The electromagnetic spectrum and Maxwell's equations: HSC Physics Module 7

Radio wavelength from $c = f \lambda$:

$\lambda = c / f = 3.00 \times 10^8 / (102.5 \times 10^6) = 2.93$ m.

X-ray photon frequency:

$f = c / \lambda = 3.00 \times 10^8 / (5.0 \times 10^{-11}) = 6.0 \times 10^{18}$ Hz.

Photon energy ($h = 6.626 \times 10^{-34}$ J s):

$E = h f = 6.626 \times 10^{-34} \times 6.0 \times 10^{18} = 4.0 \times 10^{-15}$ J.

In electron-volts ($1$ eV $= 1.602 \times 10^{-19}$ J):

$E = 4.0 \times 10^{-15} / 1.602 \times 10^{-19} = 2.5 \times 10^{4}$ eV $= 25$ keV.

Markers reward correct use of $c = f \lambda$, $E = hf$, and unit conversion to eV. The X-ray photon has roughly $10^{19}$ times the energy of the radio photon, which is why X-rays ionise tissue and radio waves do not.

Maxwell's equations unify the laws of electricity and magnetism into four field equations. The two relevant for wave prediction are:

1. Faraday's law: a changing magnetic field induces a circulating electric field.
2. The Ampere-Maxwell law: a changing electric field (the displacement current) induces a circulating magnetic field.

Together these say that a changing E-field generates a B-field, which in turn generates an E-field, and so on. Manipulating the equations yields a wave equation for both $\vec{E}$ and $\vec{B}$ with a propagation speed:

$c = 1 / \sqrt{\mu_0 \varepsilon_0}$

where $\mu_0$ and $\varepsilon_0$ are the magnetic and electric constants measured in static experiments. Substituting their measured values gives $c \approx 3.0 \times 10^8$ m/s, which matches Fizeau's and Foucault's measurements of the speed of light. Maxwell therefore concluded that light is an electromagnetic wave, and that other wavelengths of EM radiation should exist (later confirmed by Hertz's radio-wave experiments).

Markers reward the changing-field-induces-changing-field idea, the speed prediction from $\mu_0$ and $\varepsilon_0$, and the agreement with measured $c$.

Rearrange $c = f\lambda$ for wavelength: $\lambda = \dfrac{c}{f}$.

$\lambda = \dfrac{3.00 \times 10^8}{2.45 \times 10^9} = 0.122\ \text{m} = 12.2\ \text{cm}$.

Marks: one for the rearranged formula with values substituted, one for the answer stated to three significant figures with the correct unit.

Use $E = hf$ directly, since the frequency is already given.

$E = hf = (6.626 \times 10^{-34})(2.4 \times 10^9) = 1.6 \times 10^{-24}\ \text{J}$.

Marks: one for the correct formula with values substituted, one for the answer to two significant figures with the unit joule. (This is about a trillion times smaller than a visible-light photon, which is why Wi-Fi radiation cannot ionise anything.)

(a) $f = \dfrac{c}{\lambda} = \dfrac{3.00 \times 10^8}{1.0 \times 10^{-13}} = 3.0 \times 10^{21}\ \text{Hz}$.

(b) $E = hf = (6.626 \times 10^{-34})(3.0 \times 10^{21}) = 2.0 \times 10^{-12}\ \text{J}$.

Converting to $\text{MeV}$ ($1\ \text{eV} = 1.602 \times 10^{-19}\ \text{J}$): $E = \dfrac{2.0 \times 10^{-12}}{1.602 \times 10^{-19} \times 10^6} = 12\ \text{MeV}$.

Marks: one for $f = 3.0 \times 10^{21}\ \text{Hz}$, one for $E$ in joules from $E = hf$, one for the correct conversion to $\text{MeV}$. Gamma photons at this energy are far above the $\sim 10\ \text{eV}$ needed to ionise atoms.

Frequency: $f = \dfrac{c}{\lambda} = \dfrac{3.00 \times 10^8}{589 \times 10^{-9}} = 5.09 \times 10^{14}\ \text{Hz}$.

Photon energy: $E = hf = (6.626 \times 10^{-34})(5.09 \times 10^{14}) = 3.37 \times 10^{-19}\ \text{J}$.

In electron-volts: $E = \dfrac{3.37 \times 10^{-19}}{1.602 \times 10^{-19}} = 2.11\ \text{eV}$.

Marks: one for $f = 5.09 \times 10^{14}\ \text{Hz}$, one for $E = 3.37 \times 10^{-19}\ \text{J}$ from $E = hf$, one for the correct conversion to $2.11\ \text{eV}$.

(a) UV: $E_{UV} = \dfrac{hc}{\lambda} = \dfrac{(6.626 \times 10^{-34})(3.00 \times 10^8)}{250 \times 10^{-9}} = 7.95 \times 10^{-19}\ \text{J}$.

Infrared: $E_{IR} = \dfrac{(6.626 \times 10^{-34})(3.00 \times 10^8)}{1500 \times 10^{-9}} = 1.33 \times 10^{-19}\ \text{J}$.

(b) Ratio: $\dfrac{E_{UV}}{E_{IR}} = \dfrac{7.95 \times 10^{-19}}{1.33 \times 10^{-19}} = 6.0$, so the UV photon carries six times the energy of the infrared photon. A single UV photon at this wavelength carries enough energy to break molecular bonds in DNA (ionising/damaging skin cells), whereas an infrared photon only adds a small amount of vibrational (thermal) energy per absorption and cannot break a bond in one hit.

Marks: one for each correctly computed energy (two marks), one for the ratio of $6.0$, one for linking higher photon energy to bond-breaking capability versus infrared merely heating.

(a) The graph is a straight line through the origin, showing that photon energy $E$ is directly proportional to frequency $f$ ($E \propto f$), consistent with $E = hf$.

(b) Reading the two points from the graph: $(4.0 \times 10^{14}\ \text{Hz},\ 2.65 \times 10^{-19}\ \text{J})$ and $(7.0 \times 10^{14}\ \text{Hz},\ 4.64 \times 10^{-19}\ \text{J})$.

Gradient $= \dfrac{\Delta E}{\Delta f} = \dfrac{(4.64 - 2.65) \times 10^{-19}}{(7.0 - 4.0) \times 10^{14}} = \dfrac{1.99 \times 10^{-19}}{3.0 \times 10^{14}} = 6.6 \times 10^{-34}\ \text{J s}$.

(c) The gradient of an $E$-$f$ graph equals Planck's constant $h$. The calculated value $6.6 \times 10^{-34}\ \text{J s}$ agrees with the accepted value $h = 6.626 \times 10^{-34}\ \text{J s}$ to two significant figures, confirming $E = hf$.

Marks: one for identifying direct proportionality (line through the origin), one for a correctly read gradient with the unit $\text{J s}$, one for naming the gradient as Planck's constant $h$, one for a valid numerical comparison with the accepted value of $h$.

Band-6 plan. (1) State the two field laws Maxwell combined (Faraday's law, the Ampere-Maxwell law) and the mutual-induction idea. (2) State the derived wave speed $c = 1/\sqrt{\mu_0 \varepsilon_0}$ and that it matched the measured speed of light. (3) State the conclusion (light is an EM wave) and the further prediction (other wavelengths exist). (4) Evaluate significance: unified optics with electromagnetism, predicted the rest of the spectrum, confirmed experimentally by Hertz. Finish with a judgement.

Model answer. Maxwell combined the existing laws of electricity and magnetism, including Faraday's law (a changing magnetic field induces a circulating electric field) and his own addition to Ampere's law, the displacement current (a changing electric field induces a circulating magnetic field). These two statements say that a changing $\vec{E}$ field generates a changing $\vec{B}$ field, which in turn regenerates a changing $\vec{E}$ field, so the disturbance can propagate through space without any charges or currents ahead of it.

Combining the equations mathematically produces a wave equation for both $\vec{E}$ and $\vec{B}$, with the wave travelling at a speed set entirely by two constants measured in static electric and magnetic experiments: $c = \dfrac{1}{\sqrt{\mu_0 \varepsilon_0}}$. Substituting the measured values of $\mu_0$ and $\varepsilon_0$ gave a predicted speed of about $3.0 \times 10^8\ \text{m/s}$, which matched contemporary measurements of the speed of light to within experimental error. Because this agreement could not be coincidence, Maxwell concluded that light itself is an electromagnetic wave, and that electromagnetic waves of other frequencies should also exist and travel at the same speed $c$.

This prediction was highly significant: it unified two previously separate fields of physics (optics and electromagnetism) into one theory, and it predicted the existence of the entire electromagnetic spectrum beyond visible light before any of those other bands had been observed. Heinrich Hertz's 1887 experiments, generating and detecting radio waves and showing they obeyed the same wave speed $c = f\lambda$, gave direct experimental confirmation and led directly to radio communication.

Marker's note: the top band names both relevant Maxwell equations (not just "Maxwell's equations" vaguely), states the derived speed formula $c = 1/\sqrt{\mu_0\varepsilon_0}$ and its numerical agreement with the measured speed of light, and evaluates significance beyond "he predicted light is a wave" by naming the unification of optics/electromagnetism and Hertz's confirmation. A response that only restates the four laws without deriving or evaluating the speed prediction caps in the middle band.

Band-6 plan. Thesis: spectrum position (via $E = hf$) is a strong but not complete predictor of interaction. Give three worked examples across bands (radio/microwave - no ionisation, low energy; visible/UV threshold - bond-breaking and vision/skin damage; X-ray/gamma - ionising, penetrating). Note the limits (intensity, exposure time and the specific material also matter). Conclude with a judgement.

Model answer. A photon's frequency fixes its energy through $E = hf$, and this energy largely determines how the radiation interacts with matter, because most interactions (breaking a chemical bond, ejecting an electron, exciting a molecule) require a minimum energy threshold. Low-frequency radio and microwave photons carry only around $10^{-6}$ to $10^{-3}\ \text{eV}$, far below the roughly $4\ \text{eV}$ needed to break a typical chemical bond or the $\sim 10\ \text{eV}$ needed to ionise an atom; this radiation mainly causes molecules to rotate or vibrate (heating, as in a microwave oven) rather than damaging them chemically.

In the visible-to-UV range, photon energies of a few electron-volts cross important biological and chemical thresholds. Visible light near $2\ \text{eV}$ can trigger the retina's photochemistry (vision) without breaking most bonds, while UV photons above about $4\ \text{eV}$ carry enough energy to dimerise DNA bases, which is why UV causes sunburn and skin damage that visible light does not. At the high-energy end, X-ray and gamma photons (keV to MeV) exceed atomic ionisation energies by orders of magnitude, so they ionise atoms directly, penetrate soft tissue and are used both to image the body and to damage cancer cells.

However, spectral position alone does not fully predict interaction: intensity (photon number), exposure time and the specific material's absorption properties also matter. A very intense beam of low-energy infrared photons can still burn skin thermally even though no single photon can ionise a molecule, and a material transparent to one band (glass is transparent to visible light but opaque to UV) will not interact the same way as an opaque one, even at the same photon energy. Overall, $E = hf$ is the dominant factor separating ionising from non-ionising radiation and predicting threshold effects, but a complete prediction of interaction also needs the intensity and the material.

Marker's note: the top band uses $E = hf$ explicitly to compare at least three bands with a stated or implied threshold energy (bond energy $\sim 4\ \text{eV}$, ionisation $\sim 10\ \text{eV}$), and explicitly evaluates the limits of spectral position alone (intensity, exposure, material) rather than treating energy as the sole factor. A response that only lists band names and generic uses (e.g. "X-rays are used in hospitals") without linking to photon energy and thresholds caps in the middle band.