Lesson 11: Introduction to Modern Physics — Photons and Relativity

At the start of the twentieth century, physics split in two revolutionary directions: down to the very small, where light turned out to arrive in particles called photons, and up to the very fast, where Einstein showed that space and time themselves are flexible. This lesson introduces both: the photoelectric effect, which proved light is quantised, and the core results of special relativity.

Photons and the photoelectric effect

A photon's energy is E = hf, where h = 6.63 × 10⁻³⁴ J s is Planck's constant and f is the frequency; since f = c ÷ λ, we also write E = hc ÷ λ. Photon energies are tiny, so physicists use the electron-volt: 1 eV = 1.602 × 10⁻¹⁹ J. A handy shortcut is hc ≈ 1240 eV·nm, so E (in eV) ≈ 1240 ÷ λ (in nm).

When light strikes a metal, it can eject electrons — but only if each photon carries at least the metal's work function Φ, the energy needed to free an electron. Einstein's equation: hf = Φ + KE_max. Below the threshold frequency, no electrons are emitted however bright the light; above it, emission is instant and brighter light means *more* electrons, not faster ones. This all-or-nothing behaviour, inexplicable for waves, is the evidence that light arrives in particle-like packets.

Relativity adds: time dilation (moving clocks run slow by the factor γ = 1 ÷ √(1 − v² ÷ c²)), length contraction, and mass–energy equivalence E = mc². These are everyday engineering now: GPS satellite clocks gain about 38 microseconds per day from relativistic effects and must be corrected, or positions would drift by kilometres.

Worked example 1: photon energy

Find the energy of a photon of green light, λ = 550 nm, in joules and electron-volts. E = hc ÷ λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) ÷ (550 × 10⁻⁹) = 1.99 × 10⁻²⁵ ÷ 5.50 × 10⁻⁷ = 3.6 × 10⁻¹⁹ J. In eV: 3.6 × 10⁻¹⁹ ÷ 1.602 × 10⁻¹⁹ = 2.3 eV (check by shortcut: 1240 ÷ 550 = 2.25 eV ✓).

Worked example 2: photoelectric emission

A metal has work function 2.30 eV. Ultraviolet light of wavelength 400 nm shines on it. Find the maximum kinetic energy and speed of the emitted electrons. Photon energy = 1240 ÷ 400 = 3.10 eV. KE_max = hf − Φ = 3.10 − 2.30 = 0.80 eV = 0.80 × 1.602 × 10⁻¹⁹ = 1.28 × 10⁻¹⁹ J. v = √(2 × KE ÷ m) = √(2 × 1.28 × 10⁻¹⁹ ÷ 9.11 × 10⁻³¹) = √(2.81 × 10¹¹) = 5.3 × 10⁵ m/s. The metal's threshold wavelength is λ₀ = 1240 ÷ 2.30 = 539 nm: any light greener than this (longer λ) cannot eject electrons at all, no matter how intense.

In the lab: measuring Planck's constant with LEDs

An LED emits light only when the applied voltage exceeds a threshold roughly equal to the photon energy in electron-volts. Measuring the turn-on voltage for several LED colours gives:

Plotting V against 1 ÷ λ should give a straight line through the origin with gradient hc ÷ e = 1240 V·nm. From the red LED alone: eV = hc ÷ λ implies h = eVλ ÷ c = (1.8 × 1.602 × 10⁻¹⁹ × 650 × 10⁻⁹) ÷ 3.00 × 10⁸ ≈ 6.3 × 10⁻³⁴ J s — within 6% of the true value from a two-dollar component. Uncertainties: "turn-on" is judged by eye (glow vs no glow, or a fixed small current), worth ±0.1 V, and the LED's peak wavelength is only nominal. Averaging over colours via the gradient is far more robust than any single reading.

Physics in the real world

Solar panels are the photoelectric effect's industrial cousin — photons free charge carriers in silicon, and panel efficiency is a photon-energy matching problem. Night-vision and camera sensors count photons; electron microscopes exploit the de Broglie wavelength of electrons, λ = h ÷ p, which at accelerating voltages of 100 V is about 0.12 nm — small enough to image atoms. Einstein's E = mc² explains why the Sun shines: it converts 4.3 million tonnes of mass into energy every second.

Common mistakes

• Believing brighter light of the wrong colour will eventually eject electrons — intensity affects the *number* of photons, not each photon's energy.
• Confusing photon energy (depends on frequency) with intensity (depends on photon flux and energy).
• Applying time dilation backwards: the moving clock measures the *shorter* time.
• Using λ in metres with the 1240 eV·nm shortcut — the units are built into the number.
• Thinking E = mc² applies only to nuclear reactions; it applies to every energy change, usually with an immeasurably small Δm.

Exam Experience

1. In the photoelectric effect, increasing the intensity of light above the threshold frequency: A) increases the maximum kinetic energy of the electrons. B) increases the number of electrons emitted per second. C) decreases the work function. D) has no effect at all. Answer: B. Brighter light delivers more photons per second, so more electrons are freed; each photon's energy, and hence KE_max, depends only on frequency.

2. The energy of a photon is greatest for: A) red light. B) microwaves. C) ultraviolet light. D) infrared radiation. Answer: C. E = hf; of the options, ultraviolet has the highest frequency and therefore the most energetic photons (about 3–10 eV each).

3. A clock moving at 0.80c relative to an observer has γ = 1.67. If 10 years pass for the observer, the moving clock advances by: A) 16.7 years. B) 10 years. C) 6.0 years. D) 8.0 years. Answer: C. Moving clocks run slow: Δt_moving = Δt ÷ γ = 10 ÷ 1.67 = 6.0 years.

Constructed response (5 marks). Electrons are accelerated from rest through a potential difference of 100 V and used in a diffraction experiment. (a) Calculate the kinetic energy of each electron in joules. (b) Calculate the electron's momentum. (c) Calculate its de Broglie wavelength. (e = 1.602 × 10⁻¹⁹ C, m_e = 9.11 × 10⁻³¹ kg, h = 6.63 × 10⁻³⁴ J s.) Scoring guide: (a) 1 mark for KE = eV = 100 × 1.602 × 10⁻¹⁹ = 1.60 × 10⁻¹⁷ J. (b) 1 mark for p = √(2m × KE) = √(2 × 9.11 × 10⁻³¹ × 1.60 × 10⁻¹⁷); 1 mark for p = 5.4 × 10⁻²⁴ kg m/s. (c) 1 mark for λ = h ÷ p; 1 mark for λ = 6.63 × 10⁻³⁴ ÷ 5.4 × 10⁻²⁴ = 1.2 × 10⁻¹⁰ m (0.12 nm) — comparable to atomic spacing, which is why electron diffraction works. Common error: using p = mv with v from KE = ½mv² but forgetting the square root in the momentum step.

Going further: E = mc² with real numbers

Mass–energy equivalence sounds abstract until you price it. One gram of matter converted entirely to energy gives: E = mc² = 1.0 × 10⁻³ × (3.00 × 10⁸)² = 1.0 × 10⁻³ × 9.0 × 10¹⁶ = 9.0 × 10¹³ J. That is roughly the electricity a large power station generates in a full day, from a paperclip's worth of mass. The Sun performs this conversion at a rate of 4.3 million tonnes per second — and even so, it has converted only about 0.03% of its mass in 4.6 billion years, because E = mc² pays out so extravagantly per kilogram. Ordinary processes convert mass too, but imperceptibly: burning a kilogram of petrol releases 4.4 × 10⁷ J, equivalent to a mass loss of E ÷ c² ≈ 5 × 10⁻¹⁰ kg — half a microgram, undetectable on any balance. Nuclear reactions release a million times more energy per kilogram because they rearrange the *strongly bound* nuclear store, not the feeble chemical one. One more threshold worth knowing: creating an electron–positron pair from a single photon needs at least 2m_e c² = 1.02 MeV of photon energy, a wavelength of about 1.2 × 10⁻¹² m — deep gamma territory. This pair production is real physics used daily in PET scanners, where the reverse process (annihilation) creates the gamma photons that are imaged.

In focus: relativity you already rely on

GPS is relativity with a user interface. The satellites orbit at about 3.9 km/s at 20,200 km altitude, and two relativistic effects fight over their clocks. Special relativity (their speed relative to the ground) makes the satellite clocks lose about 7 microseconds per day. General relativity (gravity is weaker up there, so time runs faster) makes them *gain* about 45 microseconds per day. The net is roughly +38 microseconds per day, engineered into the system by setting the satellite clock rates slightly slow before launch. Ignore the correction and the timing errors accumulate into position errors of about 10 km per day — the system would be useless for navigation within hours of switch-on. This is worth remembering whenever relativity is dismissed as exotic: it is load-bearing infrastructure, tested continuously by every phone that reports its position. The other everyday appearance of this lesson's physics is nuclear medicine. PET scanners image the body using positrons — antimatter electrons — emitted by a tracer such as fluorine-18 (half-life 110 minutes). Each positron annihilates with an electron, converting both masses entirely into two gamma photons of 511 keV travelling in opposite directions (E = mc², itemised). Detectors time the pair's arrival to reconstruct the tracer's location — Einstein's equation and Lesson 19's radioactivity, working a hospital shift.

Formula card

Exam technique: work in electron-volts whenever photons meet work functions — the arithmetic halves in length and unit errors vanish. Save joules for the final speed calculation, where ½mv² needs SI throughout.