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Physics flashcards

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Topic 1: A. Space, time and motion

Displacement vs distance: Displacement: the change in position, a vector; it can be zero or negative. Distance: the total path length, a scalar; never negative.
A trolley moves 60 m east, then 20 m west, in 16 s. Average speed and average velocity?: Average speed = 80 ÷ 16 = 5.0 m s⁻¹. Average velocity = 40 ÷ 16 = 2.5 m s⁻¹ east.
Average vs instantaneous velocity on a displacement–time graph: Average: the gradient of the chord joining two points. Instantaneous: the gradient of the tangent at that instant.
What do the gradient and the area of a velocity–time graph give?: Gradient: acceleration. Area: displacement (area below the time axis counts as negative).
Equations for uniformly accelerated motion: s = (u + v)t / 2 v = u + at s = ut + ½at² v² = u² + 2as Valid only while the acceleration is constant.
A car at 30 m s⁻¹ brakes at 6.0 m s⁻². Stopping time and distance?: t = 30 ÷ 6.0 = 5.0 s s = 30² ÷ (2 × 6.0) = 75 m
Projectile with no fluid resistance: horizontal and vertical motion: Horizontal: zero acceleration, constant velocity uₓ = u cos θ. Vertical: acceleration g downward, starting from u_y = u sin θ. At the top only the vertical velocity is zero.
Launch at 24 m s⁻¹, 40° above level ground. Time of flight and range?: uₓ = 18.4 m s⁻¹, u_y = 15.4 m s⁻¹ T = 2u_y ÷ g = 3.15 s Range = uₓT = 57.8 m
Qualitative effects of fluid resistance on a projectile: Smaller range and maximum height; the path is no longer a parabola and is steeper on the way down; the horizontal velocity keeps falling; the acceleration is not constant.
Explain how a falling body reaches terminal speed: As speed rises, drag rises, so the resultant force and the acceleration fall. When drag equals weight the acceleration is zero and the speed is constant.
Newton’s three laws of motion: First: a body stays at rest or at constant velocity unless a resultant force acts. Second: F = ma, or generally F = Δp/Δt. Third: the force of A on B is equal and opposite to the force of B on A.
Static and dynamic friction: Static: Ff ≤ μs FN. Dynamic (sliding): Ff = μd FN.
Elastic restoring force and buoyancy: Hooke’s law: FH = −kx. Buoyancy: Fb = ρVg (ρ the fluid’s density, V the volume displaced).
Linear momentum and impulse: p = mv. Impulse J = FΔt, equal to the change in momentum. Momentum is conserved unless a resultant external force acts.
A 0.5 kg ball at 4 m s⁻¹ is hit back at 6 m s⁻¹; contact time 0.02 s. Average force?: Δp = 0.5 × 6 − 0.5 × (−4) = 5 kg m s⁻¹ F = 5 ÷ 0.02 = 250 N
Elastic vs inelastic collisions: Momentum is conserved in both (no external force). Kinetic energy is conserved only in an elastic collision.
2 kg at 3 m s⁻¹ hits a stationary 1 kg trolley; they move off together at 2 m s⁻¹. Elastic?: Momentum: 6 = 6, conserved. KE before 9 J, after 6 J, so the collision is inelastic.
Centripetal acceleration: a = v²/r = ω²r = 4π²r/T², directed towards the centre. v = 2πr/T = ωr.
What provides the centripetal force?: Not a new force: it is whichever real forces (tension, gravity, friction, normal force) give the resultant towards the centre.
Work done by a constant force: W = Fs cos θ, θ the angle between force and displacement. Negative if θ > 90°; zero at 90°.
A 20 N force acts at 30° to a 4.0 m displacement. Work done?: W = 20 × 4.0 × cos 30° = 69.3 J
Kinetic, gravitational and elastic potential energy: Ek = ½mv² = p²/2m ΔEp = mgΔh (near the Earth’s surface) Ee = ½k(Δx)²
When is mechanical energy conserved?: When no frictional or other resistive forces act. Otherwise the change in mechanical energy equals the work done by the non-conservative forces.
Work done by the resultant force: Equals the change in kinetic energy, W = ΔEk.
Two expressions for power: P = ΔW/Δt, and P = Fv for a constant force at constant velocity.
Efficiency: η = useful energy output ÷ total energy input = useful power output ÷ total power input. No unit.
Energy density of a fuel: The energy released per unit volume of fuel (for example MJ per litre). Efficiency then says how much of it becomes useful output.
Sankey diagram: A diagram of energy flow in which arrow widths are proportional to energy; the outputs add up to the input.
Torque: The turning effect of a force about an axis, τ = Fr sin θ, in N m. (HL only)
Rotational equilibrium: Zero resultant torque. The body can still rotate at constant ω; it need not be at rest. (HL only)
Moment of inertia: How the distribution of mass about an axis resists a change in rotation, in kg m². For point masses I = Σmr². (HL only)
Newton’s second law for rotation: τ = Iα, where τ is the resultant torque and α the angular acceleration. (HL only)
Angular momentum, and when it is conserved: L = Iω, in kg m² s⁻¹. Conserved if no resultant external torque acts. (HL only)
Angular impulse: τΔt, equal to the change in angular momentum. (HL only)
Rotational kinetic energy: Ek = ½Iω² = L²/2I. (HL only)
A spinning body’s I falls to a third, with no external torque. What happens to ω and Ek?: L is conserved, so ω triples, and Ek = L²/2I also triples. The extra energy comes from work done on the body. (HL only)
A hoop (I = mr²) rolls without slipping from rest down a drop of 0.80 m. Speed at the bottom?: mgh = ½mv² + ½mv² = mv², so v = √(gh) = √(9.81 × 0.80) = 2.80 m s⁻¹. (HL only)
Inertial frame: A frame of reference that is not accelerating. (HL only)
The two postulates of special relativity: The laws of physics are the same in every inertial frame. The speed of light in a vacuum is the same in all inertial frames, whatever the motion of source or observer. (HL only)
Galilean transformation: x′ = x − vt, t′ = t, u′ = u − v. It works at low speeds but fails for light. (HL only)
Lorentz factor and Lorentz transformations: γ = 1/√(1 − v²/c²) x′ = γ(x − vt) t′ = γ(t − vx/c²) (HL only)
Proper time interval: The time between two events measured in the frame where both happen at the same place. It is the shortest interval; Δt = γΔt₀. (HL only)
Proper length: The length of an object measured in the frame where it is at rest. It is the longest; L = L₀/γ along the motion. (HL only)
A clock in a spacecraft ticks every 3.0 μs; the craft moves at 0.80c. Interval measured outside?: γ = 1/√(1 − 0.64) = 1.667, so Δt = 3.0 × 1.667 = 5.0 μs. (HL only)
Relativistic velocity addition: u′ = (u − v) / (1 − uv/c²). The result is always below c. (HL only)
Space-time interval: (Δs)² = (cΔt)² − (Δx)². It is invariant: the same in all inertial frames. (HL only)
Muon evidence for special relativity: In the Earth frame, the muon lifetime is dilated. In the muon frame, the distance is contracted. Both lead to the same result. (HL only)

Topic 2: B. The particulate nature of matter

Internal energy: The sum, over all molecules, of the intermolecular potential energy and the random kinetic energy.
Temperature and average kinetic energy: Temperature (in kelvin) measures the average random kinetic energy per particle: Ek = (3/2)kB T. It is not a measure of total internal energy.
Average kinetic energy of a gas particle at 620 K? (kB = 1.38 × 10⁻²³ J K⁻¹): Ek = 1.5 × 1.38 × 10⁻²³ × 620 = 1.28 × 10⁻²⁰ J
Specific heat capacity and specific latent heat equations: Q = mcΔT (temperature change, no change of phase) Q = mL (change of phase at constant temperature)
What happens to the particles during a phase change?: The temperature and average kinetic energy stay constant; the energy supplied increases the potential energy as the particles separate.
Evaporation vs boiling: Evaporation: at the surface, at any temperature. Boiling: throughout the liquid, at the boiling point.
Rate of thermal conduction: ΔQ/Δt = kA ΔT/Δx, where k is the thermal conductivity (W m⁻¹ K⁻¹), A the area and ΔT/Δx the temperature gradient.
Convection: Heated fluid expands and becomes less dense, so it rises while cooler, denser fluid sinks to replace it, forming a convection current. It needs a fluid.
Black body: An ideal body that absorbs all radiation incident on it and emits the maximum power for its temperature.
Stefan–Boltzmann law, and the effect of doubling T: L = σAT⁴ (σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴, T in kelvin). Doubling T multiplies L by 16.
Apparent brightness and Wien’s law: b = L / (4πd²), in W m⁻². λmax T = 2.9 × 10⁻³ m K: a hotter body peaks at a shorter wavelength.
Emissivity: The power radiated per unit area by a surface divided by that of a black body at the same temperature (σT⁴). No unit. P = eσAT⁴.
Albedo: Total scattered (reflected) power divided by total incident power. No unit, between 0 and 1. Fraction absorbed = 1 − α.
Solar constant, and why the mean incoming intensity is S/4: S: the intensity of the Sun’s radiation at the planet’s distance, on a surface facing the Sun at the top of the atmosphere. The planet intercepts S × πR² but has surface area 4πR².
Equilibrium temperature of a planet with no atmosphere: Absorbed = emitted: S(1 − α)/4 = eσT⁴, so T = [S(1 − α)/(4eσ)]^(1/4).
S = 1800 W m⁻², α = 0.50, e = 1. Equilibrium temperature?: Absorbed = 1800 × 0.50 ÷ 4 = 225 W m⁻². T⁴ = 225 ÷ 5.67 × 10⁻⁸ = 3.97 × 10⁹, so T = 251 K.
Four main greenhouse gases: Carbon dioxide (CO₂), methane (CH₄), nitrous oxide (N₂O) and water vapour (H₂O).
Why do greenhouse gases absorb infrared but not sunlight?: Their molecules have vibrational energy gaps (or natural frequencies) matching infrared photons. Sunlight peaks near 500 nm and passes through; the surface’s radiation peaks near 10 μm and is absorbed.
How do greenhouse gases warm the surface?: They absorb infrared from the surface and re-emit it in all directions, so part returns to the surface. The surface warms until the power escaping to space equals the absorbed solar power.
Enhanced greenhouse effect: The extra warming caused by greenhouse gases added by human activity (fossil fuels, deforestation, livestock, fertilisers).
Pressure and amount of substance: P = F/A, F perpendicular to the surface (Pa). n = N/NA, with NA = 6.02 × 10²³ mol⁻¹.
The gas laws for a fixed mass of gas: PV/T = constant (T in kelvin). Constant T: PV = constant. Constant P: V/T = constant. Constant V: P/T = constant.
Ideal gas equation in two forms: PV = nRT (R = 8.31 J K⁻¹ mol⁻¹) PV = NkBT (kB = 1.38 × 10⁻²³ J K⁻¹)
Amount of gas in 3.1 × 10⁻³ m³ at 1.5 × 10⁵ Pa and 280 K?: n = PV/RT = (1.5 × 10⁵ × 3.1 × 10⁻³) ÷ (8.31 × 280) = 0.20 mol
Pressure from kinetic theory, and mean molecular kinetic energy: P = (1/3)ρv², where v² is the mean of the squared speeds. ½mv² = (3/2)kBT per molecule.
Internal energy of a monatomic ideal gas: U = (3/2)NkBT = (3/2)nRT: only random kinetic energy, because there are no intermolecular forces.
The temperature of a gas is quadrupled. What happens to the rms speed of its molecules?: It doubles, because rms speed ∝ √T.
When does a real gas behave most like an ideal gas?: At low pressure, low density and high temperature (well above its condensation point), where molecular volume and intermolecular forces are negligible.
A graph of V against temperature at constant pressure meets the temperature axis where?: At 0 K, which is about −273 °C.
First law of thermodynamics, with the sign convention: Q = ΔU + W. Q positive for heat supplied to the gas; W positive for work done by the gas (expansion); ΔU positive when the temperature rises. (HL only)
Closed system: Matter cannot cross the boundary, but energy can cross it in either direction, as heat or as work. (HL only)
Adiabatic: No thermal energy is transferred to or from the gas, so Q = 0 and ΔU = −W. (HL only)
Work done by a gas, and change of internal energy of a monatomic ideal gas: W = PΔV at constant pressure; in general, the area under the P–V graph. ΔU = (3/2)nRΔT. (HL only)
Isothermal vs adiabatic change of a monatomic ideal gas: Isothermal: ΔU = 0, so Q = W. Adiabatic: Q = 0, temperature changes, PV^(5/3) = constant; its curve is steeper on a P–V graph. (HL only)
Entropy: two expressions: ΔS = ΔQ/T (T in kelvin). S = kB ln Ω, where Ω is the number of microstates. (HL only)
Second law of thermodynamics: Clausius, Kelvin and entropy forms: Clausius: heat does not flow by itself from cold to hot. Kelvin: no cyclic device can turn all the energy from one reservoir into work. Entropy of an isolated system never decreases. (HL only)
750 J flows from a reservoir at 360 K to one at 250 K. Total entropy change?: −750/360 + 750/250 = −2.08 + 3.00 = +0.917 J K⁻¹. Positive, so the flow is allowed. (HL only)
Heat engine efficiency and the Carnot limit: η = W/Qh, with W = Qh − Qc. η_Carnot = 1 − Tc/Th (kelvin). For Th = 900 K and Tc = 300 K, η_Carnot = 0.667. (HL only)
Current, potential difference and emf: Current: rate of flow of charge, I = Δq/Δt. P.d.: work done per unit charge, V = W/q. Emf: energy transferred to each coulomb of charge by the cell.
Resistance, Ohm’s law and resistivity: R = V/I. Ohm’s law: at constant temperature, current is directly proportional to p.d. R = ρL/A, where ρ (Ω m) is a property of the material.
A wire of resistance 8.0 Ω has area 2.0 × 10⁻⁷ m² and length 1.5 m. Resistivity?: ρ = RA/L = 8.0 × 2.0 × 10⁻⁷ ÷ 1.5 = 1.1 × 10⁻⁶ Ω m
Origin of resistance in a metal: Electrons driven by the p.d. collide with vibrating lattice ions and transfer energy to them. Hotter means larger vibrations, more collisions and higher resistance.
Electrical power, and resistors in series and parallel: P = IV = I²R = V²/R Series: Rs = R₁ + R₂ + … Parallel: 1/Rp = 1/R₁ + 1/R₂ + …
Real cell: emf, internal resistance and terminal p.d.: ε = I(R + r), so terminal p.d. V = ε − Ir. A graph of V against I has intercept ε and gradient −r.
A 1.5 V cell with r = 0.60 Ω supplies 0.50 A. Terminal p.d.?: Lost volts Ir = 0.30 V, so V = 1.5 − 0.30 = 1.2 V
I–V graphs of a filament lamp and a thermistor: Filament lamp: bends towards the V axis; heating raises R. Thermistor (common type): bends towards the I axis as it heats; more charge carriers are freed, so R falls.
Resistance of an ideal ammeter and an ideal voltmeter: Ammeter: zero (connected in series). Voltmeter: infinite (connected in parallel).
Chemical cell vs solar cell: Both are sources of emf. A chemical cell converts stored chemical energy; a solar cell converts the energy of absorbed light, so its output depends on light intensity.

Topic 3: C. Wave behaviour

Conditions for simple harmonic motion, and its defining equation: The net force acts towards equilibrium and is proportional to displacement (F = −kx). a = −ω²x.
Period, frequency and angular frequency: T = 1/f = 2π/ω, with ω = 2πf in rad s⁻¹.
Periods of a mass–spring system and a simple pendulum: T = 2π√(m/k) T = 2π√(l/g) (small angles) Neither depends on the amplitude.
A 0.50 kg mass on a spring with k = 200 N m⁻¹. Period and ω?: T = 2π√(0.50/200) = 0.314 s; ω = 2π/T = 20 rad s⁻¹
Where in SHM are the speed and the acceleration greatest?: Speed is greatest at equilibrium, where acceleration is zero. Acceleration is greatest at the turning points, where speed is zero.
Energy changes during one oscillation (no damping): Kinetic energy is maximum at equilibrium and zero at the turning points; potential energy does the opposite; the total stays constant.
Displacement and velocity as functions of time in SHM: x = x₀ sin(ωt + φ) v = ωx₀ cos(ωt + φ) (HL only)
Speed at displacement x, and total energy, in SHM: v = ±ω√(x₀² − x²) E_T = ½mω²x₀² (HL only)
ω = 10 rad s⁻¹, amplitude 0.030 m. Speed at x = 0.020 m?: v = 10 × √(0.030² − 0.020²) = 0.224 m s⁻¹ (HL only)
Travelling wave: A disturbance that transfers energy through a medium or space without the medium as a whole being shifted.
Transverse vs longitudinal waves: Transverse: particles oscillate at right angles to the direction of travel. Longitudinal: particles oscillate along it, forming compressions and rarefactions.
Wavelength, and the wave equation: Wavelength: the distance between neighbouring points in phase. v = fλ = λ/T.
Sound vs electromagnetic waves: Sound: longitudinal, needs a medium, speed depends on the medium. EM: transverse (perpendicular electric and magnetic fields), needs no medium, all travel at c in a vacuum.
A wave has period 0.125 s and wavelength 0.90 m. Speed and frequency?: v = 0.90 ÷ 0.125 = 7.2 m s⁻¹; f = 1/T = 8.0 Hz
Electromagnetic spectrum, long wavelength to short: Radio, microwave, infrared, visible (about 4 to 7 × 10⁻⁷ m), ultraviolet, X-ray, gamma. Frequency increases in that order.
On a longitudinal wave’s displacement graph, where is a compression?: At a point of zero displacement where the particles on either side are displaced towards it (a rarefaction: displaced away from it).
Wavefront and ray: Wavefront: a line or surface joining points at the same phase. Ray: the direction of travel and energy transfer, perpendicular to the wavefronts.
Snell’s law and refractive index: n₁ sin θ₁ = n₂ sin θ₂, i.e. n₁/n₂ = sin θ₂/sin θ₁ = v₂/v₁. n = c/v.
What changes at a boundary when a wave refracts?: Speed and wavelength change (and usually direction). Frequency stays the same.
Critical angle from n = 1.45 into n = 1.20?: sin C = 1.20 ÷ 1.45 = 0.8276, so C = 55.9°. Total internal reflection if the angle of incidence exceeds C.
When is diffraction significant?: When the gap or obstacle is about the same size as the wavelength or smaller. Speed, wavelength and frequency do not change.
Principle of superposition: Where waves meet, the resultant displacement is the sum of the individual displacements, signs included.
Conditions for constructive and destructive interference (coherent sources in phase): Constructive: path difference = nλ. Destructive: path difference = (n + ½)λ. Coherent: same frequency, constant phase difference.
Young’s double slit: λ = 589 nm, d = 0.45 mm, D = 1.35 m. Fringe separation?: s = λD/d = 589 × 10⁻⁹ × 1.35 ÷ 0.45 × 10⁻³ = 1.77 × 10⁻³ m
Single-slit diffraction: angle of the first minimum: θ = λ/b (radians), b the slit width. A narrower slit gives a wider, dimmer central maximum. (HL only)
Diffraction grating equation, with an example: nλ = d sin θ. 700 lines per mm gives d = 1.43 × 10⁻⁶ m; a second order at 40.0° gives λ = d sin θ / 2 = 459 nm. (HL only)
Standing wave, node and antinode: The superposition of two matching waves travelling in opposite directions; no net energy is carried along. Node: permanently zero amplitude. Antinode: maximum amplitude.
Distances in a standing wave: Node to adjacent node: λ/2. Node to adjacent antinode: λ/4.
Allowed frequencies for a string fixed at both ends and a pipe open at both ends: f = nv/2L, all harmonics n = 1, 2, 3, …
Allowed frequencies for a pipe closed at one end: f = nv/4L, odd harmonics only (n = 1, 3, 5, …).
A string of length 0.80 m has wave speed 240 m s⁻¹. First harmonic?: f₁ = 240 ÷ (2 × 0.80) = 150 Hz
Resonance: A large-amplitude response when the driving frequency is at or very near a natural frequency of the system.
Light, critical and heavy damping: Light: several oscillations of decreasing amplitude. Critical: fastest return to equilibrium with no overshoot. Heavy: no oscillation, but a slower return than critical.
Effect of increasing damping on a resonance curve: Lower peak amplitude, a broader peak, and the peak moves slightly below the natural frequency.
Doppler effect: The change in observed frequency (and wavelength) when the source and observer are in relative motion.
Moving source vs moving observer: Moving source: wavefronts bunch up in front (shorter λ, higher f) and spread out behind. Moving observer: λ is unchanged, but wavefronts are met more or less often.
Doppler shift for light (v much less than c): Δf/f = Δλ/λ ≈ v/c. A positive Δλ (redshift) means recession; blueshift means approach.
Doppler formula for sound from a moving source: f′ = f v / (v ± us): minus when the source moves towards the observer, plus when it moves away. (HL only)
A 300 Hz source recedes at 17 m s⁻¹ (v = 340 m s⁻¹). Observed frequency?: f′ = 300 × 340 ÷ (340 + 17) = 286 Hz (HL only)
Doppler formula for sound with a moving observer: f′ = f (v ± uo) / v: plus when moving towards the source, minus when moving away. (HL only)
Two applications of the Doppler effect: Doppler ultrasound: the shifted echo from moving blood gives its speed and direction. Radar speed checks: microwaves reflected from a moving vehicle are shifted.

Topic 4: D. Fields

Kepler’s three laws: First: orbits are ellipses with the Sun at a focus. Second: the Sun–planet line sweeps out equal areas in equal times. Third: T² ∝ r³.
Newton’s law of gravitation: F = Gm₁m₂/r², attractive, along the line joining the centres (r centre to centre). G = 6.67 × 10⁻¹¹ N m² kg⁻².
Gravitational field strength: g = F/m, the force per unit mass on a small point mass (N kg⁻¹). For a point mass or uniform sphere, g = GM/r².
g at 500 km above a planet of mass 6.0 × 10²³ kg and radius 3.0 × 10⁶ m?: r = 3.5 × 10⁶ m, g = 6.67 × 10⁻¹¹ × 6.0 × 10²³ ÷ (3.5 × 10⁶)² = 3.27 N kg⁻¹
A second moon orbits at twice the radius of a moon with period 2.0 days. Its period?: T ∝ r^(3/2), so T = 2.0 × 2^(3/2) = 5.7 days
When can an extended body be treated as a point mass?: When it is spherically symmetric (such as a uniform sphere), or the separation is much larger than the size of either body.
Gravitational potential at a point: The work done per kilogram to bring a small test mass from infinity to the point: Vg = −GM/r. Zero at infinity, negative everywhere else. (HL only)
Escape speed: The speed at a point that lets a body reach infinity with no kinetic energy left: vesc = √(2GM/r), independent of the body’s mass. (HL only)
Orbital speed and total energy of a satellite in a circular orbit: v = √(GM/r). Ek = GMm/2r, Ep = −GMm/r, so E = −GMm/2r. (HL only)
Atmospheric drag acts on a satellite. Effect on its orbit?: Total energy decreases, so the radius decreases and the speed increases (Ek rises while Ep falls by twice as much). (HL only)
Coulomb’s law: F = kq₁q₂/r², k = 1/(4πε₀) ≈ 8.99 × 10⁹ N m² C⁻². Like charges repel, unlike attract.
Force between 2.0 μC and 3.0 μC point charges 0.10 m apart?: F = 8.99 × 10⁹ × 2.0 × 10⁻⁶ × 3.0 × 10⁻⁶ ÷ 0.10² = 5.4 N (repulsive)
What did Millikan’s experiment show?: Balanced charged oil drops (qE = mg) always carried whole-number multiples of e = 1.60 × 10⁻¹⁹ C, so charge is quantised.
Three ways to charge a body: Friction: electrons transfer, giving equal and opposite charges. Contact: charge is shared, same sign. Induction (with earthing): opposite sign to the charged body.
Electric field strength: E = F/q, the force per unit positive test charge (N C⁻¹ or V m⁻¹). Point charge: E = kQ/r². Parallel plates: E = V/d, uniform.
600 V across plates 1.5 cm apart. Field strength?: E = 600 ÷ 0.015 = 4.0 × 10⁴ V m⁻¹
Field inside and outside a charged conducting sphere: Charge sits on the surface; E = 0 inside; outside, the field is radial, as if the charge were a point at the centre.
Electric field lines vs magnetic field lines: Electric lines start on positive and end on negative charges. Magnetic lines form closed loops (outside a magnet from N to S).
Electric potential, and its value for a point charge: The work done per unit charge to bring a small positive test charge from infinity to the point. Ve = kQ/r, a scalar with the sign of Q. (HL only)
Electric potential energy of two point charges, and field from potential: Ep = kq₁q₂/r (negative for unlike charges). E = −ΔVe/Δr: the field points towards falling potential. (HL only)
Why is no work done moving a charge along an equipotential?: ΔVe = 0, so W = qΔVe = 0. Field lines cross equipotentials at 90°. (HL only)
Force and acceleration of a charge in an electric field: F = qE, a = qE/m. Along E for a positive charge, against it for a negative one. Crossing a uniform field at right angles gives a parabolic path.
Magnetic force on a moving charge and on a conductor: F = qvB sin θ F = BIL sin θ θ is the angle between v (or I) and B.
Why does a magnetic field not change a charged particle’s speed?: The force is always at right angles to the velocity, so it does no work. Only the direction changes.
Radius and period of a charged particle’s circular path in a magnetic field: qvB = mv²/r gives r = mv/(qB). T = 2πm/(qB), independent of speed.
Velocity selector: Crossed electric and magnetic fields: a particle goes straight through when qE = qvB, i.e. v = E/B, whatever its charge or mass.
Force between two parallel current-carrying wires: F/L = μ₀I₁I₂ / (2πr), μ₀ = 4π × 10⁻⁷ T m A⁻¹. Same-direction currents attract; opposite currents repel.
10 A and 5.0 A in the same direction, 2.0 cm apart. Force per metre?: F/L = 4π × 10⁻⁷ × 10 × 5.0 ÷ (2π × 0.020) = 5.0 × 10⁻⁴ N m⁻¹, attractive
Magnetic flux: Φ = BA cos θ, in Wb (T m²), θ the angle between B and the normal to the area. (HL only)
Faraday’s law of induction: ε = −N ΔΦ/Δt: the induced emf equals the rate of change of flux linkage. (HL only)
Emf induced in a straight conductor moving through a field: ε = BvL, with B, v and the conductor mutually perpendicular. (HL only)
Lenz’s law: The induced current opposes the change in flux that produces it, as conservation of energy requires (work must be done to produce the electrical energy). (HL only)
A 60-turn coil’s flux rises uniformly from 0 to 4.0 × 10⁻³ Wb in 0.50 s. Emf?: 60 × 4.0 × 10⁻³ ÷ 0.50 = 0.48 V (zero while the flux then stays constant). (HL only)
Emf of a coil rotating in a uniform field: ε = NBAω sin ωt, peak NBAω. Zero when the coil’s plane is perpendicular to B (flux maximum); peak when parallel. (HL only)
A generator’s rotation frequency is increased by a factor k. Effect on its output?: Peak emf × k and period ÷ k: a taller, narrower sine wave. (HL only)

Topic 5: E. Nuclear and quantum physics

Geiger–Marsden–Rutherford scattering: observations and conclusions: Nearly all alpha particles go straight through: mostly empty space. A few deflect through large angles; very few bounce back: a tiny, massive, positive nucleus.
Nucleon number, proton number and neutron number: A: protons + neutrons. Z: protons. N = A − Z. A neutral atom has Z electrons.
Continuous, emission and absorption spectra: Continuous: all wavelengths (hot, dense solid). Emission: bright lines (excited gas, electrons falling). Absorption: dark lines on a continuous spectrum (cool gas absorbing matching photons).
What do line spectra show about atoms?: Energy levels are discrete. A photon emitted or absorbed has energy equal to the gap between two levels: E = hf, λ = hc/ΔE.
Levels at −8.0, −3.0 and −1.0 eV. Is a 4.0 eV photon absorbed from the ground state?: No: the gaps are 5.0, 7.0 and 2.0 eV, and none equals 4.0 eV.
The electronvolt: 1 eV = 1.60 × 10⁻¹⁹ J, the energy gained by charge e moved through 1 V.
Nuclear radius and nuclear density: R = R₀A^(1/3) (R₀ ≈ 1.2 fm). The volume is proportional to A, so all nuclei have about the same density. (HL only)
Distance of closest approach of an alpha particle to a nucleus: All the kinetic energy becomes electric potential energy: Ek = k(2e)(Ze)/d, so d = 2kZe²/Ek. (HL only)
Bohr model energy levels of hydrogen, and the n = 3 to n = 2 photon: E = −13.6/n² eV (with mvr = nh/2π). 3 → 2: 13.6 × (1/4 − 1/9) = 1.89 eV, about 658 nm. (HL only)
Threshold frequency and work function: Threshold frequency f₀: the lowest frequency that releases photoelectrons from a given metal. Work function Φ: the least energy needed to free an electron at the surface. hf₀ = Φ. (HL only)
Einstein’s photoelectric equation: Emax = hf − Φ. One photon gives all its energy to one electron; Φ is used to free it and the rest is kinetic energy. (HL only)
Three photoelectric observations that wave theory cannot explain: A threshold frequency, whatever the intensity; Emax rises with frequency but not intensity; emission with no measurable delay, even in faint light. (HL only)
Graph of stopping potential against frequency: A straight line: gradient h/e, horizontal intercept f₀, vertical intercept −Φ/e. All metals give parallel lines. (HL only)
360 nm light needs a stopping potential of 0.80 V. Work function?: Photon energy = hc/λ = 5.52 × 10⁻¹⁹ J = 3.45 eV, so Φ = 3.45 − 0.80 = 2.65 eV. (HL only)
De Broglie wavelength, and the evidence for matter waves: λ = h/p. Electrons (and neutrons, atoms) diffract, and diffraction is a wave property. For an electron accelerated through V, p = √(2meV). (HL only)
Compton scattering: A photon scatters from an electron with a longer wavelength: Δλ = (h/mₑc)(1 − cos θ), h/mₑc = 2.43 × 10⁻¹² m. It shows photons carry momentum. (HL only)
Mass defect and binding energy: Mass defect: mass of the separated nucleons minus the mass of the nucleus. Binding energy: its energy equivalent, the energy needed to separate the nucleus into free nucleons. 1 u = 931.5 MeV/c².
Shape of the binding energy per nucleon curve: Rises steeply for light nuclides (high point at helium-4), broad peak near A = 56 to 62 at about 8.8 MeV per nucleon, then falls slowly to about 7.6 MeV at uranium.
Strong nuclear force: An attractive force between nucleons acting over only a few femtometres.
Alpha, beta-minus, beta-plus and gamma decay: α: He-4 nucleus, A − 4, Z − 2. β⁻: electron + antineutrino, Z + 1. β⁺: positron + neutrino, Z − 1. γ: photon, no change to A or Z.
Random and spontaneous decay: Random: which nucleus decays next, and when, cannot be predicted. Spontaneous: not triggered or changed by outside conditions such as temperature or pressure.
Activity vs count rate, and correcting for background: Activity: decays per unit time in the source (Bq). Count rate: what the detector registers. Subtract the background count rate before halving or plotting.
A source of 1280 Bq has a half-life of 20 min. Activity after 1.0 h?: 60 ÷ 20 = 3 half-lives, so 1280 ÷ 8 = 160 Bq
Radioactive decay law and the decay constant: N = N₀e^(−λt), A = λN, T½ = ln 2 / λ. λ is the probability per unit time that a nucleus decays. (HL only)
Why does the continuous beta spectrum show a neutrino exists?: Beta electrons have a range of energies up to a maximum, so a third particle (the neutrino or antineutrino) must share the energy. (HL only)
Spontaneous vs neutron-induced fission: Spontaneous: no trigger needed; rare. Neutron-induced: the nucleus (such as uranium-235) absorbs a neutron and then splits into two fragments, usually releasing two or three neutrons.
Why does fission release energy?: The fragments have a higher binding energy per nucleon than the parent, so the products have less mass. Energy released = Δm c², mostly as kinetic energy of the fragments.
U-235 + n → Xe-140 + Sr-94 + 2n loses 0.198264 u. Energy released?: 0.198264 × 931.5 = 185 MeV (2.96 × 10⁻¹¹ J)
Subcritical, critical and supercritical chain reactions: k = average number of neutrons per fission causing another fission. k < 1: dies away. k = 1: steady (a power plant). k > 1: grows.
Role of the moderator vs the control rods: Moderator: slows fast neutrons by elastic collisions so they can cause fission. Control rods: absorb neutrons to keep k = 1, or below 1 to shut down.
Role of the heat exchanger and of shielding: Heat exchanger: passes thermal energy from the radioactive primary coolant to a second circuit that makes steam. Shielding: absorbs neutrons and gamma rays from the core.
Why is there a critical mass, and why is uranium enriched?: Bigger pieces lose a smaller fraction of neutrons through the surface. Natural uranium is only about 0.7% uranium-235; enrichment raises this fraction.
Why is a main sequence star stable?: Outward pressure from hot gas and radiation balances inward gravity at every layer. If the core cools the star contracts and heats; if it heats it expands and cools.
Conditions needed for fusion in a star: High temperature, so nuclei can overcome their electric repulsion, and high density, so collisions are frequent.
Four hydrogen atoms (1.007825 u each) fuse to helium-4 (4.002603 u). Energy released?: Mass lost = 0.028697 u, so 0.028697 × 931.5 = 26.7 MeV (about 0.7% of the mass).
Hertzsprung–Russell diagram: axes and main regions: Luminosity up, temperature across (increasing to the left). Main sequence: diagonal band. Red giants: upper right. Super giants: top. White dwarfs: bottom left.
Parallax distance, with an example: d (pc) = 1 / p (arc-second). p = 0.0125″ gives 80 pc (about 261 ly). 1 pc = 3.26 ly = 3.09 × 10¹⁶ m.
Evolution of a low-mass star and of a high-mass star: Low mass: main sequence → red giant → outer layers shed → white dwarf. High mass: main sequence → super giant → supernova → neutron star or black hole.
Two stars have the same luminosity; one has half the surface temperature. Compare radii.: L = 4πσR²T⁴, so R ∝ 1/T² at fixed L: the cooler star has four times the radius.
Why does a star leave the main sequence?: Core hydrogen runs out, core fusion falls, gravity wins, the core contracts and heats, fusion continues in a shell, and the outer layers expand and cool.

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Written for International Baccalaureate Diploma Programme Physics, (First assessment 2025).

Cards by Marlbridge Academic Team; 12 restate definitions from the Physics revision notes. This deck was reviewed by Iftikhar Azeemi on 2026-11-05. Report an error.

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