Unit 2 High-Yield Topics
Free resource by Flowxiom — Edexcel A-level Physics
Not everything. Just what’s on the paper. High-frequency topics only — covering ~80% of exam marks.
Edexcel A-level Physics | Waves, Optics & DC Circuits | WPH11 & WPH12
Topic 1: The Photoelectric Effect (frequent 6-mark question)
Question types: Explain why wave theory fails; calculate threshold frequency or maximum kinetic energy.
Einstein’s Photoelectric Equation
\[hf = \phi + E_{k(max)}\]
| Symbol | Meaning |
|---|---|
| \(hf\) | Incident photon energy (\(h = 6.63 \times 10^{-34}\) J s) |
| \(\phi\) | Work function — minimum energy to escape the metal |
| \(E_{k(max)}\) | Maximum kinetic energy of photoelectron |
Threshold frequency: \(f_0 = \dfrac{\phi}{h}\) (at this point \(E_k = 0\))
Stopping voltage: \(E_{k(max)} = eV_s\)
Why Wave Theory Fails — Three-Point Answer Chain
| Observation | Wave theory predicts (wrong) | Photon theory explains (correct) |
|---|---|---|
| Threshold frequency exists | Sufficient intensity should eject electrons at any frequency | Single photon energy \(hf\) insufficient; more photons don’t help |
| No time delay | Weak light needs time to build up energy | One photon transfers energy to one electron instantly |
| Kinetic energy independent of intensity | Higher intensity → higher kinetic energy | Intensity increases photon count, not individual photon energy |
Common Mistakes
- ❌ Must state: single photon energy \(hf < \phi\), so electron cannot escape
- ❌ Gradient of stopping voltage vs frequency graph = \(h/e\), not \(h\)
Topic 2: Double-Slit Interference and Diffraction Grating
Question types: Fringe spacing calculation, maximum order for diffraction grating, effect of changing parameters.
Double-Slit Fringe Spacing
\[\Delta y = \frac{\lambda L}{d}\]
- \(\lambda\): wavelength, \(L\): slit-to-screen distance, \(d\): slit separation
- Wider fringes: increase \(\lambda\) / increase \(L\) / decrease \(d\)
Diffraction Grating
\[d\sin\theta = n\lambda\]
- \(d\): grating spacing (if question gives N lines per mm, then \(d = 1/N\) mm)
- Maximum order: set \(\sin\theta = 1\), \(n_{max} = d/\lambda\) (round down to integer)
Conditions for Interference
Coherent: same frequency + constant phase difference
Same type of wave
Path Difference Rule
| Path difference | Result |
|---|---|
| \(n\lambda\) (whole number of wavelengths) | Constructive interference — bright fringe |
| \((n+0.5)\lambda\) | Destructive interference — dark fringe |
Common Mistakes
- ❌ \(d\) = slit separation (centre to centre), not slit width
- ❌ For the grating: \(d\) is the separation between adjacent slits, not the slit width
Topic 3: Stationary (Standing) Waves
Question types: Resonance conditions, counting nodes and antinodes, frequency calculation.
Resonance Condition
String with both ends fixed: \(L = \dfrac{n\lambda}{2}\), \(n = 1, 2, 3…\)
- Fundamental (\(n=1\)): \(f_1 = \dfrac{v}{2L}\)
- \(n\)th harmonic: \(f_n = nf_1\)
Node vs Antinode
- Node: point of zero displacement (fixed end)
- Antinode: point of maximum displacement
Common Mistakes
- ❌ Recount nodes/antinodes by drawing the wave pattern
- ❌ Distance between adjacent nodes = \(\lambda/2\), not \(\lambda\)
Topic 4: DC Circuits (Kirchhoff’s Laws)
Question types: Series/parallel equivalent resistance, complex circuit analysis, internal resistance and terminal voltage.
Series and Parallel Rules
| Series | Parallel | |
|---|---|---|
| Current | Same throughout | Branch currents sum to total |
| Voltage | Sum of p.d.s = total voltage | Same across all branches |
| Resistance | \(R_{total} = R_1 + R_2 + …\) | \(\dfrac{1}{R_{total}} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + …\) |
Two resistors in parallel: \(R_{total} = \dfrac{R_1 R_2}{R_1 + R_2}\)
Internal Resistance and Terminal Voltage
\[\varepsilon = I(R + r) \qquad V_{terminal} = \varepsilon – Ir\]
Higher current → lower terminal voltage (more voltage lost across internal resistance).
Kirchhoff’s Laws
- KCL: Sum of currents into a junction = sum of currents out
- KVL: Sum of e.m.f.s = sum of p.d.s around any closed loop
Potential Divider
\[V_{out} = \frac{R_2}{R_1 + R_2} \times V_{in}\]
Common Mistakes
- ❌ Parallel resistance is always less than the smallest individual resistance
- ❌ Current through internal resistance = total circuit current
- ❌ Terminal voltage < e.m.f. whenever current flows
Topic 5: Power Formulae
\[P = IV = I^2R = \frac{V^2}{R}\]
- Know \(I\) and \(R\): use \(P = I^2R\)
- Know \(V\) and \(R\): use \(P = V^2/R\)
- Series circuit: same current → larger resistance dissipates more power
- Parallel circuit: same voltage → smaller resistance dissipates more power
Topic 6: Basic Wave Properties
Question types: Classify transverse/longitudinal waves, wave speed calculation, phase difference, polarisation.
Basic Equations
\[v = f\lambda \qquad T = \frac{1}{f}\]
Transverse vs Longitudinal
| Transverse | Longitudinal | |
|---|---|---|
| Oscillation direction | Perpendicular to propagation | Parallel to propagation |
| Can be polarised? | Yes | No |
| Examples | Light, EM waves, rope | Sound, ultrasound |
Polarisation only occurs in transverse waves — this is evidence that light is a transverse wave.
Phase Difference
\[\Delta\phi = \frac{2\pi\,\Delta x}{\lambda}\]
- In phase: \(\Delta\phi = 2n\pi\) (whole number of wavelengths)
- Antiphase: \(\Delta\phi = (2n+1)\pi\) (half-integer wavelengths)
Intensity Relationships
\[I \propto A^2 \qquad I \propto \frac{1}{r^2} \text{ (point source, spherical spread)}\]
Common Mistakes
- ❌ Sound is longitudinal — cannot be polarised
- ❌ Wave entering new medium: frequency unchanged; speed and wavelength change
- ❌ Path difference → phase difference: \(\Delta\phi = 2\pi\Delta x/\lambda\)
Topic 7: Refraction and Total Internal Reflection
Question types: Calculate refractive index, critical angle, conditions for TIR, optical fibre principle.
Snell’s Law
\[n_1\sin\theta_1 = n_2\sin\theta_2\]
Air (\(n\approx1\)) → medium: \(n = \dfrac{\sin i}{\sin r}\)
Physical meaning: \(n = \dfrac{c}{v}\) (speed in vacuum / speed in medium, \(n > 1\))
Two Conditions for TIR
Light travels from optically denser to optically less dense medium
Angle of incidence \(\geq\) critical angle \(c\)
\[\sin c = \frac{1}{n} \quad \text{(from medium of index }n\text{ into air)}\]
Optical Fibre Principle
Light in the glass core hits the core-cladding boundary at an angle > critical angle → TIR → light propagates along the fibre without loss.
Common Mistakes
- ❌ TIR occurs when going from denser to less dense medium — not the other way
- ❌ Formula \(\sin c = 1/n\) assumes the less dense medium is air (\(n=1\))
- ❌ Entering denser medium: speed decreases, wavelength decreases, frequency unchanged
Topic 8: Atomic Energy Levels and Spectra
Question types: Explain line spectra, calculate photon frequency from energy level diagram, emission vs absorption spectra.
Photon Energy
\[E_{photon} = hf = \frac{hc}{\lambda} = E_{high} – E_{low}\]
Transition Rules
| Direction | Result |
|---|---|
| Electron drops to lower level | Emits a photon |
| Electron rises to higher level | Absorbs a photon of specific frequency |
Electrons can only occupy discrete energy levels → only specific frequencies emitted/absorbed → line spectrum.
Emission vs Absorption Spectra
| Type | Appearance | Cause |
|---|---|---|
| Emission spectrum | Bright lines on dark background | Excited electrons fall to lower level |
| Absorption spectrum | Dark lines on continuous spectrum | Electrons absorb specific frequencies |
Same frequencies for the same element.
Ionisation
Energy levels are negative values; the ionisation energy = energy needed to move an electron from ground state to \(E = 0\).
Common Mistakes
- ❌ Energy levels are negative; \(\Delta E = E_{high} – E_{low}\) (positive result)
- ❌ One photon is absorbed by one electron only — photons cannot combine
- ❌ Each line represents a specific wavelength/frequency, not a position in space
Topic 9: Resistivity and Drift Velocity
Question types: Calculate wire resistance, derive drift velocity, explain resistance-temperature behaviour.
Resistivity Formula
\[R = \frac{\rho L}{A}\]
- \(\rho\): resistivity (unit: \(\Omega\cdot\text{m}\)) — material property, independent of shape
- \(R \propto L\) (longer wire → greater resistance)
- \(R \propto 1/A\) (larger cross-section → smaller resistance)
Drift Velocity Formula
\[I = nAve\]
| Symbol | Meaning | Unit |
|---|---|---|
| \(n\) | Number density of charge carriers | m⁻³ |
| \(A\) | Cross-sectional area | m² |
| \(v\) | Drift velocity | m/s |
| \(e\) | Elementary charge \(1.6\times10^{-19}\) | C |
Drift velocity is very slow (~\(10^{-4}\) m/s). Light turns on instantly because the electric field propagates at near light speed — not because electrons move fast.
Temperature Effect
| Material | As temperature rises | Resistance change | Reason |
|---|---|---|---|
| Metal conductor | Ions vibrate more, more collisions | Increases | \(n\) unchanged, but collision frequency rises |
| NTC thermistor (semiconductor) | More electrons gain sufficient energy | Decreases | \(n\) increases exponentially |
Common Mistakes
- ❌ Unit of resistivity: \(\Omega\cdot\text{m}\), not \(\Omega/\text{m}\)
- ❌ Resistivity \(\rho\) does not change with dimensions — it is a material property
- ❌ Same current, larger cross-section → smaller drift velocity
Practice Questions
Q1. Work function \(\phi = 3.0 \times 10^{-19}\) J, light frequency \(f = 8.0 \times 10^{14}\) Hz. Find the maximum kinetic energy of the photoelectron. (\(h = 6.63 \times 10^{-34}\) J s)
Answer
\(E_k = hf – \phi = 6.63\times10^{-34} \times 8.0\times10^{14} – 3.0\times10^{-19} = \mathbf{2.3\times10^{-19}\ \text{J}}\)
Q2. Diffraction grating: 400 lines per mm, wavelength 600 nm. What is the maximum order of bright fringe?
Answer
\(d = 1/400\ \text{mm} = 2.5\times10^{-6}\ \text{m}\)
\(n_{max} = d/\lambda = 2.5\times10^{-6} / 600\times10^{-9} = 4.17\)
Maximum order \(= \mathbf{4}\) (round down)
Q3. \(6\ \Omega\) and \(3\ \Omega\) in parallel, then in series with \(2\ \Omega\). Find total resistance.
Answer
Parallel: \(R = \dfrac{6\times3}{6+3} = 2\ \Omega\)
Series: \(R_{total} = 2 + 2 = \mathbf{4\ \Omega}\)
Q4. Glass with refractive index \(n = 1.5\). Find the critical angle.
Answer
\(\sin c = \dfrac{1}{1.5} = 0.667\)
\(c = \sin^{-1}(0.667) = \mathbf{41.8°}\)
Q5. Two energy levels in hydrogen differ by \(\Delta E = 2.55\ \text{eV}\). Find the wavelength of the emitted photon. (\(h = 6.63\times10^{-34}\) J s, \(c = 3.00\times10^8\) m/s, \(1\ \text{eV} = 1.6\times10^{-19}\ \text{J}\))
Answer
\(E = 2.55 \times 1.6\times10^{-19} = 4.08\times10^{-19}\ \text{J}\)
\(\lambda = \dfrac{hc}{E} = \dfrac{6.63\times10^{-34} \times 3.00\times10^8}{4.08\times10^{-19}} = \mathbf{4.88\times10^{-7}\ \text{m}}\) (488 nm — blue-green)
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