63.5 × 10⁻³ kg/mol, a = 3.61 × 10⁻¹⁰ m.
2. a³ = (3.61)³ × 10⁻³⁰ = 47.05 × 10⁻³⁰ = 4.705 × 10⁻²⁹ m³.
3. N_A a³ = 6.022 × 10²³ × 4.705 × 10⁻²⁹ = 2.833 × 10⁻⁵.
4. ρ = (4 × 0.0635)/(2.833 × 10⁻⁵) = 0.254/(2.833 × 10⁻⁵) ≈ 8.97 × 10³ kg/m³.

Q5. Explain Miller indices and find the spacing of (1 1 0) planes in a cubic crystal with a = 4 Å.

1. Steps: find intercepts, express in units of a, b, c; take reciprocals; reduce to smallest integers; enclose in ( ).
2. Equally spaced parallel planes share the same indices.
3. For cubic crystals, d = a/√(h² + k² + l²).
4. d₁₁₀ = 4/√(1 + 1 + 0) = 4/√2 ≈ 2.83 Å.
5. Note: d₁₀₀ = 4 Å, d₁₁₁ = 4/√3 ≈ 2.31 Å — higher indices give closer planes.

Exam tip

Learn the SC–BCC–FCC table thoroughly; most short questions and numericals come from it. Draw the three cubic unit cells and the (100), (110), (111) planes neatly.

Quick summary

Atoms in a solid are held together by bonds. The type of bond decides whether the solid is hard or soft, a conductor or an insulator, and what its melting point is. This topic covers interatomic forces and the potential-energy curve, cohesive energy, and the five types of bonding: ionic, covalent, metallic, Van der Waals and hydrogen bonds, with the lattice energy of ionic crystals.

Key points

• Between two atoms there is a long-range attraction and a short-range repulsion (due to overlap of electron clouds and the Pauli principle).
• Potential energy U(r) = −A/rᵐ + B/rⁿ, with n greater than m. At equilibrium separation r₀, dU/dr = 0 and U is minimum; the depth of this minimum is the bond energy.
• Cohesive energy: energy needed to separate a solid into free neutral atoms.
• Ionic bond: transfer of electrons (Na⁺Cl⁻). Hard, brittle, high melting point, insulators as solids, conduct when molten; soluble in water.
• Covalent bond: sharing of electron pairs (diamond, Si, Ge). Very hard, very high melting point, directional bonds, poor conductors.
• Metallic bond: positive ions in a "sea" of free electrons (Cu, Na). Good electrical and thermal conductors, malleable, ductile, shiny.
• Van der Waals (molecular) bond: weak attraction between induced dipoles (solid argon, iodine). Soft, low melting point.
• Hydrogen bond: H atom between two strongly electronegative atoms (ice, DNA). Strength between Van der Waals and covalent.
• Bond strength order (roughly): covalent ≈ ionic greater than metallic greater than hydrogen greater than Van der Waals.
• Madelung constant α accounts for the attraction and repulsion from all ions in the crystal; for NaCl α = 1.748.

Important terms

• Equilibrium spacing (r₀) – separation at which the net force between atoms is zero.
• Lattice energy – energy released when free ions come together to form one mole of an ionic crystal.
• Ionisation energy – energy needed to remove an electron from a neutral atom.
• Electron affinity – energy released when a neutral atom gains an electron.
• Born exponent (n) – power in the repulsive term, usually 6–12.

Formulas

• U(r) = −A/rᵐ + B/rⁿ
• Force F = −dU/dr; at r₀, F = 0
• Ionic crystal energy per ion pair: U(r) = −αe²/(4πε₀r) + B/rⁿ
• Equilibrium energy: U₀ = −(αe²/4πε₀r₀)(1 − 1/n)
• Energy to form Na⁺Cl⁻ pair from atoms: ΔE = (ionisation energy of Na) − (electron affinity of Cl) + Coulomb energy

5 important questions

Q1. Compare ionic and covalent bonds.

Q2. Why are metals good conductors of electricity?

In a metal, valence electrons are not tied to any one atom. They form a free electron gas that moves through the lattice of positive ions. A small electric field makes these electrons drift, so current flows easily. The same free electrons also carry heat, so metals are good thermal conductors.

Q3. Derive the expression for equilibrium spacing and bond energy from U(r) = −A/rᵐ + B/rⁿ.

1. At equilibrium dU/dr = 0: mA/r₀^(m+1) − nB/r₀^(n+1) = 0.
2. So r₀^(n−m) = nB/(mA), i.e. r₀ = (nB/mA)^(1/(n−m)).
3. Put B = (mA/n) r₀^(n−m) back: U₀ = −A/r₀ᵐ + (m/n)A/r₀ᵐ = −(A/r₀ᵐ)(1 − m/n).
4. For a stable bond, d²U/dr² must be positive at r₀, which needs n greater than m (repulsion is shorter-ranged).

Q4. Calculate the energy needed to form Na⁺ and Cl⁻ ions from neutral atoms, and the energy released when they come to 0.236 nm apart. (Ionisation energy of Na = 5.14 eV, electron affinity of Cl = 3.61 eV.)

1. Forming ions: 5.14 − 3.61 = 1.53 eV must be supplied.
2. Coulomb energy at r = 0.236 nm: e²/(4πε₀r) = 1.44 eV nm ÷ 0.236 nm ≈ 6.10 eV released.
3. Net energy released ≈ 6.10 − 1.53 = 4.57 eV, so the NaCl molecule is stable (ignoring repulsion).

Q5. Explain Van der Waals bonding and hydrogen bonding with examples.

Van der Waals: in inert-gas atoms the electron cloud fluctuates, creating a momentary dipole that induces a dipole in the neighbour; the two attract. The energy varies as 1/r⁶ and is very small (about 0.01–0.1 eV), so solid argon melts at 84 K. Hydrogen bond: a hydrogen atom covalently bonded to an electronegative atom (O, N, F) is attracted to another such atom nearby. Energy about 0.1–0.5 eV. It explains the open structure of ice (less dense than water), the high boiling point of water and the double helix of DNA.

Exam tip

Use the value e²/(4πε₀) = 1.44 eV nm to do ionic-energy numericals quickly. Always sketch the U(r) curve and mark r₀ and the bond energy on it.

Quick summary

X-rays have wavelengths (about 0.1–10 Å) similar to the spacing between atoms in crystals, so crystals act as natural 3-D gratings for them. Bragg's law explains X-ray diffraction and lets us find crystal structure. This topic also covers production of X-rays, continuous and characteristic spectra, and Moseley's law.

Key points

• X-rays were discovered by W. C. Röntgen in 1895. They are produced when fast electrons strike a heavy-metal target (Coolidge tube).
• Continuous spectrum (Bremsstrahlung, braking radiation): electrons slow down near nuclei; it has a sharp short-wavelength limit λ_min that depends only on the accelerating voltage (Duane–Hunt law).
• Characteristic spectrum: sharp lines (Kα, Kβ, Lα…) when an inner-shell vacancy is filled by an outer electron; they depend on the target element.
• Moseley's law (1913): √ν of a characteristic line is proportional to (Z − b). It made atomic number, not atomic mass, the basis of the periodic table.
• Laue (1912) first showed X-ray diffraction by a crystal.
• Bragg's law (W. H. and W. L. Bragg, 1913): reflections from successive parallel planes add up when the path difference 2d sin θ equals a whole number of wavelengths.
• θ is the glancing angle (angle with the plane, not with the normal).
• Methods: Laue method (single crystal, white X-rays), rotating crystal method, powder (Debye–Scherrer) method (powder sample, monochromatic X-rays).
• Bragg's spectrometer measures θ for successive orders to find d or λ.
• Uses: crystal structure, medical radiography, detecting flaws in metals.

Important terms

• Glancing angle (θ) – angle between incident ray and crystal plane.
• Order of reflection (n) – 1, 2, 3, … for successive maxima.
• Hard and soft X-rays – short wavelength (high energy) and long wavelength (low energy).
• Diffraction (विवर्तन) – bending and spreading of waves by obstacles or openings.

Formulas

• Bragg's law: 2d sin θ = nλ
• Duane–Hunt: λ_min = hc/(eV); λ_min (Å) ≈ 12400/V (V in volts)
• Moseley: √ν = a(Z − b); for Kα, b ≈ 1 and ν = (3/4)Rc(Z − 1)²
• Cubic spacing: d = a/√(h² + k² + l²)

5 important questions

Q1. State Bragg's law and define the glancing angle.

When X-rays of wavelength λ fall on crystal planes of spacing d, strong reflection occurs when 2d sin θ = nλ, where n is an integer. θ, the glancing angle, is the angle between the incident X-ray and the crystal plane.

Q2. Why can an ordinary optical grating not diffract X-rays?

Diffraction is noticeable only when the spacing is comparable to the wavelength. X-ray wavelengths are about 1 Å, but ruled grating spacings are about 10⁴ Å. Crystal planes (spacing ~1 Å) are suitable.

Q3. Derive Bragg's law.

1. Parallel X-rays fall at glancing angle θ on parallel planes separated by d.
2. Ray 2 reflected from the lower plane travels extra distance compared with ray 1.
3. Drop perpendiculars from the point on the upper plane; extra path = 2d sin θ (d sin θ before and after reflection).
4. Constructive interference when path difference = nλ.
5. So 2d sin θ = nλ, n = 1, 2, 3 …
6. Since sin θ ≤ 1, nλ ≤ 2d — this limits the possible orders.

Q4. X-rays of wavelength 1.54 Å are reflected by planes with d = 2.0 Å. Find the glancing angle for first-order reflection and the highest possible order.

1. sin θ = nλ/(2d) = 1 × 1.54/(2 × 2.0) = 0.385.
2. θ = sin⁻¹(0.385) ≈ 22.6°.
3. Highest order: n ≤ 2d/λ = 4.0/1.54 = 2.6, so n_max = 2.

Q5. An X-ray tube works at 30 kV. Find the shortest wavelength emitted (h = 6.63 × 10⁻³⁴ J s, c = 3 × 10⁸ m/s, e = 1.6 × 10⁻¹⁹ C).

1. λ_min = hc/(eV).
2. hc = 6.63 × 10⁻³⁴ × 3 × 10⁸ = 1.989 × 10⁻²⁵ J m.
3. eV = 1.6 × 10⁻¹⁹ × 3 × 10⁴ = 4.8 × 10⁻¹⁵ J.
4. λ_min = 1.989 × 10⁻²⁵/4.8 × 10⁻¹⁵ ≈ 4.14 × 10⁻¹¹ m = 0.414 Å.
5. Check with shortcut: 12400/30000 ≈ 0.413 Å ✓.

Exam tip

Remember θ in Bragg's law is measured from the plane, not the normal. Draw the two-plane ray diagram clearly with the path difference marked — it carries most of the marks in the derivation.

Quick summary

Atoms in a crystal are not still; they vibrate about their lattice positions. These vibrations carry heat and decide the specific heat of a solid. This topic covers the classical Dulong–Petit law and its failure at low temperature, Einstein's quantum theory, Debye's theory with the T³ law, and the idea of phonons and the dispersion relation for a one-dimensional chain.

Key points

• Dulong–Petit law (classical): molar specific heat of a solid Cv = 3R ≈ 24.9 J mol⁻¹ K⁻¹ (about 6 cal mol⁻¹ K⁻¹), the same for all solids and independent of temperature.
• Experiments show Cv falls towards zero as T → 0 K. Classical physics cannot explain this.
• Einstein (1907): each atom is a 3-D quantum oscillator with one frequency ν_E; energy levels are quantised. Cv agrees with 3R at high T and falls at low T, but it falls exponentially, faster than experiment.
• Debye (1912): the solid is a continuous elastic medium with a range of frequencies up to a maximum ν_D (Debye frequency). The total number of modes is 3N.
• Debye's result: at high T, Cv → 3R; at low T, Cv ∝ T³ (Debye T³ law), which agrees well with experiment.
• Debye temperature θ_D = hν_D/k_B. Above θ_D the solid behaves classically.
• A quantum of lattice vibration energy is called a phonon, with energy hν = ħω.
• For a 1-D monatomic chain, ω = 2√(C/m) |sin(ka/2)|: frequency has an upper limit (cut-off).
• A diatomic chain has two branches: acoustic (atoms move together) and optical (atoms move opposite; can interact with light).

Important terms

• Specific heat at constant volume (Cv) – heat to raise 1 mole by 1 K at constant volume.
• Einstein temperature (θ_E) – hν_E/k_B.
• Debye frequency – highest vibration frequency allowed in the solid.
• Dispersion relation – relation between ω and wave vector k.
• Brillouin zone – range of k from −π/a to +π/a in a 1-D chain.

Formulas

• Dulong–Petit: Cv = 3R = 3N_A k_B ≈ 24.9 J mol⁻¹ K⁻¹
• Mean energy of a quantum oscillator: ε̄ = hν/(e^(hν/kT) − 1)
• Einstein: Cv = 3R (θ_E/T)² e^(θ_E/T)/(e^(θ_E/T) − 1)²
• Debye T³ law (T much less than θ_D): Cv = (12π⁴/5) R (T/θ_D)³ ≈ 1944 (T/θ_D)³ J mol⁻¹ K⁻¹
• θ_D = hν_D/k_B
• 1-D monatomic chain: ω = 2√(C/m) |sin(ka/2)|, ω_max = 2√(C/m)

5 important questions

Q1. State the Dulong–Petit law. Why does it fail?

Every solid element has a molar specific heat of about 3R (≈ 25 J mol⁻¹ K⁻¹). It is based on classical equipartition: each atom has 3 vibrational degrees of freedom, each with average energy k_BT, so U = 3RT and Cv = 3R. It fails at low temperatures (and for light, hard solids like diamond even at room temperature) because vibrational energy is quantised; low-frequency modes alone can be excited when kT is small.

Q2. Compare Einstein's and Debye's theories.

Q3. Derive Einstein's expression for specific heat (outline).

1. Mean energy of one oscillator, using Planck's result: ε̄ = hν/(e^(hν/kT) − 1).
2. For N atoms × 3 directions: U = 3N hν/(e^(hν/kT) − 1).
3. Differentiate: Cv = dU/dT = 3Nk (hν/kT)² e^(hν/kT)/(e^(hν/kT) − 1)².
4. Put x = θ_E/T. High T (x small): e^x ≈ 1 + x, so Cv → 3R (Dulong–Petit). Low T (x large): Cv ≈ 3R x² e^(−x) → 0.

Q4. The Debye temperature of copper is 343 K. Find its molar specific heat at 10 K.

Cv = 1944 (T/θ_D)³ = 1944 × (10/343)³ = 1944 × 2.478 × 10⁻⁵ ≈ 0.048 J mol⁻¹ K⁻¹. This is about 500 times smaller than 3R, showing the strong fall at low temperature.

Q5. Obtain the dispersion relation for a one-dimensional monatomic lattice.

1. Atoms of mass m, spacing a, joined by springs of constant C. Force on atom s: m d²u_s/dt² = C(u_(s+1) + u_(s−1) − 2u_s).
2. Try u_s = A e^(i(ksa − ωt)).
3. Substituting: −mω² = C(e^(ika) + e^(−ika) − 2) = −2C(1 − cos ka) = −4C sin²(ka/2).
4. So ω = 2√(C/m) |sin(ka/2)|. ω is maximum at k = π/a (edge of the Brillouin zone), so the lattice acts as a low-pass filter. For small k, ω ≈ a√(C/m) k: the speed of sound is constant.

Exam tip

For low-temperature numericals use 1944 (T/θ_D)³ directly. In the theory answer, always state the high-T and low-T limits — that is where the marks are.

Quick summary

Free electron theory treats the valence electrons of a metal as a gas of particles moving freely inside the metal. The classical (Drude–Lorentz) model explains Ohm's law and the Wiedemann–Franz law but fails for specific heat. The quantum (Sommerfeld) model uses Fermi–Dirac statistics and fixes these failures. This topic covers drift velocity, relaxation time, conductivity, mobility, the Fermi energy and the density of states.

Key points

• Drude–Lorentz model: free electrons move randomly like gas molecules and collide with the positive ions. In an electric field they gain a small drift velocity opposite to the field.
• Conductivity σ = ne²τ/m, where n is electron density and τ is relaxation time (mean time between collisions).
• Mobility μ = v_d/E = eτ/m; σ = neμ.
• Wiedemann–Franz law: the ratio of thermal to electrical conductivity is proportional to temperature, K/σ = LT (L = Lorenz number).
• Failures of the classical model: (1) predicts electronic specific heat (3/2)R per mole, but measured is about 100 times smaller; (2) wrong temperature dependence of resistivity (predicts ρ ∝ √T, measured ρ ∝ T); (3) cannot explain why some solids are insulators.
• Sommerfeld (quantum) model: electrons obey the Pauli exclusion principle and Fermi–Dirac statistics. At 0 K they fill energy levels up to the Fermi energy E_F.
• Only electrons within about k_BT of E_F can absorb heat, so the electronic specific heat is small and ∝ T.
• E_F for metals is a few eV (Cu ≈ 7.0 eV), much larger than k_BT at room temperature (0.025 eV).

Important terms

• Relaxation time (τ) – average time between collisions, about 10⁻¹⁴ s.
• Mean free path (λ) – average distance travelled between collisions.
• Fermi energy (E_F) – highest filled energy level at 0 K.
• Fermi–Dirac distribution – probability f(E) = 1/(e^((E − E_F)/kT) + 1) that a level E is occupied.
• Density of states g(E) – number of states per unit energy per unit volume; g(E) ∝ √E.

Formulas

• Current density J = nev_d = σE
• σ = ne²τ/m; resistivity ρ = 1/σ = m/(ne²τ)
• μ = eτ/m; σ = neμ
• Wiedemann–Franz: K/σT = L; quantum value L = (π²/3)(k_B/e)² = 2.44 × 10⁻⁸ W Ω K⁻²
• Fermi energy (0 K): E_F = (ħ²/2m)(3π²n)^(2/3)
• Fermi temperature T_F = E_F/k_B; Fermi velocity v_F = √(2E_F/m)
• Electronic specific heat (quantum): C_e = (π²/2) R (T/T_F) per mole of electrons

5 important questions

Q1. Define drift velocity and mobility.

Drift velocity is the small average velocity that free electrons acquire opposite to an applied electric field, on top of their random motion. Mobility is the drift velocity per unit electric field, μ = v_d/E, in m² V⁻¹ s⁻¹.

Q2. State the Fermi–Dirac distribution. What is its value at E = E_F?

f(E) = 1/[e^((E − E_F)/k_BT) + 1]. At E = E_F, f = 1/(1 + 1) = ½ at any temperature above 0 K. At 0 K, f = 1 for E below E_F and f = 0 above it.

Q3. Derive the expression for electrical conductivity in the Drude model.

1. Force on an electron: F = −eE, so acceleration a = −eE/m.
2. After a collision the electron starts with random velocity (average zero). In average time τ it gains v_d = −eEτ/m.
3. Current density J = −nev_d = (ne²τ/m) E.
4. Comparing with J = σE (Ohm's law): σ = ne²τ/m.

Q4. Copper has n = 8.5 × 10²⁸ m⁻³ and resistivity 1.7 × 10⁻⁸ Ω m. Find the relaxation time and the drift velocity for a current of 1 A in a wire of area 1 mm².

1. τ = m/(ne²ρ) = 9.11 × 10⁻³¹ / (8.5 × 10²⁸ × (1.6 × 10⁻¹⁹)² × 1.7 × 10⁻⁸) = 9.11 × 10⁻³¹/(3.70 × 10⁻¹⁷) ≈ 2.5 × 10⁻¹⁴ s.
2. v_d = I/(neA) = 1/(8.5 × 10²⁸ × 1.6 × 10⁻¹⁹ × 10⁻⁶) = 1/(1.36 × 10⁴) ≈ 7.4 × 10⁻⁵ m s⁻¹.

The drift is very slow, but the signal travels almost at the speed of light because the field sets all electrons moving at once.

Q5. Calculate the Fermi energy of copper (n = 8.5 × 10²⁸ m⁻³).

1. 3π²n = 3 × 9.87 × 8.5 × 10²⁸ = 2.52 × 10³⁰; (2.52 × 10³⁰)^(2/3) = 1.85 × 10²⁰ m⁻².
2. ħ²/2m = (1.055 × 10⁻³⁴)²/(2 × 9.11 × 10⁻³¹) = 6.11 × 10⁻³⁹ J m².
3. E_F = 6.11 × 10⁻³⁹ × 1.85 × 10²⁰ = 1.13 × 10⁻¹⁸ J = 7.0 eV. T_F = E_F/k_B ≈ 8.2 × 10⁴ K, far above room temperature.

Exam tip

Write the failures of the classical model as numbered points — a common 4-mark question. In numericals keep powers of ten in a separate line to avoid slips.

Quick summary

In a solid, the energy levels of many atoms spread into bands. The way electrons fill these bands, and the size of the gap between them, decides whether a solid is a conductor, semiconductor or insulator. This topic covers formation of bands, the Kronig–Penney idea, Fermi level, intrinsic and extrinsic semiconductors and the Hall effect.

Key points

• In an isolated atom, electrons have sharp energy levels. When N atoms come close, each level splits into N very close levels, forming an energy band.
• Valence band: the highest band filled (fully or partly) with electrons at 0 K. Conduction band: the next higher band.
• Forbidden gap (E_g): energy range between them with no allowed states.
• Conductors: bands overlap or the band is half-filled (metals) — many free electrons.
• Insulators: large gap (diamond, about 5–6 eV); electrons cannot jump at room temperature.
• Semiconductors: small gap (Si ≈ 1.1 eV, Ge ≈ 0.7 eV); some electrons cross at room temperature. Conductivity rises with temperature.
• The free-electron model could not explain why some solids are insulators. Kronig–Penney model (periodic potential) shows that allowed energy bands are separated by forbidden gaps.
• Intrinsic semiconductor: pure; electrons = holes (n = p = nᵢ).
• Extrinsic: doped. Pentavalent impurity (P, As, Sb) gives n-type (donors, electrons majority). Trivalent impurity (B, Al, Ga, In) gives p-type (acceptors, holes majority).
• Fermi level: in intrinsic, near the middle of the gap; moves towards the conduction band in n-type and towards the valence band in p-type.
• Hall effect: a current-carrying conductor in a perpendicular magnetic field develops a transverse voltage. The sign of the Hall voltage tells the carrier type; its size gives carrier density.

Important terms

• Hole – a vacancy left in the valence band, acting as a positive charge carrier.
• Fermi energy – highest occupied level at 0 K; at T > 0 the level with ½ probability of occupation.
• Mobility (μ) – drift velocity per unit electric field, m² V⁻¹ s⁻¹.
• Doping (अपमिश्रण) – adding small, controlled impurity.

Formulas

• Fermi–Dirac function: f(E) = 1/(1 + e^((E − E_F)/kT))
• Conductivity: σ = e(nμₑ + pμₕ); intrinsic σ = nᵢe(μₑ + μₕ)
• Mass action law: np = nᵢ²
• Hall voltage: V_H = BI/(net); Hall coefficient R_H = 1/(ne) (negative for electrons)
• Mobility from Hall: μ = σR_H

5 important questions

Q1. Distinguish between conductors, semiconductors and insulators on the basis of band theory.

Conductors: valence and conduction bands overlap (or partly filled band), no gap. Semiconductors: small gap (about 1 eV), few electrons cross at room temperature. Insulators: large gap (over about 3 eV), almost no electrons cross.

Q2. What is the Fermi level? Where does it lie in an intrinsic semiconductor?

The Fermi level is the energy at which the probability of finding an electron is ½ (at T > 0). In an intrinsic semiconductor it lies near the middle of the forbidden gap.

Q3. Explain the Hall effect and derive the expression for the Hall coefficient.

1. A slab carries current I along x; magnetic field B along z.
2. Carriers (charge q, drift velocity v) feel Lorentz force qvB along y and pile up on one face.
3. This builds a Hall field E_H; equilibrium when qE_H = qvB.
4. Current density J = nqv, so E_H = JB/(nq).
5. Hall coefficient R_H = E_H/(JB) = 1/(nq); for electrons R_H = −1/(ne).
6. With width w and thickness t: V_H = E_H w = BI/(nqt).

Q4. For germanium, nᵢ = 2.5 × 10¹⁹ m⁻³, μₑ = 0.39 and μₕ = 0.19 m² V⁻¹ s⁻¹. Find its intrinsic conductivity.

1. σ = nᵢe(μₑ + μₕ).
2. = 2.5 × 10¹⁹ × 1.6 × 10⁻¹⁹ × (0.39 + 0.19).
3. = 4.0 × 0.58 = 2.32 S/m.
4. Resistivity ρ = 1/σ ≈ 0.43 Ω m.

Q5. The Hall coefficient of a sample is 3.66 × 10⁻⁴ m³/C. Find the carrier concentration.

1. n = 1/(R_H e).
2. R_H e = 3.66 × 10⁻⁴ × 1.6 × 10⁻¹⁹ = 5.856 × 10⁻²³.
3. n = 1/(5.856 × 10⁻²³) ≈ 1.71 × 10²² m⁻³.
4. If R_H is positive, the carriers are holes (p-type).

Exam tip

Draw the three band diagrams (overlap, small gap, large gap) side by side. In Hall effect answers, always mention that the sign of V_H shows whether carriers are electrons or holes.

Quick summary

Some materials lose all electrical resistance when cooled below a certain critical temperature. This state is called superconductivity, discovered by Kamerlingh Onnes in 1911 in mercury at 4.2 K. This topic covers zero resistance, the Meissner effect, critical field, Type I and Type II superconductors, the isotope effect, an outline of BCS theory, high-Tc superconductors and applications.

Key points

• Below the critical temperature Tc, resistance drops sharply to zero. A current set up in a superconducting ring can flow for years without loss.
• Meissner effect (1933): a superconductor expels magnetic flux from its interior when cooled below Tc in a field. It behaves as a perfect diamagnet (B = 0 inside, susceptibility χ = −1).
• A strong enough magnetic field destroys superconductivity. The critical field Hc depends on temperature: Hc(T) = H₀[1 − (T/Tc)²].
• A large current also destroys it, because its own field exceeds Hc (Silsbee's rule).
• Type I (soft): one critical field; sudden, complete transition (Pb, Hg, Sn). Low Hc.
• Type II (hard): two critical fields Hc1 and Hc2; between them a mixed (vortex) state where flux enters as quantised tubes (Nb₃Sn, NbTi). High Hc2, so they are used in magnets.
• Isotope effect: Tc ∝ M^(−½), showing that lattice vibrations (phonons) are involved.
• BCS theory (Bardeen, Cooper, Schrieffer, 1957): two electrons with opposite momentum and spin attract each other through the lattice and form a Cooper pair. Pairs move without scattering. An energy gap of about 3.5 k_BTc opens at the Fermi level.
• High-Tc superconductors (1986 onwards): copper-oxide ceramics such as YBa₂Cu₃O₇ with Tc ≈ 92 K, above the boiling point of liquid nitrogen (77 K).
• Other properties: specific heat jumps at Tc; entropy falls (more ordered state); thermal conductivity changes.

Important terms

• Critical temperature (Tc) – temperature below which resistance is zero.
• Persistent current – current that keeps flowing in a superconducting loop without a source.
• Cooper pair – bound pair of electrons responsible for superconductivity.
• Flux quantum – smallest unit of trapped flux, Φ₀ = h/2e = 2.07 × 10⁻¹⁵ Wb.
• Josephson effect – tunnelling of Cooper pairs through a thin insulator between two superconductors.

Formulas

• Critical field: Hc(T) = H₀[1 − (T/Tc)²]
• Critical current of a wire of radius r: Ic = 2πr Hc
• Isotope effect: Tc M^α = constant, α ≈ ½
• BCS energy gap at 0 K: Eg = 2Δ ≈ 3.52 k_BTc
• Flux quantum Φ₀ = h/(2e)

5 important questions

Q1. What is the Meissner effect? Show that a superconductor is a perfect diamagnet.

When a superconductor is cooled below Tc in a magnetic field, the field lines are pushed out of it. Inside, B = μ₀(H + M) = 0, so M = −H and the susceptibility χ = M/H = −1. A susceptibility of −1 means perfect diamagnetism. This shows that superconductivity is more than zero resistance: a perfect conductor would only trap the existing flux, not expel it.

Q2. Distinguish between Type I and Type II superconductors.

Q3. The critical field of lead is 6.5 × 10⁴ A m⁻¹ at 0 K and its Tc = 7.2 K. Find the critical field at 5 K.

Hc = H₀[1 − (T/Tc)²] = 6.5 × 10⁴ × [1 − (5/7.2)²] = 6.5 × 10⁴ × (1 − 0.482) = 6.5 × 10⁴ × 0.518 ≈ 3.4 × 10⁴ A m⁻¹.

Q4. Explain the BCS theory of superconductivity (outline).

1. An electron passing through the lattice attracts nearby positive ions, creating a region of extra positive charge.
2. A second electron is attracted to this region. So, through a phonon, the two electrons attract each other.
3. Below Tc this attraction beats their Coulomb repulsion, and electrons near the Fermi surface with opposite momenta and spins bind into Cooper pairs.
4. All pairs share one quantum state. Breaking a pair needs the gap energy 2Δ, so ordinary small scatterings cannot slow the current: resistance is zero.
5. The theory explains the isotope effect (phonons depend on ion mass) and the energy gap.

Q5. Give the applications of superconductors.

• Superconducting magnets for MRI scanners, particle accelerators (LHC) and fusion research.
• Maglev trains that float using strong magnetic repulsion.
• SQUIDs (based on the Josephson effect) that measure extremely weak magnetic fields, such as those from the brain.
• Loss-free power cables and fault-current limiters.
• Fast switching elements and qubits in quantum computers.

Exam tip

Always say that zero resistance and the Meissner effect are the two defining properties. Draw the Hc–T parabola and the M–H curves for Type I and Type II; diagrams fetch easy marks.

Quick summary

Real optical instruments use thick lenses or combinations of lenses. Their behaviour can be described simply by six cardinal points. Real lenses also form imperfect images; these defects are called aberrations. This topic covers the cardinal points of a coaxial system, the equivalent focal length of two thin lenses, and the main types of aberration with methods to reduce them, including the achromatic doublet.

Key points

• A coaxial optical system is fully described by six cardinal points: two focal points (F₁, F₂), two principal points (P₁, P₂) and two nodal points (N₁, N₂).
• Principal planes: planes of unit positive lateral magnification; a ray entering at a height h on the first principal plane leaves from the same height on the second.
• Nodal points: a ray aimed at N₁ leaves from N₂ parallel to its original direction (unit angular magnification).
• If the same medium is on both sides of the system, the nodal points coincide with the principal points.
• Two thin lenses of focal lengths f₁ and f₂ separated by d act like one lens of focal length F, with 1/F = 1/f₁ + 1/f₂ − d/(f₁f₂).
• Aberrations are of two kinds: chromatic (because μ depends on wavelength) and monochromatic (because of the lens shape; Seidel aberrations).
• Five monochromatic aberrations: spherical aberration, coma, astigmatism, curvature of field, distortion.
• Spherical aberration: marginal rays focus nearer the lens than paraxial rays. Reduced by stops, by using two lenses separated by d = f₁ − f₂, or by a plano-convex lens with its curved side facing the parallel beam.
• Chromatic aberration: violet light focuses nearer the lens than red. Removed by an achromatic doublet (crown convex + flint concave lens in contact) satisfying ω₁/f₁ + ω₂/f₂ = 0.
• Two lenses of the same glass separated by d = (f₁ + f₂)/2 are achromatic for focal length (used in the Huygens eyepiece).

Important terms

• Equivalent lens – single thin lens giving the same magnification and image as the system.
• Dispersive power (ω) – ω = (μ_v − μ_r)/(μ_y − 1).
• Longitudinal chromatic aberration – distance between the violet and red focal points, = ωf.
• Coma – off-axis point appears as a comet-shaped blur.
• Astigmatism – off-axis point gives two line images at different distances.

Formulas

• Equivalent focal length: 1/F = 1/f₁ + 1/f₂ − d/(f₁f₂)
• Position of second principal point from the second lens: β = −Fd/f₁
• Position of first principal point from the first lens: α = Fd/f₂
• Achromatism, lenses in contact: ω₁/f₁ + ω₂/f₂ = 0
• Achromatism, separated lenses of the same glass: d = (f₁ + f₂)/2
• Minimum spherical aberration, two lenses: d = f₁ − f₂
• Longitudinal chromatic aberration: f_r − f_v = ωf_y

5 important questions

Q1. Define the six cardinal points of an optical system.

Focal points: F₁, where rays parallel on the image side meet the axis on the object side, and F₂, where an incident parallel beam converges. Principal points: axial points of the two principal planes, which are conjugate planes of unit lateral magnification. Nodal points: axial conjugate points with unit angular magnification; a ray towards N₁ emerges from N₂ parallel to itself.

Q2. Two thin convex lenses of focal lengths 20 cm and 10 cm are 5 cm apart. Find the equivalent focal length and the positions of the principal points.

1/F = 1/20 + 1/10 − 5/(20 × 10) = 0.05 + 0.10 − 0.025 = 0.125, so F = 8 cm.
α = Fd/f₂ = 8 × 5/10 = 4 cm (first principal point 4 cm to the right of the first lens).
β = −Fd/f₁ = −8 × 5/20 = −2 cm (second principal point 2 cm to the left of the second lens).

Q3. Derive the condition for achromatism of two thin lenses in contact.

1. Combined power: 1/F = 1/f₁ + 1/f₂.
2. For each lens, 1/f = (μ − 1)K, so d(1/f) = K dμ = (dμ/(μ − 1))(1/f) = ω/f.
3. For F to be the same for all colours, d(1/F) = 0: ω₁/f₁ + ω₂/f₂ = 0.
4. Since ω₁, ω₂ are positive, f₁ and f₂ must have opposite signs: a convex lens of crown glass (low ω) and a concave lens of flint glass (high ω). The combination is still converging because the convex lens is stronger.

Q4. Design an achromatic doublet of focal length 50 cm using crown (ω₁ = 0.015) and flint (ω₂ = 0.030) glass.

From ω₁/f₁ + ω₂/f₂ = 0: f₂ = −(ω₂/ω₁)f₁ = −2f₁.
From 1/f₁ + 1/f₂ = 1/50: 1/f₁ − 1/(2f₁) = 1/(2f₁) = 1/50, so f₁ = 25 cm (crown convex) and f₂ = −50 cm (flint concave).
Check: 0.015/25 + 0.030/(−50) = 0.0006 − 0.0006 = 0 ✔.

Q5. What is spherical aberration? How can it be reduced?

Rays from a point on the axis meeting the lens at different heights focus at different points: marginal rays come to focus closer than paraxial rays, so the image is a blurred disc. Methods: (1) use a stop to block marginal rays (dims the image); (2) use a plano-convex lens with the curved face towards the incident parallel light, so the deviation is shared equally at both surfaces; (3) use two lenses separated by d = f₁ − f₂; (4) use a crossed lens or an aspheric lens; (5) combine a convex and a suitable concave lens (aplanatic combination).

Exam tip

In cardinal-point numericals, keep the sign convention (distances measured left to right are positive). Always check an achromatic design by putting the values back into ω₁/f₁ + ω₂/f₂ = 0.

Quick summary

Interference is the redistribution of light energy when two coherent waves overlap, giving bright and dark fringes. This topic covers coherent sources, Young's double-slit experiment, interference in thin films and wedge-shaped films, Newton's rings and the Michelson interferometer.

Key points

• Principle of superposition: the resultant displacement is the sum of the individual displacements.
• Coherent sources have the same frequency and a constant phase difference. Two independent lamps are not coherent.
• Coherent sources are made by division of wavefront (Young's slits, Fresnel's biprism) or division of amplitude (thin films, Newton's rings, Michelson interferometer).
• Constructive interference: path difference = nλ (bright). Destructive: path difference = (2n + 1)λ/2 (dark).
• Energy is not destroyed; it is only redistributed.
• Thin film (soap bubble, oil on water): rays reflected from top and bottom surfaces interfere. Reflection from a denser medium adds an extra path of λ/2 (phase change π).
• Wedge-shaped film: straight, equally spaced fringes parallel to the edge; used to test flatness and to find thin wire diameter.
• Newton's rings: a plano-convex lens on a glass plate forms an air film; circular fringes with a dark centre in reflected light. Ring diameters give λ or R or the refractive index of a liquid.
• Michelson interferometer: splits a beam into two perpendicular arms; moving one mirror by λ/2 shifts one fringe. Used to measure λ, the difference of close wavelengths and thickness of thin sheets.

Important terms

• Fringe width (β) – distance between two successive bright (or dark) fringes.
• Optical path – μ × geometric path.
• Coherence (सुसंगतता) – constant phase relation between waves.
• Visibility – contrast of fringes, (I_max − I_min)/(I_max + I_min).

Formulas

• Young's experiment: β = λD/d (D = slit–screen distance, d = slit separation)
• Thin film (reflected light): 2μt cos r = nλ → dark; 2μt cos r = (2n − 1)λ/2 → bright
• Wedge: β = λ/(2μθ)
• Newton's rings (reflected, air film): Dₙ² = 4nλR (dark); Dₙ² = 2(2n − 1)λR (bright)
• λ from rings: λ = (Dₙ₊ₚ² − Dₙ²)/(4pR)
• Liquid of index μ: μ = (Dₙ₊ₚ² − Dₙ²)_air/(Dₙ₊ₚ² − Dₙ²)_liquid
• Michelson: 2d = Nλ (d = mirror shift, N = fringes crossed)

5 important questions

Q1. What are coherent sources? Why can two separate bulbs not produce interference?

Coherent sources emit waves of the same frequency with a constant phase difference. Two bulbs emit light from many atoms with random, rapidly changing phases, so the fringe pattern shifts too fast to be seen.

Q2. Why is the centre of Newton's rings dark in reflected light?

At the point of contact film thickness t = 0. The ray reflected from the lower glass surface (denser medium) gets an extra λ/2 path. So the path difference is λ/2 and destructive interference makes the centre dark.

Q3. Derive the expression for the diameter of dark rings in Newton's rings.

1. Lens of radius R on a plate; at distance r from contact, air film thickness t.
2. Geometry: r² = (2R − t)t ≈ 2Rt, so t = r²/(2R).
3. For normal incidence in air (μ = 1, cos r = 1), dark condition: 2t = nλ.
4. 2 × r²/(2R) = nλ → rₙ² = nλR.
5. With D = 2r: Dₙ² = 4nλR, so Dₙ ∝ √n — rings get closer as n increases.

Q4. In a Newton's rings experiment, the diameters of the 5th and 15th dark rings are 0.336 cm and 0.590 cm. If R = 100 cm, find λ.

1. λ = (D₁₅² − D₅²)/(4pR), with p = 10.
2. D₁₅² = 0.590² = 0.3481 cm²; D₅² = 0.336² = 0.1129 cm².
3. Difference = 0.2352 cm².
4. λ = 0.2352/(4 × 10 × 100) = 0.2352/4000 = 5.88 × 10⁻⁵ cm.
5. λ ≈ 5880 Å.

Q5. When the movable mirror of a Michelson interferometer is moved by 0.0589 mm, 200 fringes cross the field. Find λ.

1. 2d = Nλ → λ = 2d/N.
2. d = 0.0589 mm = 5.89 × 10⁻⁵ m.
3. λ = 2 × 5.89 × 10⁻⁵/200 = 5.89 × 10⁻⁷ m.
4. λ = 5890 Å (sodium light).

Exam tip

Remember the extra λ/2 for reflection at a denser medium — it decides whether the centre is dark or bright. In Newton's rings numericals, using the difference of squares (Dₙ₊ₚ² − Dₙ²) cancels errors at the centre.

Quick summary

Diffraction is the bending of light around the edges of obstacles or openings and its spreading into the geometrical shadow. It shows that light is a wave. This topic covers Fresnel and Fraunhofer diffraction, the single slit, the plane transmission grating, and resolving power with Rayleigh's criterion.

Key points

• Diffraction is noticeable when the size of the opening or obstacle is comparable to the wavelength.
• Fresnel diffraction: source and/or screen at finite distance; wavefronts are spherical or cylindrical; no lenses needed.
• Fraunhofer diffraction: source and screen effectively at infinity; plane wavefronts; achieved using lenses. Simpler to analyse.
• Fresnel's half-period zones explain the rectilinear propagation of light and the zone plate.
• Single slit (Fraunhofer): a bright central maximum with weaker secondary maxima on both sides. Minima at a sin θ = nλ.
• The central maximum is twice as wide as each secondary maximum; intensities of secondary maxima fall quickly.
• Plane transmission grating: many equally spaced parallel slits. Gives very sharp, bright principal maxima; used to measure wavelength.
• Grating element (a + b) = width of slit + width of opaque part = 1/N (N = lines per unit length).
• Rayleigh's criterion: two close images are just resolved when the central maximum of one falls on the first minimum of the other.
• Resolving power of a grating increases with the order and the total number of lines.

Important terms

• Diffraction (विवर्तन) – bending of waves around edges.
• Resolving power – ability of an instrument to show two close objects or wavelengths separately.
• Dispersive power of grating – dθ/dλ = n/((a + b) cos θ).
• Zone plate – plate with alternate zones blocked; acts like a lens.

Formulas

• Single slit minima: a sin θ = nλ (n = 1, 2, …); intensity I = I₀(sin α/α)², α = πa sin θ/λ
• Width of central maximum on screen: 2λD/a (small angles)
• Grating: (a + b) sin θ = nλ; maximum order n_max < (a + b)/λ
• Resolving power of grating: λ/dλ = nN (N = total lines illuminated)
• Telescope: limit of resolution θ = 1.22λ/D (D = aperture)

5 important questions

Q1. Distinguish between Fresnel and Fraunhofer diffraction.

In Fresnel diffraction the source and screen are at finite distances, wavefronts are spherical and no lens is used. In Fraunhofer diffraction they are effectively at infinity, wavefronts are plane and lenses are used; the analysis is simpler.

Q2. State Rayleigh's criterion of resolution.

Two point sources or two spectral lines are just resolved when the principal maximum of one coincides with the first minimum of the other.

Q3. Light of 6000 Å falls normally on a slit 0.2 mm wide. Find the angular position of the first minimum and the width of the central maximum on a screen 1 m away.

1. sin θ = λ/a = 6 × 10⁻⁷/(2 × 10⁻⁴) = 3 × 10⁻³, so θ ≈ 3 × 10⁻³ rad.
2. Distance of first minimum from centre: y = Dθ = 1 × 3 × 10⁻³ = 3 mm.
3. Width of central maximum = 2y = 6 mm.

Q4. A grating has 5000 lines per cm. Find the angle of first-order diffraction for λ = 5893 Å and the highest order visible.

1. (a + b) = 1/5000 cm = 2 × 10⁻⁴ cm = 2 × 10⁻⁶ m.
2. sin θ = nλ/(a + b) = 5.893 × 10⁻⁷/(2 × 10⁻⁶) = 0.2947.
3. θ = sin⁻¹(0.2947) ≈ 17.1°.
4. n_max < (a + b)/λ = 2 × 10⁻⁶/5.893 × 10⁻⁷ = 3.39, so the highest order is 3.

Q5. Explain diffraction at a plane transmission grating and find the lines needed to resolve the sodium D lines (5890 Å and 5896 Å) in the first order.

1. Parallel light falls normally; each slit sends secondary waves.
2. A lens focuses waves going at angle θ; path difference between corresponding points of adjacent slits = (a + b) sin θ.
3. Principal maxima when (a + b) sin θ = nλ; between them, very weak secondary maxima, so lines are sharp.
4. Resolving power needed: λ/dλ = 5893/6 ≈ 982.
5. nN ≥ 982; for n = 1, N ≥ 982 lines must be illuminated.

Exam tip

Diffraction minima in a single slit use a sin θ = nλ, while grating maxima use (a + b) sin θ = nλ — do not mix them. Always convert lines per cm into grating element in metres.