Physics • Year 12 • Module 7 • Lesson 1
The Electromagnetic Spectrum
Build HSC Band 5–6 extended-response technique on evaluating EM evidence, designing investigations, and applying photon energy reasoning in complex scenarios.
1. Data + scenario: comparing photon energy across two satellite imaging systems (Band 5–6)
8 marks Band 5–6
Scenario. The Australian Space Agency uses two remote-sensing satellite systems to monitor the Great Barrier Reef. System A uses infrared radiation at 10.0 µm wavelength to map sea surface temperature. System B uses UV radiation at 350 nm wavelength to detect coral bleaching by measuring reflected sunlight. Both systems detect photons and count the rate of photon arrivals to construct images. The table below shows key data for each system.
| Parameter | System A (infrared) | System B (UV) |
|---|---|---|
| Wavelength | 10.0 µm | 350 nm |
| Power received by detector (W) | 8.0 × 10−12 | 8.0 × 10−12 |
| Photons per second (to calculate) | — | — |
| Photon energy in eV (to calculate) | — | — |
| Biological/material hazard | Low — mild heating only | Can damage organic material |
Illustrative data based on CSIRO satellite remote-sensing program, Great Barrier Reef monitoring.
Q1. Analyse and evaluate the data above to compare the two satellite systems. In your response you must:
- Calculate the frequency and photon energy (in eV) for each system, showing full working.
- Calculate the number of photons per second detected by each system at the given power level. Show your working.
- Explain why System B photons are described as a hazard to organic material while System A photons are not, using the photon energy equation.
- Evaluate which system would require a greater number of photons to form an image of the same brightness, and why.
- State one advantage of each system for its specific monitoring purpose.
2. Experimental design — determining Planck's constant from the photoelectric effect (Band 5–6)
7 marks Band 5–6
Research question. A Physics teacher wants students to experimentally determine a value for Planck's constant h using a set of LED lights of different colours. The teacher hypothesises that the minimum voltage needed to make an LED just emit light (the threshold voltage VT) is related to the frequency of the emitted light by the equation eVT = hf, where e = 1.60 × 10−19 C.
Available equipment: Red, orange, green, blue, and violet LEDs (wavelengths known); a variable DC power supply (0–5 V); a voltmeter; a current meter (microammeter); connecting leads.
Q2. Design the investigation and present it in the format below.
- State a hypothesis in the form: if VT is plotted against f, then the graph will show… (include what the gradient and y-intercept represent).
- Identify the independent variable, dependent variable, and at least two controlled variables.
- Describe the procedure in at least four numbered steps, including how you will determine VT precisely for each LED.
- Explain how you will use your graph to calculate h, including the formula relating gradient to h.
- State one source of systematic error and one way to reduce it.
Q1 — Sample Band 6 response (8 marks), annotated
Calculations — System A (IR, λ = 10.0 µm): f = c/λ = 3.00×108/10.0×10−6 = 3.00×1013 Hz [1]; E = hf = 6.63×10−34×3.00×1013 = 1.99×10−20 J = 0.124 eV [1].
Calculations — System B (UV, λ = 350 nm): f = 3.00×108/350×10−9 = 8.57×1014 Hz [1]; E = 6.63×10−34×8.57×1014 = 5.68×10−19 J = 3.55 eV [1]. Note: UV photon energy is 3.55/0.124 ≈ 28.6 times greater than the IR photon.
Photons per second: N = P/Ephoton. System A: N = 8.0×10−12/1.99×10−20 = 4.0×108 photons/s [1]. System B: N = 8.0×10−12/5.68×10−19 = 1.41×107 photons/s [1]. System A requires approximately 28 times more photons per second to deliver the same power.
Biological hazard explanation: System B UV photons carry 3.55 eV each, sufficient to break covalent bonds in organic molecules such as DNA base pairs (bond energies ~3–5 eV), causing molecular damage. System A IR photons carry only 0.124 eV — far below covalent bond energies — so they can only cause gentle heating via molecular vibration, not bond breakage [1].
Evaluation — which needs more photons: System A (IR) delivers the same 8 pW of power using ~4×108 photons/s because each IR photon is low energy. An image of the same brightness would require 28 times as many IR photons as UV photons. Therefore, System A requires far more photons per image, making it more susceptible to noise from photon counting statistics at low light levels [1]. (Alternatively: System A has a higher photon flux = more photons arrive per second at the same power = brighter image per unit time, so in this sense it actually forms images more easily at the same power level — full credit for either logically consistent evaluation.)
Advantages: System A advantage: IR thermal emission from the sea surface is independent of sunlight illumination; it works at night and directly measures water temperature through emission spectrum [1]. System B advantage: UV reflectance distinguishes living coral pigments (which fluoresce under UV) from bleached coral (which does not), enabling early detection of bleaching events before visible colour change occurs [1].
Marking criteria (8 marks): 1 = correct f and E for System A (both correct, full working); 1 = correct f and E for System B (both correct, full working); 1 = correct photons/s for System A; 1 = correct photons/s for System B; 1 = biological hazard explanation using photon energy values and bond energy comparison; 1 = logical evaluation of which system needs more photons with quantitative support; 1 = advantage of System A for its purpose; 1 = advantage of System B for its purpose.
Q2 — Sample Band 6 response (7 marks), annotated
Hypothesis: If VT is plotted against f, the graph will show a linear relationship passing through (or near) the origin, with gradient = h/e, where h is Planck's constant and e is the elementary charge. The y-intercept should be approximately zero (no work function correction in this simplified model). [1 — states gradient = h/e and y-intercept = 0 with reasoning]
Variables: Independent variable: frequency f of emitted light (controlled by choice of LED). Dependent variable: threshold voltage VT at which the LED just begins to emit visible light. Controlled variables: temperature of LEDs (keep at room temperature); supply voltage increase rate (slow, uniform ramp); ambient light conditions (dark room to prevent stray photon detection); same current meter for all measurements. [1 — correct IV and DV named; 1 — two valid controlled variables]
Procedure: (1) Connect the LED in series with the voltmeter (in parallel with the LED) and the microammeter (in series) to the variable DC supply. (2) Starting from 0 V, slowly increase the supply voltage while monitoring the microammeter. Record the voltage VT at which current just begins to register a consistent non-zero value AND the LED visibly emits light (confirm by eye in a darkened room). (3) Repeat three times for each LED and average the VT values to improve reliability. (4) Repeat steps 1–3 for each of the five LEDs. Record the known wavelength λ for each LED and calculate f = c/λ. Plot VT (y-axis) against f (x-axis) and draw a line of best fit. [1 — four steps with a clear method for determining VT precisely]
Graph analysis: From eVT = hf, rearranging gives VT = (h/e)f. The gradient of the VT vs f graph = h/e. Therefore h = gradient × e = gradient × 1.60×10−19 C. The units of the gradient are V/Hz = V·s = J·s/C, so multiplying by e (C) gives J·s, which is correct for h. [1 — correct formula linking gradient to h with unit analysis]
Systematic error and reduction: Systematic error: ambient light in the room may cause the LED to emit photons even below VT, making it appear to emit at a lower threshold voltage than the true value, producing a systematically low gradient and underestimating h. Reduction: conduct the experiment inside a sealed dark box, or use a light-tight enclosure and a photodiode detector rather than visual observation to determine the onset of emission more precisely. [1 — named source; 1 — specific reduction strategy]
Marking criteria (7 marks): 1 = hypothesis with gradient = h/e and y-intercept = 0 justified; 1 = correct IV (LED frequency) and DV (threshold voltage) named; 1 = two valid controlled variables; 1 = four clear procedure steps including precise VT determination method; 1 = correct formula h = gradient × e with unit analysis; 1 = valid named systematic error; 1 = specific, feasible reduction strategy.