PHYS116 Final Exam Review | Source: Chapter 25.6–25.7 lecture notes
Difficulty: Intermediate | Prerequisites: Ch. 25.5 EM wave properties, the relationship c = λf, and comfort with unit conversions (eV to joules).
Chapter 25.5 established that light is an electromagnetic wave. This section adds a second, complementary picture: EM waves also behave as streams of discrete, massless particles called photons, each carrying a specific packet of energy. The EM spectrum then organises every type of EM radiation, from radio waves to gamma rays, by frequency and wavelength. Wien's Law and the Stefan-Boltzmann relation round things out by connecting an object's temperature to the radiation it emits. This dual wave-particle nature is one of the most important ideas in modern physics.
EM waves come in discrete packets called photons, each with energy E = hf. Higher frequency means higher energy per photon. The entire EM spectrum, from AM radio to gamma rays, is the same phenomenon at different frequencies, and hotter objects radiate at shorter (more energetic) peak wavelengths.
Photon
A massless particle of light (or any EM radiation) that travels at c = 3.00 × 10⁸ m/s in vacuum. Think of it as the smallest possible "chunk" of an EM wave.
Planck's constant (h)
A fundamental constant, h = 6.63 × 10⁻³⁴ J·s, that links a photon's energy to its frequency. It sets the scale for all quantum phenomena.
Photon energy (E = hf)
The energy carried by a single photon, proportional to the wave's frequency. In simple terms, bluer (higher-frequency) light carries more energy per photon than redder light.
Electronvolt (eV)
A convenient energy unit: 1 eV = 1.60 × 10⁻¹⁹ J. Photon and atomic-scale energies are much tidier in eV than in joules.
EM spectrum
The full range of electromagnetic radiation ordered by frequency (or equivalently by wavelength or photon energy), from radio waves at the low-frequency end to gamma rays at the high-frequency end.
Wien's Law
The peak wavelength of thermal radiation shifts to shorter wavelengths as temperature increases: λ_peak = (2.9 × 10⁶ nm·K) / T. In simple terms, hotter objects glow at shorter wavelengths.
Stefan-Boltzmann Law (rate of heat transfer)
The power radiated by an object is Q/Δt = eσAT⁴, where e is emissivity, σ = 5.67 × 10⁻⁸ W/m²·K⁴, A is surface area, and T is temperature in kelvin. Hotter objects radiate dramatically more power.
Emissivity (e)
A dimensionless number between 0 and 1 that describes how well a surface radiates compared to a perfect blackbody. A perfect emitter has e = 1.
EM waves have a particle-like nature. They consist of massless units called photons, each travelling at c in vacuum.
Each photon carries energy E_photon = hf, where h = 6.63 × 10⁻³⁴ J·s (Planck's constant) and f is the frequency.
The superposition of a large number of photons reproduces the characteristics of a continuous EM wave. A single photon is quantum; billions together look classical.
To convert photon energy from joules to eV: divide by 1.60 × 10⁻¹⁹ J/eV.
Worked example: A photon with wavelength 550 nm.
Find frequency: c = λf, so f = c/λ = (3.00 × 10⁸) / (550 × 10⁻⁹) = 5.45 × 10¹⁴ Hz.
Find energy: E = hf = (6.63 × 10⁻³⁴)(5.45 × 10¹⁴) = 3.61 × 10⁻¹⁹ J = 2.26 eV.
Worked example: A 40 W lightbulb emits about 1 J of visible light per second. How many photons per second?
N = 1 J / (3.61 × 10⁻¹⁹ J/photon) ≈ 3 × 10¹⁸ photons.
All EM waves travel at c; they differ only in frequency and wavelength.
Spectrum regions from longest wavelength to shortest: AM radio, FM radio/TV, microwaves, infrared, visible light, ultraviolet, X-rays, gamma rays.
Lower-frequency radiation (radio, microwave) behaves more wave-like. Higher-frequency radiation (X-rays, gamma rays) behaves more particle-like.
An antenna generates a self-sustaining EM wave. For best reception, antenna length should be about ¼ of the wavelength.
Every object above 0 K emits thermal radiation. Hotter objects emit more intensely and at shorter peak wavelengths.
Wien's Law: λ_peak = (2.9 × 10⁶ nm·K) / T
Example: A human body at 37°C = 310 K has λ_peak = (2.9 × 10⁶) / 310 = 9,400 nm (infrared).
Rate of heat transfer (Stefan-Boltzmann): Q/Δt = eσAT⁴
e = emissivity (0 to 1), σ = 5.67 × 10⁻⁸ W/m²·K⁴
An accelerating voltage V gives electrons kinetic energy eV. When these electrons hit a target, the maximum photon energy equals that kinetic energy.
E_max = eV (in eV) or E_max = eV × (1.60 × 10⁻¹⁹) (in joules).
The corresponding minimum wavelength: λ = hc / E_max.
Worked example: An X-ray tube with accelerating voltage 30 kV.
E_max = 30 × 10³ eV = 4.8 × 10⁻¹⁵ J.
λ = hc / E = (6.63 × 10⁻³⁴)(3.00 × 10⁸) / (4.8 × 10⁻¹⁵) = 4.1 × 10⁻¹¹ m = 0.041 nm.
Photon energy
E = hf = hc / λ
where h = 6.63 × 10⁻³⁴ J·s
Energy unit conversion
1 eV = 1.60 × 10⁻¹⁹ J
Wien's Law (peak wavelength of thermal radiation)
λ_peak = (2.9 × 10⁶ nm·K) / T
Stefan-Boltzmann Law (radiated power)
Q/Δt = eσAT⁴
where σ = 5.67 × 10⁻⁸ W/m²·K⁴, e = emissivity, A = surface area, T = temperature in kelvin.
Maximum photon energy from an accelerating voltage
E_max = eV (in eV), or E_max = eV × 1.60 × 10⁻¹⁹ (in joules)
Minimum wavelength from an accelerating voltage
λ_min = hc / E_max
Wien's Law explains why a hot stove element glows red (long visible wavelength) while a welding arc glows blue-white (shorter wavelength). Medical X-ray machines work by accelerating electrons through a high voltage and converting their kinetic energy into high-energy photons that can pass through soft tissue but are absorbed by bone.
Students often think that a brighter light source produces higher-energy photons. It does not. Brightness (intensity) depends on the number of photons per second, not the energy per photon. A dim UV lamp emits more energetic photons than a bright red lamp.
A common mistake is confusing wavelength and frequency when determining photon energy. Higher frequency means higher energy, but higher wavelength means lower energy. They are inversely related through c = λf.
Students sometimes assume Wien's Law gives the only wavelength an object emits. It gives the peak, the most intense wavelength. The object radiates across a continuous spectrum.
When converting eV to joules, students occasionally multiply instead of dividing (or vice versa). Keep the conversion factor straight: multiply eV by 1.60 × 10⁻¹⁹ to get joules.
⚠️ Be ready to calculate photon energy from wavelength (two steps: find f from c = λf, then E = hf) and convert between joules and eV.
⚠️ "How many photons per second" problems are common: divide total power or energy rate by the energy per photon.
⚠️ Wien's Law problems typically give a temperature and ask for peak wavelength, or vice versa. Remember to convert °C to kelvin first.
⚠️ X-ray tube problems: the accelerating voltage sets the maximum photon energy, and from that you find the minimum wavelength via λ = hc/E.
True or false: A photon of red light has more energy than a photon of blue light. False. Blue light has a higher frequency, so each photon carries more energy.
Fill in the blank: Planck's constant h = ______ J·s. 6.63 × 10⁻³⁴
True or false: According to Wien's Law, a hotter object has a longer peak wavelength. False. Hotter objects peak at shorter wavelengths.
Fill in the blank: 1 eV = ______ J. 1.60 × 10⁻¹⁹
True or false: An X-ray tube with a 60 kV accelerating voltage can produce photons with higher energy than one with 30 kV. True. The maximum photon energy equals the accelerating voltage in eV.
Q: What is the energy of a photon with a wavelength of 550 nm, in both joules and eV?
A: f = c/λ = (3.00 × 10⁸) / (550 × 10⁻⁹) = 5.45 × 10¹⁴ Hz. E = hf = (6.63 × 10⁻³⁴)(5.45 × 10¹⁴) = 3.61 × 10⁻¹⁹ J = 2.26 eV.
Q: A 40 W bulb emits about 1 J of visible light per second. Roughly how many visible-light photons does it emit per second?
A: Using E per photon ≈ 3.61 × 10⁻¹⁹ J (for 550 nm light), N = 1 / (3.61 × 10⁻¹⁹) ≈ 3 × 10¹⁸ photons per second.
Q: What is the peak wavelength of radiation from a human body at 37°C?
A: T = 37 + 273 = 310 K. λ_peak = (2.9 × 10⁶) / 310 = 9,400 nm (infrared).
Q: An X-ray tube operates at 30 kV. What is the maximum photon energy and the minimum wavelength it can produce?
A: E_max = 30 × 10³ eV = 4.8 × 10⁻¹⁵ J. λ_min = hc/E = (6.63 × 10⁻³⁴)(3.00 × 10⁸) / (4.8 × 10⁻¹⁵) = 4.1 × 10⁻¹¹ m = 0.041 nm.
This connects directly to the photoelectric effect (Ch. 28.2), where the photon energy E = hf determines whether electrons can be ejected from a metal surface. The concept of a threshold frequency comes straight from the photon model.
It also connects back to EM wave properties (Ch. 25.5), since intensity I is related to both the wave picture (field amplitudes) and the photon picture (number of photons per second times energy per photon).
Wien's Law and thermal radiation connect to thermodynamics and heat transfer topics covered earlier in the course.
Tags: photon, photon energy, E = hf, Planck's constant, electronvolt, eV, EM spectrum, electromagnetic spectrum, radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, gamma rays, Wien's Law, peak wavelength, Stefan-Boltzmann, thermal radiation, blackbody, emissivity, X-ray tube, accelerating voltage, wave-particle duality, PHYS116, general physics, University of Washington, Chapter 25.6, Chapter 25.7