Electrostatics studies charges at rest, the forces between them and the electric field and potential they produce. Gauss's law relates the electric flux through a closed surface to the charge inside it. It makes it easy to find fields of symmetric charge distributions such as a line, a sheet or a sphere.
Key points
Coulomb's law: force between two point charges is proportional to the product of charges and inversely proportional to the square of the distance.
Electric field E = F/q₀ (force per unit positive test charge); unit N/C or V/m.
Electric potential V = work done per unit charge in bringing a test charge from infinity; E = −∇V.
Field lines start on positive and end on negative charges; they never cross.
Electric flux Φ = ∮E·dA through a surface.
Gauss's law: total flux through any closed surface = (charge enclosed)/ε₀. The surface chosen is called a Gaussian surface.
Gauss's law is true for any shape of surface, but it is useful only where symmetry makes E constant on the surface.
Inside a conductor in equilibrium, E = 0; any excess charge lies on its outer surface.
A charged spherical shell acts like a point charge outside it; inside it the field is zero.
Differential form: ∇·E = ρ/ε₀. With E = −∇V this gives Poisson's equation ∇²V = −ρ/ε₀, and in charge-free space Laplace's equation ∇²V = 0.
Electric dipole: equal and opposite charges ±q separated by 2a; dipole moment p = q × 2a, directed from −q to +q.
Important terms
Permittivity of free space ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻²; 1/(4πε₀) = 9 × 10⁹ N m² C⁻².
Surface charge density σ – charge per unit area; linear density λ – charge per unit length.
Equipotential surface – surface on which V is the same everywhere (समविभव पृष्ठ).
Gaussian surface – imaginary closed surface used to apply Gauss's law.
Formulas
Coulomb: F = (1/4πε₀) q₁q₂/r²
Point charge: E = q/(4πε₀r²), V = q/(4πε₀r)
Gauss's law: ∮E·dA = q_enc/ε₀
Infinite line charge: E = λ/(2πε₀r)
Infinite plane sheet: E = σ/(2ε₀); just outside a charged conductor: E = σ/ε₀
Spherical shell (radius R): E = q/(4πε₀r²) for r > R; E = 0 for r < R
Uniformly charged solid sphere, inside: E = qr/(4πε₀R³)
Dipole on axis: E = 2p/(4πε₀r³); on equator: E = p/(4πε₀r³) (for r ≫ a)
5 important questions
Q1. State Gauss's law in electrostatics.
The total electric flux through any closed surface equals 1/ε₀ times the total charge enclosed by that surface: ∮E·dA = q_enc/ε₀.
Q2. Two charges of 2 μC and 3 μC are 0.3 m apart in air. Find the force between them.
Q3. Using Gauss's law, find the electric field due to an infinite line charge.
Line with charge per unit length λ. Choose a coaxial cylinder of radius r and length l as Gaussian surface.
By symmetry E is radial and has the same magnitude on the curved surface.
Flux through the flat ends = 0 (E parallel to the ends).
Flux through curved surface = E × 2πrl.
Charge enclosed = λl. Gauss: E × 2πrl = λl/ε₀.
E = λ/(2πε₀r), so E ∝ 1/r.
Q4. Using Gauss's law, find the field of an infinite plane sheet of charge.
Sheet with surface charge density σ. Choose a cylinder (pill-box) crossing the sheet with end area A.
E is perpendicular to the sheet, pointing away on both sides; no flux through the curved surface.
Total flux = EA + EA = 2EA.
Charge enclosed = σA.
2EA = σA/ε₀ → E = σ/(2ε₀), independent of distance.
Q5. A point charge of 1 μC is at the centre of a cube of side 10 cm. Find the flux through the whole cube and through one face.
By Gauss's law, total flux = q/ε₀ = 10⁻⁶/(8.85 × 10⁻¹²).
Φ ≈ 1.13 × 10⁵ N m² C⁻¹ (the cube size does not matter).
By symmetry, each of the 6 faces gets one-sixth.
Flux through one face ≈ 1.13 × 10⁵/6 ≈ 1.88 × 10⁴ N m² C⁻¹.
Exam tip
In every Gauss's law derivation, draw the Gaussian surface, state the symmetry, split the flux into parts, and only then apply the law. Remember sheet σ/2ε₀ but conductor surface σ/ε₀.
Dielectrics are insulators that become polarised in an electric field. Placing a dielectric between the plates of a capacitor raises its capacitance. This topic covers polar and non-polar molecules, polarisation, the three electric vectors E, P and D, Gauss's law in a dielectric, dielectric constant and susceptibility, the capacitor with a dielectric slab, and the Clausius–Mossotti relation.
Key points
A dielectric has no free charges, but its charges shift slightly in an electric field, forming tiny dipoles.
Non-polar molecules (H₂, O₂, CO₂) have no permanent dipole; the field induces one. Polar molecules (H₂O, HCl, NH₃) have permanent dipoles that line up with the field.
Polarisation P = dipole moment per unit volume (C m⁻²). It produces bound surface charge σ_b = P·n̂ and bound volume charge ρ_b = −∇·P.
The field inside a dielectric is reduced: E = E₀/K, where K (or ε_r) is the dielectric constant.
Electric displacement D = ε₀E + P. Gauss's law in a dielectric uses only free charge: ∮D·dS = q_free, or ∇·D = ρ_free.
For a linear dielectric P = ε₀χE, so D = ε₀(1 + χ)E = εE, and K = 1 + χ.
Filling a capacitor completely with a dielectric multiplies its capacitance by K.
Clausius–Mossotti relation links the microscopic polarisability α with the measurable K.
Types of polarisation: electronic, ionic, orientational (only polar molecules; falls with temperature) and space-charge.
Dielectric strength: the largest field a dielectric can take before breakdown (air ≈ 3 × 10⁶ V m⁻¹).
Important terms
Electric susceptibility (χ) – ratio P/(ε₀E); χ = K − 1.
Polarisability (α) – induced dipole moment per unit local field, p = αE_loc.
Bound charge – charge that appears on a dielectric's surface because of polarisation; it cannot move freely.
Local field – actual field acting on a molecule, E_loc = E + P/(3ε₀) for cubic symmetry.
Dielectric loss – energy turned into heat in an alternating field.
Formulas
D = ε₀E + P = ε₀KE; P = ε₀χE = ε₀(K − 1)E
Bound charge density on the surface σ_b = P = σ_f(1 − 1/K)
Parallel-plate capacitor: C = Kε₀A/d
Partly filled (slab of thickness t): C = ε₀A/[d − t(1 − 1/K)]
Energy density u = ½ E·D = ½ε₀KE²
Clausius–Mossotti: (K − 1)/(K + 2) = Nα/(3ε₀)
5 important questions
Q1. Distinguish between polar and non-polar dielectrics.
Non-polar: the centres of positive and negative charge coincide, so there is no permanent dipole moment; dipoles are only induced by a field (O₂, N₂, benzene). Polar: the centres do not coincide, so molecules have permanent dipoles that are randomly oriented without a field and partly line up in a field (H₂O, NH₃). Polar materials usually have larger K and their K decreases with rising temperature.
Q2. A parallel-plate capacitor has A = 0.02 m² and d = 2 mm. Find C with air and with mica (K = 6) filling the gap.
C₀ = ε₀A/d = 8.85 × 10⁻¹² × 0.02/(2 × 10⁻³) = 8.85 × 10⁻¹¹ F (88.5 pF). With mica: C = KC₀ = 6 × 88.5 = 531 pF.
Q3. Derive the capacitance of a parallel-plate capacitor with a dielectric slab of thickness t.
Free charge density σ on the plates. In air, E₀ = σ/ε₀; inside the slab, E = σ/(Kε₀).
Potential difference V = E₀(d − t) + Et = (σ/ε₀)[(d − t) + t/K].
C = Q/V = σA/V = ε₀A/[d − t + t/K] = ε₀A/[d − t(1 − 1/K)].
Check: t = 0 gives ε₀A/d; t = d gives Kε₀A/d.
Q4. Derive the Clausius–Mossotti relation (outline).
Each molecule feels the local field E_loc = E + P/(3ε₀) (Lorentz field inside a spherical cavity).
With N molecules per unit volume, each of dipole moment αE_loc: P = Nα E_loc = Nα[E + P/(3ε₀)].
Network theorems simplify the analysis of electrical circuits containing many sources and resistors. This topic covers Ohm's law, Kirchhoff's laws, the superposition theorem, Thevenin's and Norton's theorems, the maximum power transfer theorem and star–delta conversion.
Key points
Ohm's law: V = IR at constant temperature.
Kirchhoff's current law (KCL): the algebraic sum of currents at a junction is zero (conservation of charge).
Kirchhoff's voltage law (KVL): the algebraic sum of EMFs and potential drops round any closed loop is zero (conservation of energy).
Ideal voltage source: zero internal resistance. Ideal current source: infinite internal resistance.
Superposition theorem: in a linear network with several sources, the current in any branch is the algebraic sum of the currents due to each source acting alone (other voltage sources shorted, current sources opened, internal resistances kept).
Thevenin's theorem: any linear two-terminal network can be replaced by a voltage source V_th in series with a resistance R_th.
V_th = open-circuit voltage across the terminals. R_th = resistance seen from the terminals with sources replaced by their internal resistances.
Norton's theorem: the same network can be replaced by a current source I_N in parallel with R_N (= R_th). I_N = short-circuit current.
Thevenin and Norton forms are related by V_th = I_N R_th.
Maximum power transfer theorem: a load gets maximum power when its resistance equals the Thevenin resistance of the source network. Efficiency at that point is only 50%.
Star–delta transformation simplifies networks that are neither series nor parallel.
Reciprocity theorem: in a linear passive network, interchanging the positions of a source and an ammeter does not change the ammeter reading.
Important terms
Linear network – parameters (R, L, C) do not change with voltage or current.
Node (संधी बिंदू) – point where two or more elements meet.
Mesh / loop – closed path in a circuit.
Load – the element that receives power from the network.
Formulas
V = IR; series R = R₁ + R₂ + …; parallel 1/R = 1/R₁ + 1/R₂ + …
Star to delta: R_AB = R_A + R_B + R_A R_B/R_C (and similarly)
5 important questions
Q1. State Thevenin's theorem.
Any linear two-terminal network of sources and resistances can be replaced by a single voltage source V_th (the open-circuit voltage across the terminals) in series with a resistance R_th (the resistance seen from the terminals with all sources replaced by their internal resistances).
Q2. State the maximum power transfer theorem.
A resistive load draws maximum power from a linear network when the load resistance equals the Thevenin (internal) resistance of the network: R_L = R_th. Then P_max = V_th²/(4R_th).
Q3. A 12 V battery (negligible internal resistance) is in series with 4 Ω, and a 12 Ω resistor is connected across the output terminals A–B. Find the Thevenin equivalent and the current in a 6 Ω load across A–B.
Check by direct method: total R = 4 + (12 ∥ 6) = 4 + 4 = 8 Ω; I = 1.5 A; voltage across A–B = 1.5 × 4 = 6 V; load current = 6/6 = 1 A ✓.
Q4. For the circuit of Q3, find the Norton equivalent and the maximum power that can be given to a load.
Short A–B: short-circuit current I_N = 12/4 = 3 A (12 Ω is shorted).
R_N = R_th = 3 Ω. Check: I_N R_N = 9 V = V_th ✓.
Maximum power when R_L = 3 Ω.
P_max = V_th²/(4R_th) = 81/12 = 6.75 W.
Q5. Explain the superposition theorem with the steps for applying it.
Applies only to linear, bilateral networks with more than one source.
Keep one source; replace other voltage sources by short circuits and current sources by open circuits (keeping internal resistances).
Find the current in the required branch due to this one source.
Repeat for each source in turn.
Add all the currents algebraically (with directions) to get the actual current.
It cannot be used directly for power, since power depends on I².
Exam tip
For Thevenin problems, draw three small diagrams: one for V_th (load removed), one for R_th (sources killed), and the final equivalent with the load. Always check your answer with Norton's form or a direct method.
Alternating current (AC) changes its direction periodically, usually as a sine wave. This topic covers mean and rms values, the behaviour of R, L and C in AC circuits, reactance and impedance, phasor diagrams, the series LCR circuit and its resonance, the quality factor, the parallel resonant circuit, and power and power factor. The j-operator (complex number) method is also used.
Key points
AC voltage: V = V₀ sin ωt, with ω = 2πf. In India the mains is 230 V (rms), 50 Hz.
rms value I_rms = I₀/√2 is the steady DC that would give the same heating. Mean value over a half cycle = 2I₀/π.
Pure R: current and voltage in phase.
Pure L: current lags voltage by 90°; inductive reactance X_L = ωL rises with frequency.
Pure C: current leads voltage by 90°; capacitive reactance X_C = 1/(ωC) falls with frequency.
Series LCR: impedance Z = √[R² + (X_L − X_C)²]; phase angle tan φ = (X_L − X_C)/R.
Series resonance at ω₀ = 1/√(LC): X_L = X_C, Z = R (minimum), current maximum. Used in radio tuning ("acceptor circuit").
Parallel resonance (LC tank): impedance is maximum and line current minimum ("rejector circuit"); dynamic resistance L/(CR).
Quality factor Q = ω₀L/R = (1/R)√(L/C): sharpness of resonance; the voltage across L or C at resonance is Q times the supply voltage.
Average power P = V_rms I_rms cos φ. cos φ is the power factor. A pure L or C draws no average power (wattless current).
In j-notation: Z = R + j(ωL − 1/ωC), where j = √(−1).
Important terms
Reactance – opposition to AC by an inductor or capacitor (ohm); stores energy instead of using it.
Impedance (Z) – total opposition to AC in a circuit.
Admittance (Y) – 1/Z.
Bandwidth – range of frequencies between the half-power points, Δf = f₀/Q.
Phasor – rotating vector representing a sinusoidal quantity.
In a circuit with only L or only C, the phase difference between voltage and current is 90°, so cos φ = 0 and average power P = V_rms I_rms cos φ = 0. The current flows but no energy is used on average; energy is stored in one quarter-cycle and returned in the next. Such current is called wattless current.
Q2. Find the rms and peak values for the Indian mains supply.
V_rms = 230 V, so V₀ = √2 × 230 ≈ 325 V. Mean value over a half cycle = 2V₀/π = 2 × 325/3.1416 ≈ 207 V.
Q3. Derive the impedance and the resonance condition of a series LCR circuit using phasors.
Same current I in all elements. Voltages: V_R = IR in phase with I; V_L = IX_L leading by 90°; V_C = IX_C lagging by 90°.
V_L and V_C are opposite, so the net reactive voltage is I(X_L − X_C), perpendicular to V_R.
V = √[V_R² + (V_L − V_C)²] = I√[R² + (X_L − X_C)²], so Z = √[R² + (X_L − X_C)²] and tan φ = (X_L − X_C)/R.
Z is least when X_L = X_C, i.e. ωL = 1/ωC → ω₀ = 1/√(LC). At resonance Z = R, φ = 0 and I = V/R is maximum.
Q4. A series circuit has R = 10 Ω, L = 0.1 H and C = 10 μF, connected to 100 V (rms). Find f₀, the current at resonance, Q and the voltage across L at resonance.
f₀ = 1/(2π√(0.1 × 10⁻⁵)) = 1/(2π × 10⁻³) ≈ 159 Hz
I₀ = V/R = 100/10 = 10 A (rms)
ω₀ = 1000 rad s⁻¹, Q = ω₀L/R = 1000 × 0.1/10 = 10
V_L = I X_L = 10 × 100 = 1000 V, i.e. Q times the supply. Such high voltages make resonance dangerous in power circuits.
Q5. A coil of resistance 30 Ω and inductance 0.127 H is connected to 230 V, 50 Hz mains. Find the impedance, current, power factor and power.
X_L = 2π × 50 × 0.127 ≈ 39.9 Ω ≈ 40 Ω
Z = √(30² + 40²) = 50 Ω
I = 230/50 = 4.6 A
cos φ = R/Z = 30/50 = 0.6 (current lags by 53°)
P = V I cos φ = 230 × 4.6 × 0.6 ≈ 635 W (check: I²R = 4.6² × 30 = 634.8 W ✔)
Exam tip
Draw the phasor diagram for every LCR question; it gives Z and φ in one step. Remember the series circuit is an acceptor (minimum Z) and the parallel circuit a rejector (maximum Z).
Magnetostatics deals with magnetic fields produced by steady currents. The Biot–Savart law gives the field of a small current element, and Ampère's circuital law gives fields of symmetric current distributions easily. This topic also covers the force on currents and charges in a magnetic field.
Key points
A steady current produces a magnetic field around it (Oersted, 1820).
Biot–Savart law: the field dB due to a current element I dl at distance r is proportional to I dl sin θ/r².
Direction of B round a straight wire: right-hand thumb rule.
Ampère's circuital law: the line integral of B round any closed path equals μ₀ times the current enclosed.
Ampère's law is useful for long straight wires, solenoids and toroids, where B is constant along the chosen path.
Inside a long solenoid the field is uniform and along its axis; outside it is nearly zero.
Lorentz force: a charge q moving with velocity v in fields E and B feels F = q(E + v × B). The magnetic part does no work.
A current-carrying conductor in a magnetic field feels F = I(l × B).
Two parallel wires with currents in the same direction attract; in opposite directions they repel. This defines the ampere.
Differential forms: ∇·B = 0 (no magnetic monopoles); ∇ × B = μ₀J.
Magnetic vector potential A: B = ∇ × A.
A current loop behaves like a magnetic dipole with moment m = NIA.
Important terms
Permeability of free space μ₀ = 4π × 10⁻⁷ T m A⁻¹ (H/m).
Magnetic field (चुंबकीय क्षेत्र) B – unit tesla (T); 1 T = 10⁴ gauss.
Solenoid – long coil of many turns; toroid – solenoid bent into a ring.
Magnetic dipole moment – NIA, unit A m².
Formulas
Biot–Savart: dB = (μ₀/4π) I dl sin θ/r²
Long straight wire: B = μ₀I/(2πr)
Centre of circular coil (N turns, radius R): B = μ₀NI/(2R)
On the axis of a coil: B = μ₀NIR²/[2(R² + x²)^(3/2)]
Ampère: ∮B·dl = μ₀I_enc
Long solenoid: B = μ₀nI (n = turns per metre); toroid: B = μ₀NI/(2πr)
Force per unit length between parallel wires: F/l = μ₀I₁I₂/(2πd)
Force on a moving charge: F = qvB sin θ; radius of circular path r = mv/(qB)
5 important questions
Q1. State Ampère's circuital law.
The line integral of the magnetic field B around any closed path equals μ₀ times the net current passing through the area bounded by the path: ∮B·dl = μ₀I_enc.
Q2. Find the magnetic field 5 cm from a long straight wire carrying 10 A.
Learn the four standard fields (straight wire, coil centre, coil axis, solenoid) with their figures. In Ampère's law derivations, always show which parts of the loop contribute zero and why.
A changing magnetic flux through a circuit produces an emf in it. This is electromagnetic induction, discovered by Faraday in 1831. This topic covers Faraday's and Lenz's laws, motional emf, the integral and differential forms of Faraday's law, self and mutual inductance, energy stored in a magnetic field, eddy currents, and growth and decay of current in an LR circuit.
Key points
Magnetic flux Φ = ∫B·dS (weber, Wb). Flux changes if B, the area or the angle changes.
Lenz's law: the induced current flows so as to oppose the change that causes it. The minus sign in Faraday's law expresses this; it follows from conservation of energy.
Motional emf: a rod of length l moving with speed v perpendicular to B has emf e = Blv.
Integral form: ∮E·dl = −(d/dt)∫B·dS. Using Stokes' theorem, differential form: ∇×E = −∂B/∂t. A changing B creates a non-conservative (curly) electric field.
Self-inductance L: a changing current in a coil induces emf in the same coil, e = −L dI/dt. Unit: henry (H).
Mutual inductance M: a changing current in one coil induces emf in a nearby coil, e₂ = −M dI₁/dt. M₁₂ = M₂₁ (reciprocity).
Energy stored in an inductor U = ½LI²; energy density in a magnetic field u = B²/(2μ₀).
Eddy currents are loops of current induced in bulk conductors. Uses: induction furnace, electromagnetic braking, damping in galvanometers, speedometers. Losses are cut by laminating transformer cores.
In an LR circuit, current grows as I = I₀(1 − e^(−t/τ)) and decays as I = I₀e^(−t/τ), with time constant τ = L/R.
Important terms
Henry – inductance of a coil in which 1 A s⁻¹ change of current induces 1 V.
Coefficient of coupling (k) – M = k√(L₁L₂), 0 ≤ k ≤ 1.
Time constant (τ = L/R) – time for current to reach 63% of its final value while growing.
Back emf – self-induced emf opposing the change of current.
Formulas
e = −N dΦ/dt; motional emf e = Blv
∮E·dl = −dΦ/dt; ∇×E = −∂B/∂t
Long solenoid: L = μ₀N²A/l = μ₀n²Al
Two coaxial solenoids: M = μ₀N₁N₂A/l
Energy U = ½LI²; u = B²/2μ₀
Series inductors with coupling: L = L₁ + L₂ ± 2M
LR growth: I = (E/R)(1 − e^(−Rt/L)); decay: I = I₀e^(−Rt/L)
5 important questions
Q1. State Faraday's laws and Lenz's law.
Faraday: (1) whenever the magnetic flux linked with a circuit changes, an emf is induced in it; (2) the emf equals the rate of change of flux linkage, e = −N dΦ/dt. Lenz: the induced current always opposes the change of flux that produces it. If a magnet's N-pole approaches a coil, the near face of the coil becomes an N-pole and repels it, so work must be done — this work appears as electrical energy.
Q2. A rod 0.5 m long moves at 4 m s⁻¹ perpendicular to a field of 0.2 T. Find the emf. If the circuit resistance is 2 Ω, find the current and the force needed to keep it moving.
e = Blv = 0.2 × 0.5 × 4 = 0.4 V; I = 0.4/2 = 0.2 A; F = BIl = 0.2 × 0.2 × 0.5 = 0.02 N. Check: mechanical power Fv = 0.08 W = electrical power I²R = 0.04 × 2 = 0.08 W ✔.
Q3. Derive the self-inductance of a long solenoid and the energy stored in it.
Field inside: B = μ₀nI, where n = N/l.
Flux linkage: NΦ = N × B × A = N μ₀(N/l)I A.
L = NΦ/I = μ₀N²A/l.
Work done against the back emf while the current rises from 0 to I: W = ∫ L I dI = ½LI².
Put L and I = Bl/(μ₀N): U = ½(μ₀N²A/l)(Bl/μ₀N)² = (B²/2μ₀)(Al). Dividing by volume Al gives u = B²/(2μ₀).
Q4. A solenoid has 1000 turns, length 50 cm and cross-section 10 cm². Find L and the energy stored when it carries 2 A.
L = μ₀N²A/l = 4π × 10⁻⁷ × 10⁶ × 10⁻³/0.5 = 4π × 10⁻⁴/0.5 ≈ 2.51 × 10⁻³ H (2.51 mH). U = ½LI² = ½ × 2.51 × 10⁻³ × 4 ≈ 5.0 × 10⁻³ J.
Q5. Derive the differential form of Faraday's law and explain its meaning.
Integral form: ∮_C E·dl = −(d/dt)∫_S B·dS.
Stokes' theorem on the left: ∮E·dl = ∫(∇×E)·dS.
For a fixed surface, the time derivative moves inside: ∫(∇×E)·dS = −∫(∂B/∂t)·dS.
This holds for every surface, so ∇×E = −∂B/∂t.
Meaning: a time-varying magnetic field produces an electric field whose lines form closed loops. Unlike the electrostatic field, this field has non-zero curl and cannot be written as −∇V alone. It is one of Maxwell's four equations.
Exam tip
Lenz's law questions need a clear statement of which way the induced current flows and why (opposing the change, not the field). In numericals, check your answer with energy conservation where possible.
Maxwell brought together the laws of electricity and magnetism into four equations and added the displacement current. From these equations he predicted electromagnetic (EM) waves travelling at the speed of light, showing that light itself is an EM wave. This topic covers displacement current, the four Maxwell equations in differential and integral form, the wave equation, properties of EM waves, the Poynting vector and the EM spectrum.
Key points
Ampere's law ∇×B = μ₀J fails for changing fields: taking the divergence gives ∇·J = 0, which contradicts the continuity equation when charge density changes (for example in a charging capacitor).
Maxwell added the displacement current density J_d = ε₀ ∂E/∂t. It is a changing electric field, not a flow of charge, but it produces a magnetic field just like a real current.
∇·E = ρ/ε₀ (Gauss's law)
∇·B = 0 (no magnetic monopoles)
∇×E = −∂B/∂t (Faraday's law)
∇×B = μ₀J + μ₀ε₀ ∂E/∂t (Ampere–Maxwell law)
The four Maxwell equations (free space with charges and currents):
In empty space (ρ = 0, J = 0) the equations give wave equations for E and B with speed c = 1/√(μ₀ε₀) = 3 × 10⁸ m s⁻¹.
EM waves are transverse: E and B are perpendicular to each other and to the direction of travel.
E and B are in phase, and E₀ = cB₀.
EM waves need no medium and carry energy and momentum. The energy flow per unit area per second is given by the Poynting vector S = (1/μ₀)(E × B).
The EM spectrum, in order of increasing frequency: radio, microwaves, infrared, visible, ultraviolet, X-rays, gamma rays. All travel at c in vacuum.
Hertz produced and detected EM waves in 1887, confirming Maxwell's prediction.
Important terms
Displacement current – term ε₀ dΦ_E/dt that completes Ampere's law.
Poynting vector – vector giving the direction and rate of energy flow in an EM wave (W m⁻²).
Intensity – time-average of |S|, I = ½cε₀E₀².
Radiation pressure – pressure exerted by EM waves, I/c for full absorption.
Impedance of free space – E₀/H₀ = √(μ₀/ε₀) ≈ 377 Ω.
Q1. What is displacement current? Why did Maxwell introduce it?
It is the term ε₀ dΦ_E/dt (or density ε₀∂E/∂t) added to the conduction current in Ampere's law. Between the plates of a charging capacitor there is no conduction current, but a magnetic field is still observed; the changing electric field there acts as a current. Without it, Ampere's law would contradict the conservation of charge (continuity equation).
Q2. Find the speed of EM waves in vacuum from μ₀ and ε₀.
c = 1/√(μ₀ε₀) = 1/√(4π × 10⁻⁷ × 8.854 × 10⁻¹²) = 1/√(1.113 × 10⁻¹⁷) = 1/(3.336 × 10⁻⁹) ≈ 3.0 × 10⁸ m s⁻¹ — exactly the measured speed of light.
Q3. Derive the wave equation for E in free space from Maxwell's equations.
Take the curl of ∇×E = −∂B/∂t: ∇×(∇×E) = −∂(∇×B)/∂t.
With J = 0: ∇×B = μ₀ε₀ ∂E/∂t, so the right side is −μ₀ε₀ ∂²E/∂t².
Use the identity ∇×(∇×E) = ∇(∇·E) − ∇²E, and ∇·E = 0 in free space.
So −∇²E = −μ₀ε₀ ∂²E/∂t², i.e. ∇²E = μ₀ε₀ ∂²E/∂t².
Comparing with the standard wave equation ∇²ψ = (1/v²)∂²ψ/∂t² gives v = 1/√(μ₀ε₀) = c. The same steps give the equation for B.
Q4. Show that EM waves are transverse.
For a plane wave E = E₀ e^(i(kx − ωt)) travelling along x, ∇·E = 0 gives ik E₀x = 0, so E₀x = 0: E has no component along the direction of travel. Similarly ∇·B = 0 gives B₀x = 0. From ∇×E = −∂B/∂t, k × E₀ = ωB₀, so B is perpendicular to both k and E, with B₀ = E₀/c.
Q5. Sunlight reaching the Earth has intensity 1400 W m⁻². Find the peak electric and magnetic fields.
Write all four Maxwell equations with their names in both forms — it is the most common 5-mark question. Mention that the wave speed comes out equal to c: this was the evidence that light is an EM wave.
All materials respond to a magnetic field, but in different ways. Based on this response they are classified as diamagnetic, paramagnetic and ferromagnetic (with antiferromagnetic and ferrimagnetic as special cases). This topic covers B, H and M, susceptibility and permeability, Curie's law, domain theory and the hysteresis loop.
Key points
Magnetising field H: field due to free (conduction) currents, unit A/m.
Magnetisation M: magnetic moment per unit volume of the material, unit A/m.
Diamagnetic (bismuth, copper, water, gold): atoms have no permanent moment; induced moment opposes the field; χ small and negative; independent of temperature; weakly repelled. A superconductor is a perfect diamagnet (χ = −1).
Paramagnetic (aluminium, platinum, oxygen): atoms have permanent moments, randomly oriented; they partly line up in a field; χ small and positive; weakly attracted.
Curie's law for paramagnets: χ = C/T.
Ferromagnetic (iron, cobalt, nickel): χ large and positive; strongly attracted; shows hysteresis.
Domain theory: a ferromagnet contains small regions (domains) that are fully magnetised. In an unmagnetised sample domains point randomly. A field grows favourable domains (wall movement) and rotates them.
Above the Curie temperature a ferromagnet becomes paramagnetic, obeying the Curie–Weiss law χ = C/(T − θ). For iron the Curie temperature is about 770 °C.
Antiferromagnetic (MnO): neighbouring moments are equal and opposite. Ferrimagnetic (ferrites): opposite moments of unequal size give net magnetisation.
Hysteresis: B lags behind H when a ferromagnet is taken through a cycle. Area of the loop = energy lost per cycle per unit volume.
Soft iron: narrow loop, low coercivity, low loss — used in transformer cores and electromagnets. Steel/alnico: wide loop, high retentivity and coercivity — used for permanent magnets.
Important terms
Retentivity (remanence) – B left when H is reduced to zero.
Coercivity – reverse H needed to make B zero.
Magnetic domain (चुंबकीय क्षेत्रक) – small region with all atomic moments aligned.
Curie temperature – temperature above which ferromagnetism disappears.
Formulas
B = μ₀(H + M) = μ₀μᵣH
χ = M/H; μᵣ = 1 + χ
Curie's law: χ = C/T
Curie–Weiss law: χ = C/(T − θ), for T > θ
Energy loss per cycle = (area of B–H loop) per unit volume
5 important questions
Q1. Distinguish between diamagnetic and paramagnetic substances.
Diamagnetic substances have no permanent atomic moment, small negative χ independent of temperature, and are weakly repelled. Paramagnetic substances have permanent atomic moments, small positive χ that decreases with temperature (χ = C/T), and are weakly attracted.
Q2. What is the Curie temperature?
It is the temperature above which a ferromagnetic material loses its ferromagnetism and becomes paramagnetic, because thermal agitation destroys the alignment within domains. For iron it is about 770 °C.
Q3. Explain the hysteresis loop of a ferromagnetic material.
Start with an unmagnetised sample; increasing H raises B along the magnetisation curve to saturation.
Reduce H to zero: B does not fall to zero — the remaining value is the retentivity.
Apply reverse H: B becomes zero at the coercivity value.
Further reverse H gives reverse saturation; bringing H back completes a closed loop.
The area of the loop gives energy lost as heat per cycle per unit volume.
Soft iron gives a thin loop (cores); steel gives a fat loop (permanent magnets).
Q4. An iron rod with μᵣ = 500 is placed in a field H = 1000 A/m. Find B, M and χ.
Q5. The susceptibility of a paramagnetic salt is 2.8 × 10⁻⁴ at 300 K. Find it at 350 K.
Curie's law: χ ∝ 1/T, so χ₁T₁ = χ₂T₂.
χ₂ = χ₁T₁/T₂ = 2.8 × 10⁻⁴ × 300/350.
χ₂ = 2.8 × 10⁻⁴ × 0.857 ≈ 2.4 × 10⁻⁴.
Susceptibility falls as temperature rises.
Exam tip
Make a comparison table of dia-, para- and ferromagnetism (sign of χ, size, temperature dependence, examples) — it fits many short and long questions. Draw the hysteresis loop with retentivity and coercivity labelled.
A p-n junction diode allows current mainly in one direction. This one-way property is used to convert AC into DC (rectification). This topic covers the formation of a p-n junction, its characteristics, the Zener diode as a voltage regulator, half-wave and full-wave rectifiers, and filter circuits.
Key points
When p-type and n-type regions are joined, electrons and holes diffuse across and recombine, leaving a depletion layer of fixed ions with an internal field.
This creates a barrier potential: about 0.7 V for silicon and 0.3 V for germanium.
Forward bias (p to +, n to −) lowers the barrier; current rises sharply after the knee (cut-in) voltage.
Reverse bias widens the depletion layer; only a tiny reverse saturation current flows due to minority carriers.
At a large reverse voltage, breakdown occurs: Zener breakdown (heavily doped, strong field pulls electrons from bonds, low voltage) or avalanche breakdown (lightly doped, carriers ionise atoms by collision, higher voltage).
A Zener diode works in reverse breakdown, keeping its voltage nearly constant — used as a voltage regulator.
Half-wave rectifier: one diode; conducts only in one half-cycle.
Full-wave centre-tap rectifier: two diodes and a centre-tapped transformer; both halves used.
Bridge rectifier: four diodes, no centre tap needed; lower PIV per diode than centre-tap.
Ripple: AC part left in the rectified output. Filters (capacitor, choke, π-filter) smooth it.
Special diodes: LED (light emitting), photodiode (light sensing, reverse biased), solar cell, varactor.
Important terms
Knee voltage – forward voltage above which current increases rapidly.
PIV (peak inverse voltage) – maximum reverse voltage a diode faces in a circuit.
Ripple factor (γ) – rms value of AC component ÷ DC value.
Rectifier (दिष्टकारी) – circuit converting AC to DC.
Formulas
Diode equation: I = I₀(e^(V/ηV_T) − 1), V_T = kT/e ≈ 26 mV at room temperature
It is the region near the p-n junction that has no free charge carriers, only fixed positive ions (n side) and negative ions (p side). It forms because carriers diffuse across and recombine, and it creates the barrier potential.
Q2. Distinguish between Zener and avalanche breakdown.
Zener breakdown occurs in heavily doped junctions at low reverse voltage (below about 5–6 V); the strong field breaks covalent bonds directly. Avalanche breakdown occurs in lightly doped junctions at higher voltage, where accelerated carriers knock out more carriers by collision.
Q3. Derive the efficiency of a half-wave rectifier.
Output current i = I_m sin ωt for 0 to π, zero for π to 2π.
I_dc = (1/2π) ∫₀^π I_m sin θ dθ = I_m/π.
I_rms = √[(1/2π) ∫₀^π I_m² sin²θ dθ] = I_m/2.
DC output power = I_dc² R_L = I_m²R_L/π². AC input power = I_rms²(R_L + r_f) = (I_m²/4)(R_L + r_f).
Q5. A Zener regulator has V_in = 15 V, V_z = 10 V, R_s = 100 Ω and R_L = 1 kΩ. Find the currents.
Current through R_s: I_s = (15 − 10)/100 = 0.05 A = 50 mA.
Load current: I_L = 10/1000 = 0.01 A = 10 mA.
Zener current: I_z = I_s − I_L = 50 − 10 = 40 mA.
If the load changes, I_z adjusts so the output stays at 10 V.
Exam tip
Draw rectifier circuits with input and output waveforms; examiners give most marks for neat diagrams. Memorise the comparison table: V_dc, efficiency, ripple factor and PIV for half-wave, centre-tap and bridge.
Optoelectronic and Special Diodes: LED, Photodiode, Solar Cell, Varactor
Quick summary
Besides ordinary rectifier diodes, many p-n junction devices are designed for special jobs: giving out light, sensing light, turning sunlight into electricity, or acting as a voltage-controlled capacitor. This topic covers the LED, photodiode, solar cell, varactor diode and tunnel diode — their principle, construction, characteristics and uses. (The Zener diode and rectifiers are covered in the previous topic.)
Key points
LED (light emitting diode): a forward-biased p-n junction of a direct band-gap semiconductor (GaAs, GaAsP, GaN). Electrons and holes recombine at the junction and the energy comes out as light (electroluminescence).
The colour of an LED depends on the band gap: λ = hc/Eg. GaAs (1.43 eV) gives infrared; GaAsP gives red to yellow; GaN gives blue.
Silicon and germanium have an indirect band gap, so they give heat, not light, on recombination.
Photodiode: works in reverse bias. Light falling on the junction creates electron–hole pairs, increasing the reverse current in proportion to light intensity. Fast response; used in light meters, optical-fibre receivers, smoke detectors.
Solar cell: an unbiased p-n junction with a large area. Light creates electron–hole pairs; the junction field separates them, giving an open-circuit voltage (about 0.5–0.6 V for Si) and a short-circuit current. It works in the fourth quadrant of the I–V graph.
Solar cell figures of merit: fill factor FF = (V_m I_m)/(V_oc I_sc) and efficiency η = (V_m I_m)/(P_in).
Varactor (varicap): a reverse-biased diode used as a variable capacitor. Junction capacitance falls as reverse voltage rises because the depletion layer widens. Used in TV/radio tuners and FM modulators.
Tunnel diode: very heavily doped junction with a very thin depletion layer. Electrons tunnel through it, giving a negative resistance region. Used in high-frequency oscillators.
Important terms
Electroluminescence – emission of light when current passes through a material.
Direct band gap – conduction band minimum and valence band maximum at the same momentum; good for light emission.
Dark current – small reverse current of a photodiode with no light.
Open-circuit voltage (V_oc) – voltage across a solar cell when no current is drawn.
Negative resistance – region where current falls as voltage rises.
Formulas
LED wavelength λ = hc/Eg; λ (nm) ≈ 1240/Eg (eV)
Photodiode current I = I_dark + I_photo, with I_photo ∝ light intensity
Solar cell: FF = V_m I_m/(V_oc I_sc); η = V_m I_m/P_in × 100%
Junction (depletion) capacitance C = εA/W; for an abrupt junction C ∝ 1/√(V_R + V_bi)
5 important questions
Q1. Why is a photodiode operated in reverse bias?
In reverse bias the dark current is very small, so the extra current caused by light is easy to measure. The reverse current is almost proportional to light intensity, and the wide depletion layer collects the light-generated carriers quickly. In forward bias the large diffusion current would swamp the small photocurrent.
Q2. An LED is made of a material with band gap 1.9 eV. Find the wavelength and colour of the light.
λ = 1240/Eg = 1240/1.9 ≈ 653 nm, which is red light.
Q3. Explain the construction and working of a solar cell with its I–V characteristic.
Construction: a thin n-layer on a p-type silicon wafer, with a large junction area facing the light; metal fingers on top and a full metal contact at the back; an anti-reflection coating.
Working: photons with energy greater than Eg create electron–hole pairs near the junction. The built-in field pushes electrons to the n-side and holes to the p-side, so the n-side becomes negative and the p-side positive: an emf appears.
Characteristic: drawn in the fourth quadrant, current I_sc at V = 0 and voltage V_oc at I = 0. The maximum power point (V_m, I_m) is the corner of the largest rectangle under the curve.
Uses: satellites, solar panels for homes, calculators, street lights.
Q4. A solar cell has V_oc = 0.6 V, I_sc = 3 A, V_m = 0.5 V and I_m = 2.8 A. Light of 1000 W m⁻² falls on its area of 100 cm². Find the fill factor and efficiency.
FF = (0.5 × 2.8)/(0.6 × 3) = 1.4/1.8 ≈ 0.78
P_in = 1000 × 100 × 10⁻⁴ = 10 W
η = 1.4/10 × 100 = 14%
Q5. What is a varactor diode? Explain how its capacitance changes.
A varactor is a p-n junction diode designed to be used as a voltage-controlled capacitor in reverse bias. The depletion layer acts as the dielectric and the p and n regions as plates. Increasing the reverse voltage widens the depletion layer W, so C = εA/W falls. For an abrupt junction, C ∝ (V_R + V_bi)^(−½). Uses: electronic tuning in radios and TVs, frequency multipliers, automatic frequency control.
Exam tip
Learn λ (nm) = 1240/Eg (eV) — it makes LED numericals one-liners. Draw the I–V graphs: LED in the first quadrant, photodiode in the third, solar cell in the fourth.
A bipolar junction transistor (BJT) has three regions — emitter, base and collector — forming two p-n junctions. A small base current controls a much larger collector current, so a transistor can amplify signals or act as a switch. This topic covers transistor action, configurations, characteristics, α and β, biasing, load line and the CE amplifier.
Key points
Types: n-p-n and p-n-p. The emitter is heavily doped, the base is thin and lightly doped, and the collector is moderately doped and largest.
For normal (active) operation: emitter–base junction forward biased, collector–base junction reverse biased.
Transistor action (n-p-n): electrons from the emitter enter the thin base; very few recombine; most are swept into the collector. So I_C is almost equal to I_E.
Current relation: I_E = I_B + I_C.
Three configurations: common base (CB), common emitter (CE) and common collector (CC).
CE is the most used amplifier: high current, voltage and power gain; output is 180° out of phase with the input.
CB: low input and high output resistance; current gain slightly less than 1. CC (emitter follower): voltage gain about 1; high input and low output resistance; used for impedance matching.
Regions of operation: active (amplifier), saturation (switch ON), cut-off (switch OFF).
Biasing fixes a stable operating point (Q-point). Fixed bias is simple but unstable with temperature; voltage-divider bias gives good stability.
DC load line: straight line on output characteristics joining (V_CC, 0) and (0, V_CC/R_C). The Q-point should lie near its middle for undistorted amplification.
h-parameters model a transistor for small signals: h_ie (input resistance), h_fe (current gain), h_re, h_oe.
Important terms
α (current gain in CB) = I_C/I_E, slightly less than 1.
β (current gain in CE) = I_C/I_B, typically 20–300.
Q-point (कार्यबिंदू) – DC operating values (I_C, V_CE).
Thermal runaway – rise in temperature increases I_C, which further heats the transistor.
CE voltage gain (approx.): A_v = −β R_L/r_in (minus sign = phase reversal)
5 important questions
Q1. Why is the base of a transistor made thin and lightly doped?
So that very few carriers coming from the emitter recombine in the base. Almost all of them reach the collector, giving a large collector current and high current gain.
Q2. If α = 0.98, find β. If I_B = 50 μA, find I_C and I_E for β = 100.
V_CE is well above saturation (~0.2 V), so the transistor is in the active region.
Q5. Explain the working of a CE amplifier.
Voltage-divider bias sets the Q-point; coupling capacitors pass the AC signal; emitter bypass capacitor keeps gain high.
Input signal adds to the base-emitter bias, changing I_B slightly.
I_C changes β times more, so the drop across R_C changes a lot.
Output V_CE = V_CC − I_C R_C: when input rises, I_C rises and output falls — 180° phase reversal.
Gives high voltage and power gain; frequency response curve shows mid-band gain with fall at low and high frequencies.
Exam tip
Draw input and output characteristics for CE with the active, saturation and cut-off regions marked. In biasing numericals, always check V_CE at the end to confirm the active region.
An operational amplifier (op-amp) is a high-gain, direct-coupled differential amplifier made as an integrated circuit (the 741 is the classic example). With a few external resistors and capacitors it can amplify, add, subtract, integrate and differentiate signals. This topic covers the ideal op-amp, the concepts of virtual ground and CMRR, the inverting and non-inverting amplifiers, the voltage follower, the summing amplifier, the integrator and the differentiator.
Key points
An op-amp has two inputs — inverting (−) and non-inverting (+) — and one output. Output V_o = A(V₂ − V₁), where A is the open-loop gain.
Ideal op-amp: infinite open-loop gain, infinite input impedance, zero output impedance, infinite bandwidth, infinite CMRR, zero offset.
Real 741: A ≈ 2 × 10⁵, input impedance about 2 MΩ, output impedance about 75 Ω, slew rate 0.5 V/μs.
With negative feedback, the two inputs are held at almost the same voltage. If the + input is grounded, the − input is at 0 V without being connected to ground: a virtual ground.
No current enters the input terminals (infinite input impedance).
Inverting amplifier: gain = −R_f/R₁ (output is 180° out of phase).
Non-inverting amplifier: gain = 1 + R_f/R₁ (in phase).
Voltage follower: gain 1, very high input impedance; used as a buffer.
Summing amplifier adds input voltages; used in audio mixers and D/A converters.
Integrator (capacitor in the feedback path) gives output proportional to the integral of the input; a square wave input gives a triangular output.
Differentiator (capacitor at the input) gives output proportional to the rate of change of the input.
Important terms
CMRR – common-mode rejection ratio, A_d/A_c; how well the op-amp rejects signals common to both inputs (often in dB).
Slew rate – maximum rate of change of the output voltage.
Offset voltage – small input voltage needed to make the output zero.
Closed-loop gain – gain with feedback connected.
Virtual ground – a point at ground potential that carries no current to ground.
Infinite open-loop voltage gain; infinite input impedance (no input current); zero output impedance; infinite bandwidth; infinite CMRR; zero offset voltage and drift; infinite slew rate.
Q2. An inverting amplifier has R₁ = 10 kΩ and R_f = 100 kΩ. Find the gain and the output for an input of 0.2 V.
Q5. Explain the op-amp integrator and find its output for a constant input.
A resistor R carries the input to the − input (virtual ground); a capacitor C is in the feedback path.
Current I = V_in/R flows through R and charges C: I = −C dV_o/dt.
So dV_o/dt = −V_in/(RC), giving V_o = −(1/RC)∫V_in dt.
Example: V_in = +1 V constant, R = 100 kΩ, C = 1 μF (RC = 0.1 s): the output falls at 10 V per second, i.e. a ramp, reaching −5 V after 0.5 s (unless it saturates first). A square-wave input therefore gives a triangular output. A large resistor across C is used in practice to stop drift to saturation.
Exam tip
Always start op-amp derivations with the two golden rules: no current into the inputs, and the inputs are at the same voltage. Draw each circuit neatly — diagrams carry half the marks.
Around 1900, classical physics failed to explain several experiments involving light and atoms. New ideas — energy quanta, photons, matter waves and the uncertainty principle — led to quantum mechanics. This topic covers blackbody radiation (briefly), the photoelectric effect, the Compton effect, de Broglie waves, the Davisson–Germer experiment and Heisenberg's uncertainty principle.
Key points
Failures of classical physics: blackbody spectrum (ultraviolet catastrophe), photoelectric effect, Compton effect, line spectra and stability of atoms, specific heat of solids at low temperature.
Planck (1900): energy is emitted and absorbed in quanta E = hν.
Photoelectric effect: light ejects electrons from a metal. Observations: no emission below a threshold frequency; maximum KE depends on frequency, not intensity; number of electrons depends on intensity; emission is instantaneous.
Einstein (1905) explained it: light consists of photons of energy hν; one photon ejects one electron: hν = φ + KE_max.
Compton effect (1923): X-rays scattered by electrons have a longer wavelength. Explained by treating the photon as a particle with momentum h/λ colliding elastically with an electron.
de Broglie hypothesis (1924): every moving particle has a wave of wavelength λ = h/p.
Davisson–Germer experiment (1927): electrons scattered from a nickel crystal showed diffraction, confirming matter waves (strong peak for 54 eV electrons at a 50° scattering angle).
Wave–particle duality: light and matter show both wave and particle nature, but not in the same experiment.
Heisenberg's uncertainty principle (1927): position and momentum cannot both be known exactly at the same time.
Consequences: an electron cannot exist inside a nucleus; a confined particle has non-zero minimum (zero-point) energy.
Important terms
Work function (कार्यफलन) φ – minimum energy needed to free an electron from a metal surface.
Stopping potential – reverse voltage that stops the fastest photoelectrons: eV₀ = KE_max.
Compton wavelength – h/(m₀c) ≈ 2.43 × 10⁻¹² m (0.0243 Å) for an electron.
Photon – quantum of light with energy hν and momentum h/λ.
Electron accelerated through V volts: λ = h/√(2meV) ≈ 12.27/√V Å
Uncertainty: Δx Δp ≥ ħ/2; ΔE Δt ≥ ħ/2 (ħ = h/2π)
Useful: hc ≈ 1240 eV nm
5 important questions
Q1. State Einstein's photoelectric equation and the meaning of each term.
hν = φ + KE_max, where hν is the photon energy, φ = hν₀ is the work function of the metal (ν₀ = threshold frequency) and KE_max = ½mv²_max is the maximum kinetic energy of the emitted electrons.
Q2. State Heisenberg's uncertainty principle.
It is impossible to measure the position and momentum of a particle at the same time with unlimited accuracy. The product of their uncertainties is at least ħ/2: Δx Δp ≥ ħ/2.
Q3. Light of wavelength 300 nm falls on a metal with work function 2.3 eV. Find the maximum KE of the photoelectrons and the stopping potential.
Photon energy E = hc/λ = 1240 eV nm/300 nm ≈ 4.13 eV.
KE_max = E − φ = 4.13 − 2.3 = 1.83 eV.
Stopping potential V₀ = KE_max/e = 1.83 V.
Threshold wavelength λ₀ = 1240/2.3 ≈ 539 nm; light longer than this ejects no electrons.
Q4. Find the de Broglie wavelength of an electron accelerated through 100 V, and the Compton shift for X-rays scattered at 90°.
λ = 12.27/√V Å = 12.27/√100 = 1.227 Å.
(This is of the order of atomic spacings, so crystals diffract electrons.)
Compton shift Δλ = (h/m₀c)(1 − cos 90°) = 0.0243 Å × (1 − 0).
Δλ = 0.0243 Å = 2.43 × 10⁻¹² m, independent of the X-ray wavelength.
Q5. Describe the Davisson–Germer experiment and its result.
An electron gun accelerates electrons through a known voltage V.
The beam falls normally on a nickel crystal; a movable detector measures the scattered electrons at different angles.
The intensity is plotted against the scattering angle for different voltages.
A sharp peak appears at 54 V and a scattering angle of 50°.
Using the known spacing of nickel atoms, the diffraction peak gives λ ≈ 1.65 Å, very close to de Broglie's value h/p ≈ 1.67 Å for 54 eV electrons.
Conclusion: electrons behave as waves — matter waves are real.
Exam tip
Keep hc = 1240 eV nm and λ = 12.27/√V Å ready; they turn long numericals into one line. For Compton and photoelectric effects, always mention why classical wave theory fails.
The Schrödinger equation is the basic equation of quantum mechanics. It tells how the wave function of a particle behaves. This topic covers the meaning of the wave function, its conditions, operators, the time-dependent and time-independent Schrödinger equations, expectation values, and the particle in a one-dimensional box.
Key points
The state of a particle is described by a wave function ψ(x, t), which can be complex.
Born interpretation: |ψ|² = ψ*ψ is the probability density; |ψ|² dx is the probability of finding the particle between x and x + dx.
Normalisation: ∫|ψ|² dx over all space = 1 (the particle is somewhere).
An acceptable (well-behaved) ψ must be single-valued, finite, continuous, and have a continuous first derivative (where the potential is finite).
Observables are represented by operators: momentum p → −iħ ∂/∂x; energy E → iħ ∂/∂t.
The time-dependent Schrödinger equation describes how ψ changes in time.
When the potential does not depend on time, separation of variables gives the time-independent (steady-state) equation, an eigenvalue equation Hψ = Eψ.
Allowed energies (eigenvalues) are often discrete — quantisation comes out naturally from boundary conditions.
Expectation value ⟨A⟩ is the average result of many measurements.
Particle in a 1-D box (infinite walls at x = 0 and L): ψ = 0 outside; inside, energies are Eₙ = n²h²/(8mL²), n = 1, 2, 3 …
The lowest energy E₁ is not zero (zero-point energy), consistent with the uncertainty principle. Energy levels are spaced in the ratio 1 : 4 : 9 …
Classical physics is recovered for large n or large mass (correspondence principle).
Important terms
Wave function (तरंग फलन) – mathematical function describing the quantum state.
Eigenvalue / eigenfunction – if Âψ = aψ, a is the eigenvalue and ψ the eigenfunction.
Hamiltonian operator H – −(ħ²/2m)∇² + V, the energy operator.
Stationary state – state with definite energy whose |ψ|² does not change with time.
Q1. Give the physical meaning of the wave function.
ψ itself has no direct physical meaning, but |ψ|² gives the probability density. |ψ(x)|² dx is the probability of finding the particle between x and x + dx at that time.
Q2. What conditions must an acceptable wave function satisfy?
It must be single-valued, finite everywhere, continuous, have a continuous first derivative (for finite potentials), and be normalisable (∫|ψ|² dx finite).
Q3. Derive the energy eigenvalues of a particle in a one-dimensional box.
V = 0 for 0 < x < L, V = ∞ outside, so ψ = 0 outside and at the walls.
Inside: d²ψ/dx² + k²ψ = 0, with k² = 2mE/ħ².
General solution: ψ = A sin kx + B cos kx.
ψ(0) = 0 gives B = 0. ψ(L) = 0 gives sin kL = 0, so kL = nπ (n = 1, 2, 3 …; n = 0 gives ψ = 0 everywhere).
k = nπ/L, so E = ħ²k²/(2m) = n²π²ħ²/(2mL²) = n²h²/(8mL²).
In eV: 6.03 × 10⁻¹⁸/1.6 × 10⁻¹⁹ ≈ 37.7 eV. Then E₂ = 4E₁ ≈ 151 eV.
Q5. Derive the time-independent Schrödinger equation from the time-dependent one.
Time-dependent: iħ ∂ψ/∂t = −(ħ²/2m) ∂²ψ/∂x² + V(x)ψ, with V independent of t.
Try ψ(x, t) = φ(x) f(t).
Substitute and divide by φf: iħ (1/f)(df/dt) = −(ħ²/2m)(1/φ)(d²φ/dx²) + V.
Left side depends only on t, right only on x, so both equal a constant E.
Time part: f = e^(−iEt/ħ). Space part: −(ħ²/2m) d²φ/dx² + Vφ = Eφ.
Rearranged: d²φ/dx² + (2m/ħ²)(E − V)φ = 0.
Exam tip
The particle-in-a-box derivation is the most frequently asked long question — practise applying the two boundary conditions and normalisation step by step. Remember Eₙ ∝ n² and ∝ 1/L².
When a particle meets a sudden change in potential energy — a step or a barrier — quantum mechanics gives results that classical physics cannot: a particle can be reflected even when it has enough energy to pass, and it can pass through a barrier even when it does not have enough energy. This topic covers the potential step, the rectangular barrier, reflection and transmission coefficients, quantum tunnelling and its applications such as alpha decay, the tunnel diode and the scanning tunnelling microscope.
Key points
Solve the time-independent Schrödinger equation −(ħ²/2m)ψ″ + Vψ = Eψ in each region and join the solutions using the boundary conditions: ψ and dψ/dx must be continuous.
E greater than V₀: classically all particles go on; quantum mechanically some are reflected. R = [(k₁ − k₂)/(k₁ + k₂)]², T = 4k₁k₂/(k₁ + k₂)², R + T = 1.
E less than V₀: total reflection (R = 1), but the wave function enters the step and decays as e^(−αx) — a finite chance of finding the particle in the classically forbidden region.
Potential step (V = 0 for x less than 0, V = V₀ for x greater than 0):
Rectangular barrier of height V₀ and width a, with E less than V₀: the wave decays inside the barrier and a small wave emerges on the far side. The particle "tunnels" through.
For a wide or high barrier (αa much greater than 1), T ≈ 16(E/V₀)(1 − E/V₀) e^(−2αa), with α = √(2m(V₀ − E))/ħ.
Tunnelling probability falls exponentially with barrier width and with √(mass × (V₀ − E)), so it matters only for light particles and very thin barriers.
Applications: alpha decay (Gamow, 1928), tunnel diode, scanning tunnelling microscope (STM), nuclear fusion in the Sun, flash memory, field emission.
Important terms
Reflection coefficient (R) – ratio of reflected to incident probability current.
Transmission coefficient (T) – ratio of transmitted to incident probability current.
Classically forbidden region – region where E is less than V, so classical kinetic energy would be negative.
Penetration depth – distance 1/α over which ψ falls by a factor e inside the step.
Probability current density – J = (ħk/m)|A|² for a plane wave.
α = √(2m(V₀ − E))/ħ (inside the barrier, E less than V₀)
Step, E greater than V₀: R = [(k₁ − k₂)/(k₁ + k₂)]², T = 4k₁k₂/(k₁ + k₂)²
Barrier, thick: T ≈ 16(E/V₀)(1 − E/V₀) e^(−2αa)
Penetration depth δ = 1/α
5 important questions
Q1. What is quantum tunnelling? Give two examples.
It is the passage of a particle through a potential barrier whose height is greater than the particle's energy. Classically this is impossible, but the wave function does not become zero inside the barrier; it decays and leaves a small amplitude on the other side. Examples: alpha particles escaping from nuclei; electrons crossing the thin junction of a tunnel diode; the STM current between tip and surface.
Q2. For a potential step with E = 2V₀, find R and T.
k₂/k₁ = √((E − V₀)/E) = √(1/2) = 0.707. R = [(1 − 0.707)/(1 + 0.707)]² = (0.293/1.707)² = (0.1716)² ≈ 0.029. T = 4 × 0.707/(1.707)² = 2.828/2.914 ≈ 0.971. Check: R + T = 1 ✔. About 3% of particles are reflected even though they have twice the step energy.
Q3. Derive R and T for a potential step with E greater than V₀.
Region I (x less than 0): ψ₁ = Ae^(ik₁x) + Be^(−ik₁x) (incident + reflected).
Region II (x greater than 0): ψ₂ = Ce^(ik₂x) (transmitted only).
At x = 0: ψ continuous → A + B = C; ψ′ continuous → k₁(A − B) = k₂C.
If the width doubles to 1 nm, T falls to about 4 × e^(−10.24) ≈ 1.4 × 10⁻⁴: tunnelling is very sensitive to width.
Q5. Explain alpha decay using tunnelling.
Inside the nucleus an alpha particle is held by the strong nuclear force; outside, the Coulomb repulsion forms a barrier about 25–30 MeV high. Alpha particles have only 4–9 MeV, so classically they could never escape. Quantum mechanically, they tunnel through the barrier with a tiny probability per collision with the wall. Since they hit the wall about 10²¹ times per second, escape eventually happens. A small increase in energy makes the barrier thinner and the decay much faster: this explains the Geiger–Nuttall law, linking half-life (seconds to billions of years) with alpha energy.
Exam tip
State the boundary conditions clearly before solving any step or barrier problem. In tunnelling numericals, compute α first and remember that the exponent 2αa dominates the answer.
Lagrangian mechanics reformulates Newton's mechanics using energies instead of forces. With generalised coordinates and the Lagrangian L = T − V, the equations of motion of complicated constrained systems can be found easily. This topic covers constraints, degrees of freedom, D'Alembert's principle, Lagrange's equations, cyclic coordinates and an introduction to the Hamiltonian.
Key points
Constraints restrict motion. Holonomic: expressible as equations among coordinates and time (bead on a wire, rigid body). Non-holonomic: not expressible so (gas in a container; rolling with inequalities).
Scleronomic constraints do not depend on time; rheonomic constraints do (bead on a moving wire).
Degrees of freedom = number of independent coordinates needed = 3N − k (N particles, k holonomic constraints).
Generalised coordinates q₁, q₂, … are any independent quantities (lengths, angles) that fix the configuration.
Constraint forces (like tension, normal reaction) do no virtual work in ideal cases, so they drop out.
D'Alembert's principle: Σ(Fᵢ − ṗᵢ)·δrᵢ = 0; Lagrange's equations follow from it.
Lagrangian L = T − V (kinetic minus potential energy).
Lagrange's equation for each coordinate: d/dt(∂L/∂q̇) − ∂L/∂q = 0.
Generalised momentum: p = ∂L/∂q̇.
Cyclic (ignorable) coordinate: one that does not appear in L; its generalised momentum is conserved.
Hamiltonian H = Σpq̇ − L; for scleronomic conservative systems H = T + V (total energy).
Hamilton's equations: q̇ = ∂H/∂p, ṗ = −∂H/∂q.
Advantages: works with scalars (energies), handles constraints easily, valid in any coordinates.
Important terms
Virtual displacement – imaginary instantaneous small displacement consistent with constraints (आभासी विस्थापन).
Configuration space – space of generalised coordinates.
Generalised force – Qⱼ = ΣFᵢ·∂rᵢ/∂qⱼ.
Formulas
Degrees of freedom f = 3N − k
L = T − V
d/dt(∂L/∂q̇ⱼ) − ∂L/∂qⱼ = 0
pⱼ = ∂L/∂q̇ⱼ; ∂L/∂qⱼ = 0 ⇒ pⱼ = constant
H = Σpⱼq̇ⱼ − L; q̇ⱼ = ∂H/∂pⱼ, ṗⱼ = −∂H/∂qⱼ
5 important questions
Q1. What are holonomic and non-holonomic constraints? Give an example of each.
Holonomic constraints can be written as equations f(r₁, r₂, …, t) = 0, e.g. a simple pendulum with a rigid rod (x² + y² = l²). Non-holonomic constraints cannot, e.g. gas molecules inside a container (r ≤ R, an inequality).
Q2. What is a cyclic coordinate? What is its importance?
A coordinate that does not appear in the Lagrangian (only its velocity may appear). Since ∂L/∂q = 0, Lagrange's equation gives d/dt(∂L/∂q̇) = 0, so the corresponding generalised momentum is conserved.
Q3. Using Lagrange's equation, obtain the equation of motion of a simple pendulum.
Generalised coordinate: angle θ; bob mass m, length l.
T = ½ml²θ̇²; V = −mgl cos θ (taking the suspension point as zero).
L = ½ml²θ̇² + mgl cos θ.
∂L/∂θ̇ = ml²θ̇, so d/dt(∂L/∂θ̇) = ml²θ̈; ∂L/∂θ = −mgl sin θ.
Equation: ml²θ̈ + mgl sin θ = 0 → θ̈ + (g/l) sin θ = 0.
For small θ: θ̈ + (g/l)θ = 0, SHM with T = 2π√(l/g).
Q4. Find the acceleration of Atwood's machine using the Lagrangian method.
Masses m₁ and m₂ on a light string over a frictionless pulley; x = distance m₁ has descended; m₂ is then at (l − x) below the pulley.
T = ½(m₁ + m₂)ẋ².
V = −m₁gx − m₂g(l − x).
L = ½(m₁ + m₂)ẋ² + m₁gx + m₂g(l − x).
Lagrange: (m₁ + m₂)ẍ − (m₁ − m₂)g = 0.
ẍ = (m₁ − m₂)g/(m₁ + m₂). The tension never had to be found.
Q5. Write the Hamiltonian of a one-dimensional harmonic oscillator and obtain its equations of motion.
L = ½mẋ² − ½kx²; p = ∂L/∂ẋ = mẋ, so ẋ = p/m.
H = pẋ − L = p²/m − (p²/2m − ½kx²) = p²/(2m) + ½kx² (= total energy).
Follow the same four steps every time: choose generalised coordinates, write T and V, form L, apply Lagrange's equation for each coordinate. State the number of degrees of freedom at the start.
A central force always acts along the line joining a body to a fixed centre, and its size depends only on the distance from that centre. Gravity and the Coulomb force are the main examples. This topic covers the reduction of the two-body problem to a one-body problem, conservation of angular momentum, the effective potential, the differential equation of the orbit, and the derivation of Kepler's laws from the inverse-square law.
Key points
A central force has the form F = f(r) r̂. Its torque about the centre is zero, so angular momentum L is conserved.
Because L is constant in direction, the motion stays in one plane.
Because L is constant in size, areal velocity dA/dt = L/(2m) is constant (Kepler's second law).
A central force is conservative, so total energy E = ½m(ṙ² + r²θ̇²) + V(r) is conserved.
Two-body problem: two masses m₁ and m₂ interacting with each other can be replaced by one body of reduced mass μ = m₁m₂/(m₁ + m₂) moving about a fixed centre, plus free motion of the centre of mass.
Effective potential: V_eff(r) = V(r) + L²/(2μr²). The extra term is the "centrifugal barrier", which keeps the body away from r = 0.
For the gravitational potential V = −k/r, the energy decides the orbit: E less than 0 → ellipse (or circle at the minimum of V_eff); E = 0 → parabola; E greater than 0 → hyperbola.
Orbit equation (with u = 1/r): d²u/dθ² + u = −(μ/L²u²) f(1/u). For the inverse-square force this gives a conic section.
Kepler's third law T² = (4π²/GM) a³ follows from the areal velocity and the area of the ellipse.
Important terms
Reduced mass (μ) – m₁m₂/(m₁ + m₂); for the Earth–Sun system it is almost the Earth's mass.
Eccentricity (e) – shape of the conic: e = 0 circle, 0 less than e less than 1 ellipse, e = 1 parabola, e greater than 1 hyperbola.
Turning points – values of r where ṙ = 0 (perihelion and aphelion).
Centrifugal barrier – the L²/(2μr²) term in the effective potential.
Formulas
L = μr²θ̇ = constant; dA/dt = L/(2μ)
E = ½μṙ² + L²/(2μr²) + V(r)
μ = m₁m₂/(m₁ + m₂)
Orbit equation: d²u/dθ² + u = −(μ/(L²u²)) f(1/u), u = 1/r
Inverse-square force f = −k/r²: 1/r = (μk/L²)(1 + e cos θ)
e = √(1 + 2EL²/(μk²))
Kepler III: T² = 4π²a³μ/k = 4π²a³/(G(m₁ + m₂)) (≈ 4π²a³/GM for a planet)
5 important questions
Q1. Show that motion under a central force is confined to a plane.
τ = r × F = r × f(r) r̂ = 0, since r and r̂ are parallel. So dL/dt = 0 and L = r × p is constant. The position vector r is always perpendicular to the fixed vector L, so r always lies in the plane perpendicular to L passing through the centre. The motion is planar.
Q2. Find the reduced mass of the hydrogen atom (m_p = 1836 m_e).
μ = m_e m_p/(m_e + m_p) = m_e × 1836/1837 ≈ 0.99946 m_e. Using μ instead of m_e changes the Rydberg constant slightly (R_H = R_∞ μ/m_e), which explains the small difference between hydrogen and deuterium lines.
Q3. Derive the differential equation of the orbit.
In polar form the radial equation is μ(r̈ − rθ̇²) = f(r), with L = μr²θ̇.
Put u = 1/r. Then ṙ = −(L/μ) du/dθ and r̈ = −(L²u²/μ²) d²u/dθ².
Also rθ̇² = L²u³/μ².
Substituting: −(L²u²/μ)(d²u/dθ² + u) = f(1/u).
So d²u/dθ² + u = −(μ/L²u²) f(1/u).
Q4. Using the orbit equation, show that an inverse-square attractive force gives a conic section.
For f = −k/r² = −ku²: the right side becomes −(μ/L²u²)(−ku²) = μk/L².
d²u/dθ² + u = μk/L², a constant.
Solution: u = μk/L² + C cos θ, i.e. 1/r = (μk/L²)(1 + e cos θ) with e = CL²/(μk).
This is the polar equation of a conic with the force centre at a focus. Bound orbits (e less than 1) are ellipses — Kepler's first law.
Q5. Using the effective potential, find the radius and speed of a circular orbit for V = −k/r, and solve for a satellite.
V_eff = −k/r + L²/(2μr²). For a circular orbit dV_eff/dr = 0: k/r² − L²/(μr³) = 0 → r₀ = L²/(μk). Equivalently the force balance k/r² = μv²/r gives v = √(k/(μr)). For a satellite around the Earth, k/μ = GM = 3.99 × 10¹⁴ m³ s⁻². At r = 7.0 × 10⁶ m: v = √(3.99 × 10¹⁴/7.0 × 10⁶) = √(5.70 × 10⁷) ≈ 7.55 km s⁻¹; period T = 2πr/v = 2π × 7.0 × 10⁶/7550 ≈ 5.8 × 10³ s (about 97 min).
Exam tip
State the two conservation laws (angular momentum and energy) at the start of any central-force answer. In the orbit-equation derivation, show the change of variable to u = 1/r clearly; it is where most students lose marks.
Atoms emit and absorb light only at certain wavelengths, giving line spectra. Bohr's model explains the hydrogen spectrum, but the fine structure of lines and the effect of magnetic fields need the vector atom model, which adds electron spin and treats angular momenta as vectors. This topic covers hydrogen series, quantum numbers, spin, coupling schemes, spectral terms, sodium D lines and the Zeeman effect.
Key points
Bohr model: electrons move in fixed orbits with angular momentum nħ; a photon of energy hν = E₂ − E₁ is emitted when an electron jumps down.
Hydrogen series: Lyman (to n = 1, ultraviolet), Balmer (to n = 2, visible), Paschen (n = 3), Brackett (n = 4), Pfund (n = 5) — the last three in infrared.
Bohr's model fails for many-electron atoms, line intensities, fine structure and the Zeeman effect.
Vector atom model: orbital (l) and spin (s) angular momenta are vectors that combine to give total angular momentum j.
Quantum numbers: principal n (1, 2, 3 …), orbital l (0 to n − 1), magnetic mₗ (−l to +l), spin s = ½ with mₛ = ±½, total j = l ± ½ (for one electron), and mⱼ.
Space quantisation: an angular momentum vector can take only certain directions with respect to a field.
Stern–Gerlach experiment (1922): a beam of silver atoms split into two in a non-uniform magnetic field, showing space quantisation and (later understood) electron spin.
Spin: proposed by Uhlenbeck and Goudsmit (1925).
Pauli exclusion principle: no two electrons in an atom can have all four quantum numbers (n, l, mₗ, mₛ) the same.
L–S (Russell–Saunders) coupling: for lighter atoms, all l's add to L, all s's add to S, then J = L + S. j–j coupling: for heavy atoms, each electron's l and s combine first to j.
Spectral term: ²ˢ⁺¹L_J, with L = 0, 1, 2, 3 written S, P, D, F.
Sodium D lines: the 3P level splits into ²P₃/₂ and ²P₁/₂ by spin–orbit interaction; transitions to ²S₁/₂ give D₂ (5890 Å) and D₁ (5896 Å).
Selection rules: Δl = ±1, Δj = 0, ±1 (j = 0 to j = 0 not allowed).
Zeeman effect: splitting of spectral lines in a magnetic field. Normal (lines split into three — Lorentz triplet; spin plays no part, S = 0). Anomalous (more components; depends on Landé g factor). In very strong fields: Paschen–Back effect.
Important terms
Fine structure (सूक्ष्म संरचना) – splitting of a spectral line into close components.
Landé g factor – factor relating magnetic moment to total angular momentum.
No two electrons in an atom can have the same set of four quantum numbers n, l, mₗ and mₛ. So an orbital can hold at most two electrons, with opposite spins.
Q2. Write the possible j values and spectral terms for a single electron with l = 2.
j = l ± s = 2 ± ½ = 5/2 or 3/2. With S = ½ (doublet), the terms are ²D₅/₂ and ²D₃/₂.
Q3. Calculate the Landé g factor for the ²P₃/₂ and ²P₁/₂ states.
For both, l = 1, s = ½; l(l + 1) = 2, s(s + 1) = 3/4.
Draw the energy-level diagram of sodium with term symbols and the two transitions labelled. Practise the Landé g formula — it is a quick, high-scoring numerical.
Molecules have energy not only in their electrons but also in rotation and vibration of the nuclei. Changes in these energies give molecular spectra: rotational spectra in the microwave region, vibrational (vibration–rotation) spectra in the infrared, and electronic band spectra in the visible and ultraviolet. This topic covers the rigid rotator, the harmonic and anharmonic oscillator, selection rules, P and R branches, and how bond length and force constant are found from spectra.
Key points
Order of energy spacings: electronic (a few eV) much greater than vibrational (about 0.1 eV) much greater than rotational (about 0.001 eV).
Molecular spectra are band spectra (many close lines), unlike the line spectra of atoms.
Rigid rotator model of a diatomic molecule: energy E_J = (ħ²/2I) J(J + 1) = hcB J(J + 1), J = 0, 1, 2 …
Rotational constant B = h/(8π²Ic) (in cm⁻¹); I = μr² where μ is the reduced mass and r the bond length.
Selection rule for pure rotation: ΔJ = ±1, and the molecule must have a permanent dipole moment (HCl yes; H₂, N₂ no).
Rotational lines are equally spaced by 2B; line positions ν̄ = 2B(J + 1).
Harmonic oscillator: E_v = (v + ½)hν₀, v = 0, 1, 2 …; even at v = 0 there is zero-point energy ½hν₀. Selection rule Δv = ±1, so there is one fundamental line.
Real molecules are anharmonic (Morse potential): levels get closer as v rises, overtones (Δv = ±2, ±3) appear weakly, and the molecule can dissociate.
Vibration–rotation spectrum: each vibrational line splits into a P branch (ΔJ = −1) and an R branch (ΔJ = +1), with a gap at the band centre (no Q branch for most diatomics).
The Franck–Condon principle explains the intensity of electronic band lines: electronic transitions are so fast that the nuclei do not move during them.
Important terms
Moment of inertia (I) – μr² for a diatomic molecule.
Wave number (ν̄) – 1/λ, in cm⁻¹.
Force constant (k) – stiffness of the bond, ν₀ = (1/2π)√(k/μ).
Anharmonicity constant (x_e) – measures departure from SHM.
Dissociation energy – energy needed to break the molecule from its lowest level.
Formulas
Rotational energy: E_J = hcB J(J + 1); B = h/(8π²cI) (cm⁻¹ if c in cm s⁻¹)
Rotational lines: ν̄ = 2B(J + 1); spacing = 2B
I = μr², μ = m₁m₂/(m₁ + m₂)
Vibrational energy (harmonic): E_v = (v + ½)hν₀; ν₀ = (1/2π)√(k/μ)
Anharmonic: E_v = (v + ½)hν̄_e c − (v + ½)² x_e hν̄_e c
Vibration–rotation: R branch ν̄ = ν̄₀ + 2B(J + 1); P branch ν̄ = ν̄₀ − 2BJ
5 important questions
Q1. Why do H₂ and N₂ show no pure rotational spectrum?
A molecule can absorb or emit microwave radiation during rotation only if it has a permanent electric dipole moment that rotates with it. Homonuclear molecules like H₂, N₂ and O₂ are symmetric and have no dipole moment, so they are microwave-inactive. (Their rotation can be studied by the Raman effect instead.)
Q2. The rotational lines of CO are spaced by 3.86 cm⁻¹. Find B and the moment of inertia.
2B = 3.86 cm⁻¹ → B = 1.93 cm⁻¹. I = h/(8π²cB) = 6.626 × 10⁻³⁴/(8 × 9.87 × 3 × 10¹⁰ × 1.93) = 6.626 × 10⁻³⁴/(4.572 × 10¹²) ≈ 1.45 × 10⁻⁴⁶ kg m².
Q3. For CO, μ = 1.14 × 10⁻²⁶ kg. Using Q2, find the bond length.
r = √(I/μ) = √(1.45 × 10⁻⁴⁶/1.14 × 10⁻²⁶) = √(1.272 × 10⁻²⁰) ≈ 1.13 × 10⁻¹⁰ m = 0.113 nm, which matches the known C–O bond length.
Q4. Derive the rotational energy levels and show that the lines are equally spaced.
Classical rotational energy E = L²/(2I).
In quantum mechanics L² = J(J + 1)ħ², so E_J = J(J + 1)ħ²/(2I) = hcB J(J + 1).
Keep c in cm s⁻¹ (3 × 10¹⁰) when B and ν̄ are in cm⁻¹ — mixing units is the commonest mistake. Remember: rotation → microwave, vibration → infrared, electronic → visible/UV.
When monochromatic light passes through a transparent substance, most of the scattered light has the same frequency as the incident light, but a very small part has slightly changed frequencies. This is the Raman effect, discovered by C. V. Raman in Kolkata on 28 February 1928 (Nobel Prize 1930). This topic covers Rayleigh and Raman scattering, Stokes and anti-Stokes lines, the quantum and classical explanations, selection rules, the experimental set-up, and applications of Raman spectroscopy.
Key points
Rayleigh scattering: scattered light has the same frequency ν₀ as the incident light (elastic scattering). It is much stronger than Raman scattering.
Raman lines: weak lines at ν₀ ± Δν appear on both sides of the Rayleigh line. The shift Δν is a property of the scattering molecule and does not depend on the exciting frequency.
Stokes lines (ν₀ − Δν): the molecule takes energy from the photon (goes to a higher vibrational or rotational level). Stronger, because most molecules are in the ground state.
Anti-Stokes lines (ν₀ + Δν): the molecule gives energy to the photon. Weaker; their intensity rises with temperature.
Raman shifts equal the vibrational or rotational frequencies of the molecule, so the Raman spectrum gives the same kind of information as IR spectra, but using visible light.
Quantum picture: a photon hν₀ collides with a molecule; energy conservation gives hν′ = hν₀ ± ΔE.
Classical picture: the field induces a dipole p = αE. If the polarisability α changes as the molecule vibrates, the induced dipole contains frequencies ν₀ and ν₀ ± ν_vib.
Selection rule: a vibration is Raman active if it changes the polarisability; it is IR active if it changes the dipole moment. So H₂, N₂ and O₂, which are IR-inactive, give Raman spectra.
Rule of mutual exclusion: in molecules with a centre of symmetry (CO₂), modes active in IR are inactive in Raman and vice versa.
Rotational Raman selection rule ΔJ = 0, ±2, giving lines spaced by 4B (first line at 6B).
Important terms
Raman shift – difference in wave number between the incident and Raman lines, Δν̄ = ν̄₀ − ν̄_s.
Polarisability (α) – ease with which the electron cloud of a molecule is distorted by an electric field.
Raman active – a mode that appears in the Raman spectrum.
Virtual state – short-lived intermediate state in the scattering process, not a real energy level.
Formulas
hν_s = hν₀ ± hν_m (Stokes −, anti-Stokes +)
Raman shift Δν̄ = |ν̄₀ − ν̄_s| = ν̄_m (cm⁻¹)
Induced dipole p = αE₀ cos 2πν₀t, with α = α₀ + α₁ cos 2πν_m t → frequencies ν₀, ν₀ ± ν_m
Intensity ratio anti-Stokes/Stokes ≈ e^(−hν_m/kT)
Rotational Raman lines: Δν̄ = ±(4B J + 6B)
5 important questions
Q1. Distinguish between Rayleigh scattering and Raman scattering.
Rayleigh
Raman
Elastic: frequency unchanged
Inelastic: frequency changes
Strong
Very weak (about 10⁻⁶ of incident light)
No information on molecular levels
Gives vibrational and rotational levels
One line at ν₀
Stokes and anti-Stokes lines at ν₀ ∓ Δν
Q2. Why are Stokes lines more intense than anti-Stokes lines?
Stokes lines come from molecules in the ground vibrational state, and anti-Stokes lines from molecules already in an excited state. By the Boltzmann law, at room temperature far more molecules are in the ground state, so Stokes lines are much stronger. Heating the sample increases the anti-Stokes intensity.
Q3. Light of wavelength 435.8 nm gives a Stokes line at 444.7 nm. Find the Raman shift and the wavelength of the anti-Stokes line.
Q4. Explain the Raman effect on the classical and quantum theories.
Quantum: an incident photon hν₀ raises the molecule to a short-lived virtual state. It falls back either to the original level (Rayleigh, hν₀), to a higher vibrational level (Stokes, h(ν₀ − ν_m)) or, if it started excited, to a lower level (anti-Stokes, h(ν₀ + ν_m)). Classical: E = E₀ cos 2πν₀t induces p = αE. If α = α₀ + α₁ cos 2πν_m t, then p = α₀E₀ cos 2πν₀t + ½α₁E₀[cos 2π(ν₀ − ν_m)t + cos 2π(ν₀ + ν_m)t]. The oscillating dipole radiates at ν₀ (Rayleigh) and ν₀ ± ν_m (Raman). If α₁ = 0 (polarisability does not change), there is no Raman line — the selection rule. The classical model cannot explain why Stokes lines are stronger.
Q5. Describe the experimental set-up and give applications of Raman spectroscopy.
Set-up: a monochromatic source (Raman used a mercury arc with filters; today a laser) is focused into the liquid in a tube. Scattered light is collected at 90° to the beam (to avoid the direct beam), passed into a spectrograph and recorded on a plate or CCD. A dark background (black-painted tube end) cuts stray light. Applications: identifying molecules and their structure (fingerprint spectra), studying symmetric molecules that are IR-inactive, analysing minerals and drugs, detecting stress in semiconductors, non-destructive testing of art and gems, studying biological samples in water (water is a weak Raman scatterer).
Exam tip
Remember the date: 28 February is celebrated as National Science Day in India because of the Raman effect. Convert wavelengths to wave numbers (cm⁻¹) before subtracting — never subtract wavelengths directly.