Class 12 Physics

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Class 12 – Physics

Master 30 advanced topics covering electrostatics, current electricity, magnetism, electromagnetic induction, alternating current, optics, modern physics and semiconductor electronics. Every topic now includes detailed theory, key formulas, a fully worked numerical, common mistakes and practice questions.

📘 How to use this page

Read the theory first, then study the formula box, and only then attempt the worked example on your own before looking at the solution. Use the “Common mistake” boxes to avoid typical board-exam errors. Finish each topic by solving the practice questions without notes. Physics is learned by solving problems, so aim to do at least five numericals per topic.

⚡ Electrostatics and Capacitance

1

Electric Charge Distribution and Quantisation

Electric charge is a fundamental property of matter that comes in two kinds, positive and negative. Like charges repel and unlike charges attract. Three properties define how charge behaves: it is additive (total charge is the algebraic sum), conserved (it can be transferred but never created or destroyed in an isolated system), and quantised (any observable charge is an integral multiple of the elementary charge e = 1.6 × 10⁻¹⁹ C, so q = ne).

At the macroscopic scale, the charge on an object is so large compared with e that quantisation goes unnoticed and charge appears continuous. This is why we can describe a charged body using smooth charge densities. When charge is spread along a line, over a surface, or through a volume, we use linear (λ), surface (σ) and volume (ρ) charge density respectively. For non-uniform distributions the density varies with position and charge is found by integration, for example Q = ∫λ dl.

Ways of charging

  • Friction: electrons transfer from one body to another (glass rubbed with silk becomes positive).
  • Conduction: contact with a charged body shares charge; both end up with the same sign.
  • Induction: a charged body brought near a conductor redistributes its charge; earthing then leaves the conductor with a charge of opposite sign.
Key formulas:
q = ±ne  |  λ = Q/L (C m⁻¹)  |  σ = Q/A (C m⁻²)  |  ρ = Q/V (C m⁻³)
Worked example: A thin wire of length 2 m carries a uniform charge of 5 μC. Find λ and the number of excess electrons if the charge is negative.
λ = Q/L = 5 × 10⁻⁶ / 2 = 2.5 × 10⁻⁶ C m⁻¹. Number of electrons n = Q/e = 5 × 10⁻⁶ / 1.6 × 10⁻¹⁹ ≈ 3.1 × 10¹³.
Common mistake: Saying charge is “created” when a rod is rubbed. Rubbing only transfers electrons; the total charge of the system stays the same.
Practice:
  1. Compare linear, surface and volume charge density and write their SI units.
  2. A metal sphere of radius 10 cm carries 4 μC uniformly on its surface. Find σ.
  3. How many electrons together carry a charge of 1 C?
Remember: Electric charge is conserved, additive and quantised. Only electrons (and ions) move in solids; protons stay fixed in the nucleus.
2

Coulomb’s Law in Multi-Charge Systems

Coulomb’s law states that the electrostatic force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. The force acts along the line joining the charges, is attractive for unlike charges and repulsive for like charges, and obeys Newton’s third law.

When more than two charges are present, the principle of superposition applies: the force on any charge is the vector sum of the forces due to each of the other charges taken one at a time. The presence of other charges does not change the pairwise force. Vector addition therefore becomes the key skill: resolve each force into components, add components, then find magnitude and direction.

In a medium of relative permittivity (dielectric constant) K, the force reduces by the factor K. Compared with gravity, the electrostatic force between an electron and a proton is about 10³⁹ times stronger, which is why electrical forces dominate atomic and molecular structure.

Key formulas:
F = kq₁q₂/r² = q₁q₂ / (4πε₀r²), k = 9 × 10⁹ N m² C⁻²  |  In a medium: F = F₀/K  |  Vector form: F₁₂ = k q₁q₂ r̂₁₂ / r²
Worked example: Four equal charges q sit at the corners of a square of side a. Find the force on one corner charge.
Two adjacent charges each give kq²/a², at right angles; their resultant is √2 kq²/a² along the diagonal. The opposite corner (distance a√2) gives kq²/(2a²) in the same direction. Net force = (√2 + ½) kq²/a² ≈ 1.91 kq²/a², directed away from the centre.
Common mistake: Adding force magnitudes directly. Forces are vectors, so directions must be accounted for before adding.
Practice:
  1. Two 2 μC charges are 3 cm apart in air. Find the force between them (answer: 40 N).
  2. Draw force vectors for three charges arranged at the vertices of a triangle.
  3. Three charges +q, −2q and +q lie on a line with equal spacing d. Find the net force on the middle charge.
Remember: Calculate each pairwise force separately, resolve into components, then add vectors.
3

Electric Field Due to Continuous Charge Distributions

The electric field at a point is the force per unit positive test charge placed there, E = F/q₀. For a point charge, E = kq/r² directed radially outward for positive charge and inward for negative. Field lines begin on positive charges, end on negative charges, never intersect, and are denser where the field is stronger.

For a continuous distribution, divide the body into small elements dq, write the field dE of each element, and integrate. The skill lies in choosing dq well and exploiting symmetry so that components cancel. For a rod use dq = λ dx, for a ring use dq = λ R dθ, and for a disc use thin concentric rings of radius r and width dr, with dq = σ 2πr dr.

Standard results:
Ring on axis: E = kQx / (x² + R²)3/2 (maximum at x = R/√2; zero at centre)
Disc on axis: E = (σ/2ε₀) [1 − x/√(x² + R²)]
Infinite line: E = λ / (2πε₀ r)  |  Infinite sheet: E = σ/2ε₀  |  Two opposite sheets: E = σ/ε₀ between, 0 outside
Worked example: A ring of radius 3 cm carries 10 nC. Find E on its axis 4 cm from the centre.
x² + R² = 25 cm², so (x² + R²)3/2 = (0.05)³ = 1.25 × 10⁻⁴ m³. E = (9 × 10⁹)(10⁻⁸)(0.04) / 1.25 × 10⁻⁴ = 2.88 × 10⁴ N C⁻¹, directed along the axis away from the ring.
Common mistake: Integrating the magnitude of dE when only one component survives. Always find which components cancel by symmetry first.
Practice:
  1. Identify a suitable differential charge element dq for a charged rod of length L.
  2. Show that the field at the centre of a uniformly charged ring is zero.
  3. Find the field of a disc at a large distance and compare with a point charge.
Remember: Match the coordinate system to the symmetry of the charge distribution.
4

Electric Dipoles in External Electric Fields

An electric dipole is a pair of equal and opposite charges ±q separated by a small distance 2a. Its strength is the dipole moment p = q × 2a, a vector pointing from the negative to the positive charge. Many molecules such as H₂O and HCl are natural dipoles.

In a uniform field, the net force on a dipole is zero because the forces on the two charges are equal and opposite, but a torque acts that tends to rotate it into alignment with the field. In a non-uniform field there is also a net force, which draws the dipole toward the region of stronger field. This explains why small pieces of paper are attracted to a charged comb.

Key formulas:
τ = p × E, τ = pE sinθ  |  U = −p·E = −pE cosθ
Work done in rotating from θ₁ to θ₂: W = pE (cosθ₁ − cosθ₂)
Axial point (far): E = 2kp/r³  |  Equatorial point (far): E = kp/r³, directed opposite to p
Potential: V = kp cosθ / r² (zero on the equatorial line)
Worked example: A dipole of moment 4 × 10⁻⁹ C m lies at 30° to a uniform field of 2 × 10⁴ N C⁻¹. Find the torque and the work needed to turn it to 90°.
τ = pE sinθ = 4 × 10⁻⁹ × 2 × 10⁴ × 0.5 = 4 × 10⁻⁵ N m. Work = pE (cos30° − cos90°) = 8 × 10⁻⁵ × 0.866 ≈ 6.9 × 10⁻⁵ J.
Common mistake: Mixing up axial and equatorial fields. The axial field is twice the equatorial one in magnitude and points along p, whereas the equatorial field points against p.
Practice:
  1. Determine the orientations corresponding to maximum and zero torque.
  2. Which orientation is stable equilibrium and what is the potential energy there?
  3. Why does a dipole in a uniform field experience no net force?
Remember: Stable equilibrium occurs when dipole moment and electric field are aligned (θ = 0); θ = 180° is unstable equilibrium.
5

Electric Flux and Applications of Gauss’s Law

Electric flux measures how much electric field passes through a surface: Φ = ∮E·dA. It is a scalar, with sign taken positive for field lines leaving a closed surface and negative for entering. Gauss’s law states that the total flux through any closed surface equals the net enclosed charge divided by ε₀, regardless of the surface’s shape or where inside the charge sits.

The law is always true, but it is useful for finding E only when symmetry (spherical, cylindrical or planar) lets us pull E out of the integral. The method: choose a Gaussian surface on which E is constant and perpendicular to the surface (or parallel, contributing zero), calculate the flux, and equate to qenc/ε₀. Only enclosed charge matters for the total flux, but E itself at the surface is due to all charges, inside and outside.

Key results:
Φ = ∮E·dA = qenc/ε₀
Uniformly charged thin shell: E = kQ/r² (r ≥ R), E = 0 (r < R)
Solid non-conducting sphere: E = kQr/R³ (inside), kQ/r² (outside)
Infinite wire: E = λ/2πε₀r  |  Infinite plane sheet: E = σ/2ε₀  |  Conductor surface: E = σ/ε₀
Worked example: A charge of 8.85 nC lies at the centre of a cube. Find the total flux and the flux through one face.
Φ = q/ε₀ = 8.85 × 10⁻⁹ / 8.85 × 10⁻¹² = 1000 N m² C⁻¹. By symmetry each face receives 1000/6 ≈ 167 N m² C⁻¹.
Common mistake: Using Gauss’s law for E when no symmetry exists, or forgetting that charges outside the surface still contribute to the local field though not to the net flux.
Practice:
  1. Choose a suitable Gaussian surface for an infinitely long charged wire and derive E.
  2. Why is the field inside a hollow charged conductor zero?
  3. Find the flux through a hemispherical surface of radius R placed in a uniform field E with its axis along E.
Remember: Gauss’s law is most useful when symmetry makes the electric field simple.
6

Electric Potential and Equipotential Surfaces

Electric potential at a point is the work done per unit positive charge in bringing it from infinity to that point, V = W/q. Potential difference between two points is the work per unit charge moving between them. Potential is a scalar, so contributions from several charges simply add algebraically, which is often much easier than adding vector fields.

Field and potential are linked by E = −dV/dr: the field points in the direction of fastest decrease of potential and its magnitude equals the potential gradient. An equipotential surface is one on which V is constant. No work is done along it, so the field is always perpendicular to equipotential surfaces. The surface of a conductor in electrostatic equilibrium is an equipotential, and so is the entire interior.

Key formulas:
V = kq/r  |  V = Σ kqᵢ/rᵢ  |  WAB = q (VB − VA)  |  E = −dV/dr
Uniform field: E = V/d  |  Charged shell: V = kQ/R inside and on surface
Worked example: Find the potential 30 cm from a 2 nC point charge and the work to bring a 5 nC charge from infinity to that point.
V = 9 × 10⁹ × 2 × 10⁻⁹ / 0.3 = 60 V. W = qV = 5 × 10⁻⁹ × 60 = 3 × 10⁻⁷ J.
Common mistake: Assuming that zero potential means zero field (or vice versa). At the midpoint between equal and opposite charges V = 0 yet E ≠ 0; at the midpoint between two equal charges E = 0 yet V ≠ 0.
Practice:
  1. Sketch field lines and equipotential surfaces around a positive charge.
  2. Why can two equipotential surfaces never intersect?
  3. A uniform field of 500 V m⁻¹ exists. What is the potential difference between two points 4 cm apart along the field?
Remember: Electric field points toward decreasing electric potential and is always perpendicular to equipotential surfaces.
7

Electrostatic Potential Energy of Charge Systems

Electrostatic potential energy is the work required to assemble a configuration of charges by bringing them in from infinity. It is stored in the arrangement of charges and is independent of the path taken to build it. For two charges the energy is positive if they are alike (work must be done against repulsion) and negative if unlike (the system is bound).

For three or more charges, add the interaction energy of every distinct pair exactly once. A charge q in an external potential V has energy U = qV. The work done by an external agent in moving a charge slowly equals the change in potential energy, while the work done by the field is its negative.

Key formulas:
U = kq₁q₂/r  |  Utotal = k(q₁q₂/r₁₂ + q₂q₃/r₂₃ + q₁q₃/r₁₃)  |  U = qV  |  1 eV = 1.6 × 10⁻¹⁹ J
Energy conservation: ½mv² + qV = constant
Worked example: An electron is accelerated from rest through a potential difference of 100 V. Find its kinetic energy and speed.
KE = eV = 100 eV = 1.6 × 10⁻¹⁷ J. v = √(2eV/m) = √(2 × 1.6 × 10⁻¹⁷ / 9.1 × 10⁻³¹) ≈ 5.9 × 10⁶ m s⁻¹.
Common mistake: Counting a pair twice when finding total energy. Three charges have three pairs, not six.
Practice:
  1. Predict the sign of potential energy for like and unlike charges.
  2. Find the energy to assemble three equal charges q at the vertices of an equilateral triangle of side a (answer: 3kq²/a).
  3. A proton and an alpha particle are accelerated through the same potential difference. Compare their kinetic energies.
Remember: Potential energy belongs to the configuration of the system, not to any single charge.
8

Capacitors and Dielectric Materials

A capacitor consists of two conductors separated by an insulator and stores energy in the electric field between them. Its capacitance C = Q/V depends only on the geometry of the conductors and the medium between them, never on the charge or voltage. The SI unit is the farad (F); practical values are in μF, nF and pF.

Inserting a dielectric places polarisable molecules in the field. They align (or are distorted) so that induced surface charges produce an opposing field, reducing the net field within the dielectric to E₀/K. For the same charge, the potential difference falls by K and capacitance rises by K. A conducting slab behaves like K → ∞.

Key formulas:
Parallel plate: C = ε₀A/d; with dielectric C = Kε₀A/d
Partly filled (slab thickness t): C = ε₀A / (d − t + t/K); conducting slab: C = ε₀A/(d − t)
Isolated sphere: C = 4πε₀R  |  Spherical capacitor: C = 4πε₀ ab/(b − a)
Worked example: Plates of area 0.01 m² are 1 mm apart. Find C in air and with K = 5.
C₀ = 8.85 × 10⁻¹² × 0.01 / 10⁻³ = 8.85 × 10⁻¹¹ F ≈ 88.5 pF. With dielectric C = 5 × 88.5 = 442.5 pF.

Battery connected vs disconnected

QuantityBattery stays connected (V constant)Battery removed (Q constant)
Charge QIncreases K timesUnchanged
Voltage VUnchangedDecreases K times
Electric field EUnchangedDecreases K times
Energy UIncreases K timesDecreases K times
Common mistake: Saying capacitance doubles when charge doubles. Charge and voltage change together; their ratio is fixed by geometry.
Practice:
  1. Explain how capacitance changes when plate area increases.
  2. A 4 μF capacitor charged to 50 V is isolated and a K = 4 dielectric is inserted. Find the new voltage.
  3. Why is air breakdown (about 3 × 10⁶ V m⁻¹) a limit on capacitor voltage?
Remember: Capacitance depends on geometry and medium, not directly on stored charge.
9

Combination and Energy Storage of Capacitors

In a series combination the same charge flows onto every capacitor and the voltages add, so reciprocals of capacitance add. The equivalent capacitance is smaller than the smallest individual one. In a parallel combination every capacitor has the same potential difference and charges add, so capacitances add directly.

The energy stored equals the work done in charging, which is the area under the V–q graph. It resides in the electric field itself, with energy density u = ½ε₀E² (or ½Kε₀E² in a dielectric). When a charged capacitor is connected to an uncharged one, charge redistributes until potentials are equal, and some energy is always lost as heat and radiation even though charge is conserved.

Key formulas:
Series: 1/C = 1/C₁ + 1/C₂ + …  |  Parallel: C = C₁ + C₂ + …
U = ½CV² = ½QV = Q²/2C  |  u = ½ε₀E²
Common potential after sharing: V = (C₁V₁ + C₂V₂)/(C₁ + C₂); energy loss = C₁C₂(V₁ − V₂)² / 2(C₁ + C₂)
Worked example: A 2 μF and a 3 μF capacitor are joined in series across 100 V. Find equivalent capacitance, charge and total energy.
C = (2 × 3)/(2 + 3) = 1.2 μF. Q = CV = 120 μC on each. U = ½CV² = ½ × 1.2 × 10⁻⁶ × 10⁴ = 6 mJ. The voltages are 60 V across 2 μF and 40 V across 3 μF.
Common mistake: Applying the parallel formula to series capacitors (or vice versa). Always check whether charge or voltage is shared.
Practice:
  1. Compare two identical capacitors C connected in series and in parallel.
  2. A 6 μF capacitor charged to 10 V is connected to an uncharged 3 μF capacitor. Find the common voltage and energy lost.
  3. Find the energy density at a point where E = 10⁴ V m⁻¹ in air.
Remember: Series capacitors carry equal charge; parallel capacitors have equal voltage.

🔋 Current Electricity and Circuit Analysis

10

Drift Velocity and Microscopic Theory of Current

Inside a metal, free electrons move randomly at high thermal speeds (about 10⁵ m s⁻¹) but with zero net displacement. When an electric field is applied, each electron is accelerated between collisions with lattice ions, giving a very small average drift superimposed on the random motion. The mean time between collisions is the relaxation time τ.

The drift speed is surprisingly small (a fraction of a millimetre per second) yet a bulb lights almost instantly because the electric field is established throughout the wire at nearly the speed of light, setting all electrons in motion together. Mobility μ = vd/E measures how easily carriers drift, and microscopic Ohm’s law links current density to field: J = σE.

Key formulas:
I = nAevd  |  vd = eEτ/m  |  μ = vd/E = eτ/m  |  J = I/A = nevd  |  J = σE
Worked example: A copper wire of cross-section 1 mm² carries 1 A. With n = 8.5 × 10²⁸ m⁻³, find the drift speed.
vd = I/(nAe) = 1 / (8.5 × 10²⁸ × 10⁻⁶ × 1.6 × 10⁻¹⁹) ≈ 7.4 × 10⁻⁵ m s⁻¹, or about 0.07 mm s⁻¹.
Common mistake: Believing electrons travel from battery to bulb in an instant. The signal travels fast; electrons themselves drift slowly.
Practice:
  1. Explain why conventional current is opposite to electron drift in a metal.
  2. If the temperature of a conductor rises, what happens to τ and vd at fixed E?
  3. Two wires of the same material have radii in the ratio 1:2 and carry equal current. Compare drift speeds.
Remember: Random thermal motion alone does not produce a net current.
11

Electrical Resistivity and Temperature Dependence

Ohm’s law states that for a conductor at constant temperature V = IR. Resistance R depends on length, area and material: R = ρL/A. The property ρ is the resistivity, a characteristic of the material. Microscopically, ρ = m/(ne²τ), so it increases when collisions become more frequent (smaller τ).

In metals, heating increases lattice vibrations, reducing τ, so resistivity rises roughly linearly with temperature. In semiconductors and insulators, heating releases many more carriers, so resistivity decreases (negative temperature coefficient). Alloys like manganin and constantan have very small temperature coefficients and are used in standard resistors. Some materials become superconducting, with zero resistance below a critical temperature.

Key formulas:
R = ρL/A  |  ρ = m/(ne²τ)  |  RT = R₀[1 + α(T − T₀)]
Series: R = ΣRᵢ  |  Parallel: 1/R = Σ1/Rᵢ  |  P = VI = I²R = V²/R
Worked example: A wire of resistance 10 Ω is uniformly stretched to twice its length. Find its new resistance.
Volume is constant, so area halves. R′ = ρ(2L)/(A/2) = 4ρL/A = 40 Ω. In general, stretching to n times the length multiplies resistance by n².
Common mistake: Doubling resistance when length doubles for a stretched wire. Area also changes when a wire is drawn out.
Practice:
  1. Predict the effect of doubling wire length while keeping area fixed.
  2. A copper coil has 50 Ω at 20 °C. Find its resistance at 100 °C if α = 4 × 10⁻³ °C⁻¹.
  3. Why are nichrome and manganin used for heaters and standard resistors respectively?
Remember: Resistance describes a component; resistivity characterises its material.
12

Kirchhoff’s Rules and Complex Electrical Circuits

Some circuits cannot be simplified using series and parallel rules. Kirchhoff’s two rules solve any network. The junction rule says that total current entering a junction equals total leaving it (conservation of charge). The loop rule says that the algebraic sum of potential changes around any closed loop is zero (conservation of energy).

Procedure: assign current directions arbitrarily, mark each element’s polarity, pick loops, and write the loop rule with a consistent sign convention. Crossing a resistor in the direction of current is a drop (−IR); against the current is a rise (+IR). Crossing a cell from − to + is a rise (+ε). Write as many independent equations as there are unknowns. Remember also that a real cell has internal resistance r, so its terminal voltage is V = ε − Ir while discharging.

Key relations:
ΣIin = ΣIout  |  ΣΔV = 0 around any loop
Terminal voltage: V = ε − Ir (discharging), V = ε + Ir (charging)
n cells in series: εeq = nε, req = nr  |  m identical cells in parallel: εeq = ε, req = r/m
Worked example: A 12 V cell of internal resistance 1 Ω drives a 5 Ω resistor. Find the current and terminal voltage.
I = ε/(R + r) = 12/6 = 2 A. V = ε − Ir = 12 − 2 = 10 V (also = IR = 10 V). In a second case, 10 V and 4 V cells opposing in series through 3 Ω give I = (10 − 4)/3 = 2 A in the direction set by the larger cell.
Common mistake: Changing sign convention midway around a loop. Choose one direction and stay consistent.
Practice:
  1. Write independent equations for a two-loop circuit with two cells and three resistors.
  2. Find the current through a cell when it is short-circuited.
  3. When is maximum power delivered to an external load? (Answer: R = r.)
Remember: A negative calculated current means its actual direction is opposite to the assumed direction.
13

Wheatstone Bridge and Electrical Measurement

A Wheatstone bridge has four resistors P, Q, R, S in a diamond with a galvanometer across one diagonal and a cell across the other. When the bridge is balanced, the two junctions connected to the galvanometer are at the same potential, so no current flows through it. This gives P/Q = R/S, which lets an unknown resistance be found without needing the cell’s emf or internal resistance, a strong advantage for accurate measurement.

The metre bridge is a practical form: a uniform wire of length 100 cm acts as the ratio arm. The potentiometer is another null-method device. It compares emfs (ε₁/ε₂ = l₁/l₂) and measures internal resistance without drawing current from the cell at balance, making it more accurate than a voltmeter.

Key formulas:
Balance: P/Q = R/S  |  Metre bridge: X = R(100 − l)/l
Potentiometer: ε₁/ε₂ = l₁/l₂  |  Internal resistance: r = R(l₁ − l₂)/l₂
Worked example: In a metre bridge, a known resistance R = 10 Ω is in the left gap and the null point is at 40 cm. Find the unknown X.
X = R(100 − l)/l = 10 × 60/40 = 15 Ω.
Common mistake: Placing the null point near the ends of the wire, which increases percentage error. Adjust R so that balance lies near the middle.
Practice:
  1. Explain why galvanometer current becomes zero at bridge balance.
  2. Why is a potentiometer preferred over a voltmeter for measuring emf?
  3. Why are copper connecting strips used at the metre-bridge ends?
Remember: The balanced bridge is a null-measurement method.

🧲 Magnetism and Electromagnetic Induction

14

Motion of Charged Particles in Magnetic Fields

A magnetic field exerts the Lorentz force F = qv × B on a moving charge. Because this force is always perpendicular to velocity, it changes the direction but never the speed or kinetic energy of the particle; the field does no work. A charge at rest feels no magnetic force.

If v is perpendicular to B, the particle moves in a circle because the magnetic force supplies the centripetal force. If v makes an angle with B, it follows a helix: circular motion due to the perpendicular component and uniform motion along the field due to the parallel component. If v is parallel to B, the particle goes undeflected. The time period of circular motion is independent of speed, the principle behind the cyclotron. When both fields exist, F = q(E + v × B), and a velocity selector passes only particles with v = E/B.

Key formulas:
F = qvB sinθ  |  r = mv/(qB) = p/(qB)  |  T = 2πm/(qB)  |  f = qB/(2πm)
Helix pitch: p = v cosθ · T  |  Velocity selector: v = E/B
Worked example: A proton moves at 2 × 10⁶ m s⁻¹ perpendicular to a 0.1 T field. Find the radius and period.
r = mv/(eB) = (1.67 × 10⁻²⁷ × 2 × 10⁶)/(1.6 × 10⁻¹⁹ × 0.1) ≈ 0.21 m. T = 2πm/(eB) ≈ 6.6 × 10⁻⁷ s.
Common mistake: Using the right-hand rule for a negative charge without reversing the result. For electrons the force is opposite to that on a positive charge.
Practice:
  1. Use the right-hand rule to predict force direction on a positive charge moving east in a field pointing north.
  2. An electron and a proton enter a field with equal momentum. Compare their radii.
  3. Why does the cyclotron frequency not change as the particle speeds up (non-relativistic)?
Remember: A magnetic field alone does no work on a charged particle.
15

Magnetic Force on Current-Carrying Conductors

A conductor carrying current in a magnetic field feels a force because the field acts on each moving charge carrier and the force is transmitted to the lattice. For a straight wire, F = IL × B. Two parallel conductors exert forces on each other: parallel currents attract and antiparallel currents repel. This effect defines the SI unit of current, the ampere.

A current-carrying loop in a uniform field experiences zero net force but a torque τ = m × B with magnetic moment m = NIA. This is the working principle of the moving-coil galvanometer, where a restoring spring torque balances magnetic torque so that deflection is proportional to current. A galvanometer becomes an ammeter when a small shunt resistance is placed in parallel, and a voltmeter when a large resistance is placed in series.

Fields produced by currents

Long straight wire: B = μ₀I/2πr. Centre of a circular loop: B = μ₀I/2R (for N turns, multiply by N). Solenoid: B = μ₀nI inside, where n is turns per unit length. Here μ₀ = 4π × 10⁻⁷ T m A⁻¹.

Key formulas:
F = BIL sinθ  |  τ = NIAB sinθ  |  F/L = μ₀I₁I₂/2πd = 2 × 10⁻⁷ I₁I₂/d
Shunt: S = IgG/(I − Ig)  |  Series resistance: R = V/Ig − G
Worked example: A 0.5 m wire carrying 4 A lies perpendicular to a 0.2 T field. Find the force. Then find the force per metre between two parallel wires 1 m apart each carrying 1 A.
F = BIL = 0.2 × 4 × 0.5 = 0.4 N. Between the wires: F/L = 2 × 10⁻⁷ × 1 × 1 / 1 = 2 × 10⁻⁷ N m⁻¹.
Common mistake: Placing an ammeter in parallel or a voltmeter in series. An ammeter needs low resistance in series; a voltmeter needs high resistance in parallel.
Practice:
  1. Find the orientation that gives maximum magnetic force on a straight wire.
  2. A galvanometer of resistance 50 Ω gives full-scale deflection at 2 mA. Convert it into a 0–1 A ammeter.
  3. Why is a radial magnetic field used in a moving-coil galvanometer?
Remember: Force is zero when current and magnetic field are parallel.
16

Magnetic Dipoles and Properties of Magnetic Materials

A current loop is equivalent to a magnetic dipole with moment m = NIA, directed by the right-hand rule. A bar magnet behaves similarly, and magnetic field lines form continuous closed loops, so no isolated magnetic monopole exists and the net magnetic flux through a closed surface is always zero (Gauss’s law for magnetism). In a uniform field a dipole has torque τ = mB sinθ and energy U = −m·B.

Materials respond to an applied field through their magnetisation M and magnetic susceptibility χ. Diamagnetic substances have small negative χ and are slightly repelled (bismuth, copper, water). Paramagnetic substances have small positive χ and are weakly attracted (aluminium, sodium); their susceptibility follows Curie’s law χ ∝ 1/T. Ferromagnetic substances (iron, cobalt, nickel) have very large positive χ due to domains, and become paramagnetic above the Curie temperature. Their magnetisation traces a hysteresis loop, which defines retentivity and coercivity.

Key formulas:
m = NIA  |  Axial field of bar magnet: B = μ₀ 2m / 4πr³  |  Equatorial: B = μ₀ m / 4πr³
B = μ₀(H + M)  |  M = χH  |  μr = 1 + χ
PropertyDiamagneticParamagneticFerromagnetic
Susceptibility χSmall, negativeSmall, positiveLarge, positive
Relative permeabilitySlightly < 1Slightly > 1≫ 1
In a fieldWeakly repelledWeakly attractedStrongly attracted
Worked example: A circular coil of 100 turns, area 0.02 m², carries 2 A. Find its magnetic moment and the maximum torque in a 0.5 T field.
m = NIA = 100 × 2 × 0.02 = 4 A m². τmax = mB = 4 × 0.5 = 2 N m.
Common mistake: Choosing steel for transformer cores. Cores need soft iron (low coercivity, narrow loop); permanent magnets need steel or alnico (high retentivity and coercivity).
Practice:
  1. Compare diamagnetic, paramagnetic and ferromagnetic behaviour.
  2. Why do magnetic field lines form closed loops?
  3. What does the area of a hysteresis loop represent?
Remember: Magnetic behaviour depends on microscopic magnetic moments and their alignment.
17

Electromagnetic Induction and Faraday’s Law

Faraday discovered that a changing magnetic flux through a circuit induces an emf. Magnetic flux is Φ = B·A = BA cosθ, with unit weber (Wb). Flux can change in three ways: by changing B, by changing the area A, or by changing the orientation angle θ. A steady flux, however large, induces nothing.

Faraday’s law states that the induced emf equals the rate of change of flux linkage: ε = −N dΦ/dt. A conductor of length l moving with speed v perpendicular to a uniform field has motional emf ε = Blv. The induced current through a closed circuit of resistance R is I = ε/R, and the charge that flows depends only on the total flux change, q = ΔΦ/R, not on how quickly it happens. Practical applications include AC generators, transformers, induction cooktops and magnetic braking by eddy currents.

Key formulas:
Φ = BA cosθ  |  ε = −N dΦ/dt  |  ε = Blv  |  q = NΔΦ/R
Rotating coil: ε = NBAω sin ωt, with ε₀ = NBAω
Worked example: A coil of 200 turns and area 0.05 m² sits perpendicular to a field that rises uniformly from 0 to 0.4 T in 0.2 s. Find the induced emf.
ε = NAΔB/Δt = 200 × 0.05 × 0.4 / 0.2 = 20 V. For a 0.5 m rod moving at 4 m s⁻¹ across a 0.3 T field, ε = Blv = 0.3 × 0.5 × 4 = 0.6 V.
Common mistake: Thinking a strong magnetic field itself induces emf. Only a change in flux does.
Practice:
  1. List three different ways of changing magnetic flux through a loop.
  2. A 20 cm square loop of resistance 2 Ω is pulled out of a field of 0.5 T in 0.1 s. Find the induced current.
  3. Explain how eddy currents are reduced by laminating a core.
Remember: A changing flux, not merely the presence of a magnetic field, produces induced emf.
18

Lenz’s Law and Induced Electric Fields

Lenz’s law gives the direction of the induced current: it flows so as to oppose the change in magnetic flux that caused it. This is not an extra rule but a consequence of conservation of energy. If the induced current aided the change, flux would grow without limit and energy would be created from nothing. The minus sign in Faraday’s law expresses exactly this opposition.

To use it: (1) decide whether flux through the loop is increasing or decreasing, (2) state the direction of the induced field that would oppose that change, (3) use the right-hand rule to find the current producing that field. Mechanical work done against the opposing force is converted into electrical energy and then heat. When flux changes in a region without a circuit, it induces a non-conservative electric field with ∮E·dl = −dΦ/dt, whose field lines are closed loops.

For a rod sliding on rails (length l, speed v, resistance R, field B):
ε = Blv  |  I = Blv/R  |  Magnetic drag F = B²l²v/R  |  Power = Fv = B²l²v²/R = I²R
Worked example: A rod of length 0.4 m moves at 5 m s⁻¹ along rails in a 0.5 T field; the circuit resistance is 2 Ω.
ε = 0.5 × 0.4 × 5 = 1 V; I = 0.5 A; F = BIl = 0.1 N; mechanical power = Fv = 0.5 W, equal to the heating I²R = 0.25 × 2 = 0.5 W. Energy is conserved.
Common mistake: Saying the induced field opposes the original field. It opposes the change; when flux decreases, the induced field points the same way as the original.
Practice:
  1. Predict induced current direction when a north pole is withdrawn from a coil.
  2. Explain why a magnet falls slowly through a copper pipe.
  3. Why is the induced electric field non-conservative?
Remember: Lenz’s law opposes the change in flux, not necessarily the flux itself.
19

Self-Inductance and Mutual Inductance

When current in a coil changes, its own flux changes and induces an emf opposing that change. This property is self-inductance L, defined by Φ = LI, with unit henry (H). It plays the role of electrical “inertia”. Two nearby coils show mutual inductance M: a changing current in one induces an emf in the other, ε₂ = −M dI₁/dt.

An inductor stores energy in its magnetic field. In an LR circuit, the current does not jump instantly but grows exponentially with time constant τ = L/R, reaching about 63% of its final value after one time constant. On breaking such a circuit, a large back emf can produce sparks across the switch.

Key formulas:
ε = −L dI/dt  |  Solenoid: L = μ₀N²A/l = μ₀n²Al  |  U = ½LI²
Mutual (coaxial solenoids): M = μ₀N₁N₂A/l  |  Growth: I = I₀(1 − e−t/τ), τ = L/R
Worked example: A solenoid has 500 turns, length 0.5 m and cross-section 4 cm². Find L and energy stored at 2 A.
L = μ₀N²A/l = 4π × 10⁻⁷ × (500)² × 4 × 10⁻⁴ / 0.5 ≈ 2.5 × 10⁻⁴ H (0.25 mH). U = ½LI² = ½ × 2.5 × 10⁻⁴ × 4 = 5 × 10⁻⁴ J.
Common mistake: Saying an inductor opposes current. It opposes change in current; steady DC passes through an ideal inductor freely.
Practice:
  1. Explain why an inductor opposes a sudden change in current.
  2. How does inserting an iron core affect L, and why?
  3. An L = 2 H, R = 10 Ω circuit is connected to a 20 V cell. Find the time constant and final current.
Remember: Inductors store energy in magnetic fields; capacitors store it in electric fields.

〰️ Alternating Current and Electromagnetic Waves

20

Alternating Current and Phasor Analysis

Alternating current (AC) varies sinusoidally with time, i = I₀ sin ωt, where ω = 2πf. Its average over a full cycle is zero, so the useful measure is the RMS value, defined so that it produces the same heating as an equal DC value. Household mains of 220 V is an RMS value; its peak is about 311 V.

Different components shift the phase between voltage and current. In a resistor, they are in phase. In an inductor, voltage leads current by 90° and the opposition is inductive reactance XL = ωL. In a capacitor, current leads voltage by 90° and the opposition is XC = 1/ωC. In a series LCR circuit a phasor diagram gives the impedance Z and phase angle φ. Average power depends on the power factor cosφ, so a pure inductor or capacitor consumes no power (wattless current).

Key formulas:
Irms = I₀/√2 ≈ 0.707 I₀  |  Vrms = V₀/√2  |  XL = ωL  |  XC = 1/ωC
Z = √[R² + (XL − XC)²]  |  tanφ = (XL − XC)/R  |  P = VrmsIrms cosφ
Worked example: A 220 V rms supply feeds R = 30 Ω in series with an inductor of reactance 40 Ω. Find Irms, power factor and power.
Z = √(30² + 40²) = 50 Ω; Irms = 220/50 = 4.4 A; cosφ = R/Z = 0.6; P = 220 × 4.4 × 0.6 ≈ 581 W, which equals I²R = 19.36 × 30 = 581 W.
Common mistake: Reading an AC ammeter as peak current. Meters read RMS values.
Practice:
  1. Compare phase relations in pure R, L and C circuits.
  2. Find XL of a 0.1 H inductor at 50 Hz.
  3. Why is the average power in a pure capacitor zero?
Remember: RMS values connect AC heating effects with equivalent DC values.
21

LCR Circuits and Electrical Resonance

In a series LCR circuit, XL increases with frequency while XC decreases. At one special frequency they are equal and cancel, so impedance becomes purely resistive and minimal (Z = R), and current is at its maximum. This is resonance, occurring at ω₀ = 1/√(LC). Voltage and current are then in phase and the power factor is 1.

The sharpness of resonance is described by the quality factor Q. A high Q (low resistance) gives a tall, narrow current peak and good frequency selectivity, while a low Q gives a broad, flat response. Radio tuning uses this: varying a capacitor changes the resonant frequency to select one station out of many. Note that individual voltages across L and C at resonance can far exceed the supply voltage, each being Q times larger.

Key formulas:
ω₀ = 1/√(LC)  |  f₀ = 1/(2π√(LC))  |  Imax = V/R
Q = ω₀L/R = 1/(ω₀RC) = (1/R)√(L/C)  |  Bandwidth Δω = R/L  |  Q = ω₀/Δω
Worked example: L = 2 mH, C = 0.5 μF, R = 10 Ω. Find resonant frequency and Q.
LC = 10⁻⁹ s², so ω₀ = 1/√(10⁻⁹) ≈ 3.16 × 10⁴ rad s⁻¹ and f₀ ≈ 5.03 kHz. Q = ω₀L/R = 3.16 × 10⁴ × 2 × 10⁻³ / 10 ≈ 6.3.
Common mistake: Using f = 1/(√LC) without the 2π, which gives angular frequency rather than frequency.
Practice:
  1. Explain why resonance is useful in radio tuning.
  2. What happens to resonant frequency if C is made four times larger?
  3. At resonance, what is the phase difference between supply voltage and current?
Remember: Resonance occurs at a particular frequency determined by L and C only, not by R.
22

Transformers and AC Power Transmission

A transformer consists of two coils wound on a common soft-iron core. An alternating current in the primary produces a changing flux in the core that links the secondary and induces an emf by mutual induction. With turns Np and Ns, the voltage ratio equals the turns ratio. A step-up transformer (Ns > Np) raises voltage and lowers current; a step-down transformer does the reverse. Energy is conserved, so an ideal transformer does not increase power.

Real transformers lose energy by copper loss (I²R heating of windings), eddy currents (reduced by laminating the core), hysteresis (reduced by using soft iron), and flux leakage. Efficiencies of 95–99% are common. Power is transmitted at very high voltage because for fixed power P = VI, higher V means lower current, and the line loss I²R falls as the square of the current.

Key formulas:
Vs/Vp = Ns/Np = Ip/Is (ideal)  |  Efficiency η = (Pout/Pin) × 100%
Line loss: Ploss = I²Rline = (P/V)²Rline
Worked example: A step-down transformer converts 2200 V to 220 V with 1000 primary turns. The secondary delivers 10 A. Find Ns and Ip.
Ns = 1000 × 220/2200 = 100. Ip = Is × Ns/Np = 1 A. Line loss: 100 kW sent through 5 Ω at 2000 V has I = 50 A and loss = 12.5 kW; at 20,000 V, I = 5 A and loss = 125 W, a hundred times smaller.
Common mistake: Connecting a transformer to a DC source. Without changing flux there is no induced emf, and the primary may burn out.
Practice:
  1. Explain why electrical power is transmitted at very high voltage.
  2. List four energy losses in a transformer and how each is reduced.
  3. A 90%-efficient transformer takes 2 kW. How much power reaches the load?
Remember: Ordinary transformers require changing current and operate with AC.
23

Electromagnetic Waves and Their Applications

Maxwell realised that Ampere’s law was incomplete: a changing electric field also produces a magnetic field, as though it were a current. This displacement current Id = ε₀ dΦE/dt completes the four Maxwell equations and predicts self-sustaining electromagnetic waves. In these waves E and B oscillate in phase, perpendicular to each other and to the direction of propagation (transverse), and need no medium. In vacuum all travel at c = 1/√(μ₀ε₀) ≈ 3 × 10⁸ m s⁻¹, and E₀/B₀ = c.

The waves carry energy and momentum. They are produced by accelerating charges, with radio waves generated by oscillating circuits. The spectrum is ordered by frequency, all being the same kind of wave.

RegionApprox. wavelengthTypical use
Radio waves> 0.1 mBroadcasting, communication
Microwaves1 mm – 0.1 mRadar, ovens, satellite links
Infrared700 nm – 1 mmRemote controls, thermal imaging
Visible400 – 700 nmVision, photography
Ultraviolet1 – 400 nmSterilisation, forensics
X-rays10⁻³ – 1 nmMedical imaging, crystallography
Gamma rays< 10⁻³ nmCancer therapy, nuclear studies
Key formulas:
c = fλ = E₀/B₀ = 1/√(μ₀ε₀)  |  Id = ε₀ dΦE/dt
Energy densities: uE = ½ε₀E², uB = B²/2μ₀ (equal in a wave)  |  Intensity I = ½ε₀cE₀²
Worked example: The electric field amplitude of a wave is 60 V m⁻¹. Find B₀. A radio station broadcasts at 100 MHz; find the wavelength.
B₀ = E₀/c = 60/(3 × 10⁸) = 2 × 10⁻⁷ T. λ = c/f = 3 × 10⁸ / 10⁸ = 3 m.
Common mistake: Calling sound an electromagnetic wave. Sound is a mechanical, longitudinal wave that needs a medium.
Practice:
  1. Arrange the main electromagnetic regions in increasing frequency.
  2. Why can electromagnetic waves travel through vacuum while sound cannot?
  3. Why do ozone layers matter for ultraviolet radiation?
Remember: Electromagnetic waves can travel through vacuum, are transverse, and all travel at speed c in vacuum.

🔭 Optics

24

Ray Optics and Advanced Optical Instruments

Ray optics treats light as straight rays that obey the laws of reflection and refraction (Snell’s law n₁ sinθ₁ = n₂ sinθ₂). Use the Cartesian sign convention: distances are measured from the pole or optical centre, positive along the direction of incident light. When light travels from a denser to a rarer medium at an angle above the critical angle (sin ic = 1/n), it is totally reflected. This effect drives optical fibres, prisms in binoculars, and the sparkle of diamonds.

Thin lenses form images by refraction, with focal length given by the lens maker’s formula. The power of a lens is the reciprocal of focal length in metres, measured in dioptres. In a compound microscope, a short-focus objective forms a magnified real image that the eyepiece magnifies further. In an astronomical telescope, a large-aperture objective collects light from a distant object and the eyepiece magnifies the image. Eye defects are corrected with lenses: myopia with a concave lens, hypermetropia with a convex lens.

Key formulas:
Mirror: 1/v + 1/u = 1/f  |  Lens: 1/v − 1/u = 1/f  |  m = v/u (lens), m = −v/u (mirror)
Lens maker: 1/f = (n − 1)(1/R₁ − 1/R₂)  |  P = 1/f (m)  |  Combination: P = P₁ + P₂
Prism: n = sin[(A + δm)/2] / sin(A/2)  |  Microscope (image at near point): M = (L/fo)(1 + D/fe)
Telescope (normal adjustment): M = fo/fe, length = fo + fe
Worked example: An object is 30 cm from a convex lens of focal length 20 cm. Find image position and magnification.
u = −30, f = +20: 1/v = 1/f + 1/u = 1/20 − 1/30 = 1/60, so v = +60 cm (real image beyond the lens). m = v/u = −2, so it is inverted and twice the size. For a telescope with fo = 100 cm and fe = 5 cm, M = 20 and tube length = 105 cm.
Common mistake: Ignoring signs. Always substitute u as negative for real objects, then interpret the sign of v and m.
Practice:
  1. Draw a labelled ray diagram for an astronomical telescope.
  2. Find the critical angle for glass of refractive index 1.5.
  3. A myopic person has a far point of 80 cm. Find the power of the corrective lens.
Remember: Total internal reflection requires travel from denser to rarer medium and an incidence angle above the critical angle.
25

Wave Optics, Interference and Diffraction

Huygens’ principle treats every point on a wavefront as a source of secondary wavelets; the new wavefront is their common tangent. It explains reflection and refraction and shows that light slows down in denser media. Interference occurs when two coherent waves (constant phase difference) superpose, giving bright fringes where path difference is nλ and dark fringes where it is (n + ½)λ.

In Young’s double-slit experiment, fringes of equal width β = λD/d appear on a screen. Diffraction is the bending of light around obstacles or through narrow openings of size comparable to the wavelength. In single-slit diffraction a broad central maximum is flanked by weaker, narrower maxima, with minima at a sinθ = nλ. Unlike interference fringes, diffraction fringes have unequal widths and intensities. Polarisation demonstrates that light is transverse: unpolarised light passing through a polariser emerges with half its intensity, and Malus’s law governs a second polariser.

Key formulas:
Fringe width β = λD/d  |  Position of nth bright fringe: y = nλD/d
I = 4I₀ cos²(φ/2)  |  Imax/Imin = (a₁ + a₂)²/(a₁ − a₂)²
Single slit minima: a sinθ = nλ; central maximum width = 2λD/a
Malus: I = I₀ cos²θ  |  Brewster: tanθB = n
Worked example: In Young’s experiment λ = 600 nm, slit separation d = 0.3 mm, D = 1.5 m. Find the fringe width.
β = λD/d = (600 × 10⁻⁹ × 1.5)/(3 × 10⁻⁴) = 3 × 10⁻³ m = 3 mm. If the whole setup is immersed in water (n = 1.33), the fringes shrink to β/n ≈ 2.25 mm.
Common mistake: Treating two separate bulbs as coherent sources. Interference needs sources with a fixed phase relationship, usually obtained from one source split in two.
Practice:
  1. State the path-difference conditions for constructive and destructive interference.
  2. What happens to fringe width if slit separation is doubled?
  3. Light of intensity I₀ passes through two polaroids with axes at 60°. Find the final intensity (assume I₀ is polarised light).
Remember: Interference is strong evidence for the wave nature of light; polarisation shows it is transverse.

⚛️ Modern Physics and Electronics

26

Photoelectric Effect and Photon Behaviour

When light of sufficiently high frequency strikes a metal surface, electrons are ejected. Experiments show that (1) there is a threshold frequency below which no emission occurs, however intense the light, (2) the maximum kinetic energy of electrons depends on frequency, not intensity, (3) the photocurrent is proportional to intensity, and (4) emission is essentially instantaneous. The wave theory cannot explain these results.

Einstein explained them by proposing that light consists of quanta, called photons, each of energy E = hν. One photon gives all its energy to one electron. Part of this, the work function φ, is needed to escape; the remainder becomes kinetic energy. Increasing intensity increases the number of photons, hence the number of electrons, while increasing frequency increases the energy per photon. The stopping potential V₀ is the retarding voltage needed to stop even the fastest electrons.

Key formulas:
E = hν = hc/λ  |  hν = φ + Kmax  |  Kmax = eV₀  |  ν₀ = φ/h
V₀–ν graph: slope = h/e  |  Photon momentum p = h/λ  |  hc ≈ 1240 eV nm
Worked example: Light of wavelength 400 nm falls on a metal with work function 2.0 eV. Find Kmax, stopping potential and threshold wavelength.
E = 1240/400 = 3.1 eV. Kmax = 3.1 − 2.0 = 1.1 eV, so V₀ = 1.1 V. λ₀ = 1240/2.0 = 620 nm; light longer than 620 nm will not eject electrons.
Common mistake: Thinking brighter light of low frequency can eject electrons if you wait long enough. Below threshold frequency, no electron is emitted at any intensity or time.
Practice:
  1. Distinguish the effects of increasing intensity and increasing frequency.
  2. Why does the wave theory fail to explain the absence of time lag?
  3. What quantity does the slope of a V₀–ν graph give, and what does the intercept on the ν axis give?
Remember: Below threshold frequency, photoemission does not occur regardless of intensity.
27

Matter Waves and Wave-Particle Duality

De Broglie extended the dual nature of light to matter: every moving particle has an associated wave with wavelength λ = h/p. For everyday objects this wavelength is far too small to observe, but for electrons, neutrons and atoms it is comparable to atomic spacings, giving measurable diffraction. Davisson and Germer confirmed this by scattering electrons from a nickel crystal, observing diffraction peaks that matched λ calculated from de Broglie’s relation.

Wave-particle duality means that radiation and matter show wave-like properties in propagation (interference, diffraction) and particle-like properties in interaction (photoelectric effect, collisions). De Broglie’s idea also explains Bohr’s quantisation: a stable orbit must hold a whole number of wavelengths, 2πr = nλ. Matter waves are not mechanical or electromagnetic waves; their intensity gives the probability of finding the particle.

Key formulas:
λ = h/p = h/mv = h/√(2mK)  |  Electron through V volts: λ = 1.227/√V nm
Charged particle: λ = h/√(2mqV)  |  h = 6.63 × 10⁻³⁴ J s
Worked example: Find the de Broglie wavelength of an electron accelerated through 150 V, and of a 0.15 kg ball moving at 30 m s⁻¹.
Electron: λ = 1.227/√150 ≈ 0.100 nm, comparable to atomic spacing. Ball: λ = 6.63 × 10⁻³⁴ / (0.15 × 30) ≈ 1.5 × 10⁻³³ m, undetectably small.
Common mistake: Using λ = h/v instead of h/p. Wavelength depends on momentum, which includes mass.
Practice:
  1. Compare the wavelengths of two particles with different momenta.
  2. An electron and a proton have equal kinetic energy. Which has the longer wavelength?
  3. Why do we not observe wave behaviour for a moving car?
Remember: Wave-particle duality applies to both matter and radiation. Increasing a particle’s momentum decreases its de Broglie wavelength.
28

Atomic Energy Levels and Spectral Transitions

Rutherford’s alpha-scattering experiment showed that an atom has a tiny, dense, positive nucleus with electrons around it. Classical physics predicts such an atom should collapse and emit a continuous spectrum, neither of which happens. Bohr’s model for hydrogen-like atoms fixes this with postulates: electrons move in certain stationary orbits without radiating, with angular momentum quantised as mvr = nh/2π, and radiation is emitted or absorbed only when an electron jumps between orbits, with photon energy equal to the energy difference.

The energy of level n is negative, indicating a bound state; the lowest level (n = 1) is the ground state. Spectral lines fall in series according to the lower level: Lyman (to n = 1, ultraviolet), Balmer (n = 2, visible), Paschen (n = 3, infrared), then Brackett and Pfund. Bohr’s model works well for hydrogen but cannot explain multi-electron atoms, fine structure or relative intensities, which require quantum mechanics.

Key formulas (hydrogen):
En = −13.6/n² eV  |  rn = 0.529 n² Å  |  vn ≈ 2.19 × 10⁶/n m s⁻¹
hν = Ei − Ef  |  1/λ = R(1/n₁² − 1/n₂²), R = 1.097 × 10⁷ m⁻¹
For hydrogen-like ions: En = −13.6 Z²/n² eV
Worked example: Find the wavelength of the first Balmer line (n = 3 → 2).
ΔE = 13.6(1/4 − 1/9) = 13.6 × 5/36 ≈ 1.89 eV. λ = 1240/1.89 ≈ 656 nm, a red line in the visible region. The ionisation energy of hydrogen from the ground state is 13.6 eV.
Common mistake: Forgetting the negative sign of energy levels, which leads to negative photon energies. Photon energy = Ehigher − Elower, a positive number.
Practice:
  1. Explain why atomic spectra contain specific lines rather than every wavelength.
  2. Find the shortest wavelength in the Lyman series.
  3. How many different spectral lines can arise when electrons fall from n = 4 to lower levels?
Remember: Spectral lines reflect quantised energy differences.
29

Nuclear Binding Energy and Radioactive Processes

The nucleus consists of protons and neutrons (nucleons) bound by the short-range, charge-independent strong nuclear force. Nuclear radius follows R = R₀A1/3 with R₀ ≈ 1.2 fm, giving a density of about 2.3 × 10¹⁷ kg m⁻³ that is nearly the same for all nuclei. The mass of a nucleus is less than the sum of its nucleons; this mass defect corresponds to the binding energy, via E = mc². The binding energy per nucleon peaks near 8.75 MeV around iron (A ≈ 56), so energy is released both when heavy nuclei split (fission) and when light nuclei combine (fusion).

Unstable nuclei decay spontaneously. In alpha decay the nucleus loses ⁴He (A − 4, Z − 2). In beta-minus decay a neutron converts to a proton (Z + 1, A unchanged), emitting an electron and an antineutrino. Gamma decay releases a photon with no change in A or Z. Decay is random for individual nuclei but follows an exponential law for large numbers.

Key formulas:
Δm = Zmp + (A − Z)mn − M  |  BE = Δm × 931.5 MeV/u
N = N₀e−λt = N₀(½)t/T  |  T½ = 0.693/λ  |  τ = 1/λ  |  Activity A = λN
Worked example: For ⁴He, Δm = 0.0304 u. Find binding energy per nucleon. Also, how much of an 80 g sample with half-life 5 days remains after 15 days?
BE = 0.0304 × 931.5 ≈ 28.3 MeV, so BE/A ≈ 7.07 MeV per nucleon. Time = 3 half-lives, so remaining = 80 × (½)³ = 10 g.
Common mistake: Saying a sample is completely gone after two half-lives. After two half-lives, one quarter remains.
Practice:
  1. Compare changes in mass number and atomic number during alpha and beta decay.
  2. ²³⁸U emits one alpha particle. Write the daughter nucleus.
  3. Explain, using the binding energy curve, why both fission and fusion release energy.
Remember: Radioactive decay is spontaneous and probabilistic for individual nuclei.
30

Semiconductor Devices and Modern Electronic Circuits

In solids, atomic levels merge into energy bands. In metals the valence and conduction bands overlap; in insulators a wide gap (> 3 eV) separates them; in semiconductors the gap is small (Si ≈ 1.1 eV, Ge ≈ 0.7 eV), so thermal energy promotes some electrons. An intrinsic semiconductor has equal electron and hole concentrations. Doping with pentavalent atoms (P, As) adds donor levels and makes n-type material; trivalent atoms (B, Al, In) create acceptor levels and make p-type material. In equilibrium nenh = ni².

Joining p and n regions creates a p-n junction with a depletion layer and a barrier potential (about 0.7 V in silicon, 0.3 V in germanium). In forward bias the barrier shrinks and current flows readily; in reverse bias the barrier grows and only a tiny leakage current flows, so a diode conducts in one direction and works as a rectifier. Special diodes include the Zener (voltage regulation in reverse breakdown), LED (emits light in forward bias), photodiode (light detector in reverse bias) and solar cell (converts light to electrical energy). Logic gates (AND, OR, NOT, NAND, NOR) form the basis of digital electronics; NAND and NOR are universal gates.

Key facts:
nenh = ni²  |  Half-wave rectifier ripple frequency = fin  |  Full-wave ripple frequency = 2fin
Logic: AND (Y = A·B), OR (Y = A + B), NOT (Y = Ā), NAND (Y = (A·B)‾), NOR (Y = (A + B)‾)
Worked example: Silicon with ni = 1.5 × 10¹⁶ m⁻³ is doped so that ne = 5 × 10²² m⁻³. Find the hole concentration.
nh = ni²/ne = (2.25 × 10³²)/(5 × 10²²) = 4.5 × 10⁹ m⁻³, far smaller than ne, so electrons are the majority carriers. For a 50 Hz input, a full-wave rectifier gives 100 Hz output ripple.
Common mistake: Saying an n-type semiconductor is negatively charged. It is electrically neutral overall; “n” only indicates the majority carriers are electrons.
Practice:
  1. Compare intrinsic, n-type and p-type semiconductors.
  2. Explain how a p-n junction diode acts as a rectifier with a neat circuit diagram.
  3. Draw the truth table of a NAND gate and show it can be used as a NOT gate.
Remember: Doping controls the concentration and type of majority charge carriers.

📋 Constants and Exam Tips

Important Constants

QuantitySymbolValue
Speed of lightc3 × 10⁸ m s⁻¹
Elementary chargee1.6 × 10⁻¹⁹ C
Coulomb constantk = 1/4πε₀9 × 10⁹ N m² C⁻²
Permittivity of free spaceε₀8.85 × 10⁻¹² C² N⁻¹ m⁻²
Permeability of free spaceμ₀4π × 10⁻⁷ T m A⁻¹
Planck’s constanth6.63 × 10⁻³⁴ J s
Electron massmₑ9.1 × 10⁻³¹ kg
Proton massmₚ1.67 × 10⁻²⁷ kg
Atomic mass unit1 u931.5 MeV/c² = 1.66 × 10⁻²⁷ kg
Rydberg constantR1.097 × 10⁷ m⁻¹
Electron volt1 eV1.6 × 10⁻¹⁹ J

Exam strategy

  • Start with a diagram. Ray diagrams, circuit diagrams and force vectors earn marks and prevent sign errors.
  • Write the formula first, then substitute. Marks are awarded for the correct formula even when arithmetic slips.
  • Check units and powers of ten. Convert μC, cm, mm and nF to SI before calculating.
  • Use limiting cases to test your answer. A formula that gives nonsense at x → 0 or x → ∞ is likely wrong.
  • Learn derivations with their diagrams. Typical long-answer questions include Gauss’s law applications, capacitor with dielectric, Biot–Savart results, resonance in LCR, Young’s experiment and Bohr’s model.
  • Revise definitions precisely. Terms such as “work function”, “mass defect” and “drift velocity” need exact wording.
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