EM spectrum — all bands and applications: Worked examples

Every symbol used here was earned in the parent note. We reuse them, we do not redefine them, but we do re-anchor units at every step, because unit slips are where marks die.

The scenario matrix

Think of a "case" as a combination of two things: which direction you compute (do you know $ν$ and want $λ$ ? or know $E$ and want $λ$ ?) and what regime the numbers live in (huge radio wavelengths, tiny gamma wavelengths, degenerate/zero-like limits).

#	Cell (case class)	What makes it tricky	Hit by
A	$ν \to λ$ , low frequency (radio)	huge $λ$ , engineering meaning	Ex 1
B	$ν \to λ$ , high frequency (gamma)	tiny $λ$ , negative exponents	Ex 2
C	$λ \to E$ , energy in eV (visible)	two conversions chained	Ex 3
D	$E \to λ$ , energy given in eV (X-ray)	invert the formula, convert first	Ex 4
E	Ratio / no-numbers comparison	cancel constants, no calculator	Ex 5
F	Limiting / degenerate case ( $ν \to 0$ , $ν \to \infty$ )	what happens at the edges	Ex 6
G	Word problem (microwave oven, real device)	translate physics into a number	Ex 7
H	Boundary between bands (where does one band end?)	a threshold wavelength	Ex 8
I	Exam twist — mixed units, wrong-looking data	spot the trap, unit hygiene	Ex 9
J	Doppler-free consistency check — self-verify a band	sanity-check your own answer	Ex 10

Constants used everywhere: $c = 3 \times 1 0^{8} m/s, h = 6.63 \times 1 0^{- 34} J\cdots, 1 eV = 1.6 \times 1 0^{- 19} J .$

Figure — EM spectrum — all bands and applications

The picture above is the map. Notice the inverse tilt: as you slide right (frequency up), wavelength slides down. Every example below is just picking a point on this line and reading off a coordinate.

Worked examples

Worked example Ex 3 — Cell C: visible,

$λ \to E$ , answer in eV Red light has $λ = 650 nm$ . Find the photon energy in eV.

Forecast: Red is the low-energy end of visible; green was $2.25$ eV. Guess: a bit less than 2.25 eV, roughly $1.9$ eV.

$E = \frac{h c}{λ}$ . Why this step? We know $λ$ , want $E$ ; the parent chain $E = h ν = h c / λ$ skips computing $ν$ .
Convert $λ$ : $650 nm = 650 \times 1 0^{- 9} m = 6.5 \times 1 0^{- 7} m$ . Why this step? "nano" is $1 0^{- 9}$ ; SI metres required.
$E = \frac{( 6.63 \times 1 0 ^{- 34} ) ( 3 \times 1 0 ^{8} )}{6.5 \times 1 0 ^{- 7}} = \frac{1.989 \times 1 0 ^{- 25}}{6.5 \times 1 0 ^{- 7}} = 3.06 \times 1 0^{- 19} J$ . Why this step? Numerator first ( $6.63 \times 3 = 19.89$ , exponent $- 34 + 8 = - 26$ , so $1.989 \times 1 0^{- 25}$ ), then divide.
To eV: $E = \frac{3.06 \times 1 0 ^{- 19}}{1.6 \times 1 0 ^{- 19}} \approx 1.91 eV$ . Why this step? Divide joules by joules-per-eV to change units.

Verify: $1.91 eV < 2.25 eV$ (green) ✓ — red really is less energetic. Units: $\frac{J\cdots \cdot m/s}{m} = J$ . ✓

Worked example Ex 4 — Cell D: X-ray,

$E \to λ$ , energy given in eV A diagnostic X-ray photon has $E = 25 keV$ . Find $λ$ .

Forecast: keV is thousands of times the visible eV-scale, so $λ$ must be thousands of times smaller than 500 nm — around $0.05 nm$ , near atomic spacing.

Convert energy to joules first: $E = 25 \times 1 0^{3} eV \times 1.6 \times 1 0^{- 19} J/eV = 4.0 \times 1 0^{- 15} J$ . Why this step? $h$ and $c$ are in SI; you cannot mix eV into $h c / λ$ .
Invert the formula: $λ = \frac{h c}{E}$ . Why this step? We know $E$ , want $λ$ ; algebraically flip $E = h c / λ$ .
$λ = \frac{( 6.63 \times 1 0 ^{- 34} ) ( 3 \times 1 0 ^{8} )}{4.0 \times 1 0 ^{- 15}} = \frac{1.989 \times 1 0 ^{- 25}}{4.0 \times 1 0 ^{- 15}} = 4.97 \times 1 0^{- 11} m$ . Why this step? Same numerator as before; exponent $- 25 - (- 15) = - 10$ .

Verify: $4.97 \times 1 0^{- 11} m \approx 0.05 nm$ — inside the $0.01$ – $10$ nm X-ray band. ✓ Matches forecast. This near-atomic size is exactly why X-rays do Bragg diffraction.

Worked example Ex 5 — Cell E: ratio comparison, no calculator

Photon P has wavelength $λ_{P}$ ; photon Q has $λ_{Q} = 4 λ_{P}$ . Which has more energy, and by what factor?

Forecast: Longer wavelength ⇒ lower energy (inverse). Q has the longer $λ$ , so P wins, and probably by exactly 4.

Write both energies: $E_{P} = \frac{h c}{λ _{P}}$ , $E_{Q} = \frac{h c}{λ _{Q}}$ . Why this step? Same formula, keep it symbolic so constants cancel.
Take the ratio: $\frac{E _{P}}{E _{Q}} = \frac{h c / λ _{P}}{h c / λ _{Q}} = \frac{λ _{Q}}{λ _{P}}$ . Why this step? $h$ and $c$ divide out — no numbers needed; energy scales like $1/ λ$ .
Substitute $λ_{Q} = 4 λ_{P}$ : $\frac{E _{P}}{E _{Q}} = 4$ . Why this step? The $λ_{P}$ cancels, leaving the factor.

Verify: $E_{P} = 4 E_{Q}$ : P is 4× more energetic, and it has the shorter wavelength — consistent with "energy $\propto 1/ λ$ ." ✓ No calculator, no units — a pure structure argument.

Worked example Ex 6 — Cell F: limiting / degenerate cases

What happens to $λ$ and $E$ as (a) $ν \to 0$ and (b) $ν \to \infty$ ? Is $ν = 0$ a real photon?

Forecast: Tiny $ν$ ⇒ giant $λ$ , near-zero energy. Huge $ν$ ⇒ shrinking $λ$ , exploding energy. $ν = 0$ should be a "no wave at all" edge.

From $λ = c / ν$ : as $ν \to 0^{+}$ , $λ = c / ν \to \infty$ . Why this step? Dividing a fixed $c$ by something shrinking to zero blows up. Picture the leftmost point of the figure sliding off to infinity.
From $E = h ν$ : as $ν \to 0^{+}$ , $E \to 0$ . Why this step? Energy is directly proportional to $ν$ ; zero frequency means zero energy — no oscillation, no photon.
As $ν \to \infty$ : $λ = c / ν \to 0$ and $E = h ν \to \infty$ . Why this step? The mirror limit — the right end of the map runs to infinitely energetic, infinitely short waves.

Verify (degenerate check): Set $ν = 0$ : $E = h \cdot 0 = 0$ and $c = 0 \cdot λ$ has no finite $λ$ solution — so $ν = 0$ is not a physical photon, it's the empty limit. ✓ Both limits respect $c = ν λ$ at every finite point. (These edges are why the spectrum is an open-ended continuum, not a bounded list — see Maxwell's Equations allowing any $ν$ .)

Worked example Ex 7 — Cell G: word problem (microwave oven)

A domestic microwave oven runs at $ν = 2.45 GHz$ . (a) Find $λ$ . (b) Find the energy of one such photon in eV, and comment on whether it can break a chemical bond ( $\sim$ a few eV).

Forecast: GHz is microwave, so $λ$ should be a few centimetres. The photon energy will be microscopic — far below a bond, confirming the parent note's "not a resonance, it's heating" story.

$λ = c / ν = \frac{3 \times 1 0 ^{8}}{2.45 \times 1 0 ^{9}}$ . Why this step? $ν$ known, want $λ$ ; GHz $= 1 0^{9}$ Hz.
$λ = 0.1224 m \approx 12.2 cm$ . Why this step? $3/2.45 = 1.224$ , exponent $1 0^{8} /1 0^{9} = 1 0^{- 1}$ .
Photon energy: $E = h ν = (6.63 \times 1 0^{- 34}) (2.45 \times 1 0^{9}) = 1.62 \times 1 0^{- 24} J$ . Why this step? Direct $E = h ν$ ; no need for $λ$ here.
In eV: $E = \frac{1.62 \times 1 0 ^{- 24}}{1.6 \times 1 0 ^{- 19}} = 1.02 \times 1 0^{- 5} eV$ . Why this step? Convert to compare against the "few eV" bond scale.

Verify: $λ \approx 12 cm$ sits in the microwave band ✓. And $1 0^{- 5} eV ≪ 1 eV$ — a single microwave photon is roughly $100, 000 \times$ too weak to break a bond. ✓ So heating cannot be one-photon bond-breaking; it must be the bulk dielectric relaxation the parent note described.

Worked example Ex 8 — Cell H: boundary between two bands

The visible band ends and UV begins near $λ = 400 nm$ . What frequency and photon energy (eV) mark this violet/UV boundary?

Forecast: Shorter than red, so higher energy than the 1.91 eV of Ex 3 — around 3 eV, and $ν$ close to $1 0^{15}$ Hz.

$ν = c / λ = \frac{3 \times 1 0 ^{8}}{400 \times 1 0 ^{- 9}} = \frac{3 \times 1 0 ^{8}}{4 \times 1 0 ^{- 7}} = 7.5 \times 1 0^{14} Hz$ . Why this step? Boundary is a wavelength; convert to frequency to place it on the map.
$E = \frac{h c}{λ} = \frac{1.989 \times 1 0 ^{- 25}}{4 \times 1 0 ^{- 7}} = 4.97 \times 1 0^{- 19} J$ . Why this step? Same $h c$ numerator; get raw energy.
In eV: $E = \frac{4.97 \times 1 0 ^{- 19}}{1.6 \times 1 0 ^{- 19}} \approx 3.11 eV$ . Why this step? Convert to the atomic-transition scale.

Verify: $3.11 eV > 1.91 eV$ (red, Ex 3) ✓ — violet is more energetic than red, as it must be. $7.5 \times 1 0^{14} Hz \approx 1 0^{15}$ ✓, matching the forecast and Atomic spectra scales.

Worked example Ex 9 — Cell I: exam twist (mixed units, spot the trap)

"A UV lamp emits at $λ = 250$ . Its photons carry $8 \times 1 0^{- 19}$ J. Is the data self-consistent?" (Note: the wavelength unit is deliberately omitted — a classic exam trap.)

Forecast: UV energy of $8 \times 1 0^{- 19} J = 5 eV$ is plausible for UV, so the missing unit is almost certainly nm, not m or μm. Let's test.

Assume nm: $λ = 250 \times 1 0^{- 9} m = 2.5 \times 1 0^{- 7} m$ . Why this step? The exam expects you to supply the physically sensible unit; test the hypothesis.
Predicted energy: $E = \frac{h c}{λ} = \frac{1.989 \times 1 0 ^{- 25}}{2.5 \times 1 0 ^{- 7}} = 7.96 \times 1 0^{- 19} J$ . Why this step? Compute what the energy should be if $λ = 250 nm$ .
Compare to the stated $8 \times 1 0^{- 19} J$ : they agree to within $0.5%$ . Why this step? Consistency check — a match confirms the unit assumption.

Verify: $7.96 \times 1 0^{- 19} \approx 8 \times 1 0^{- 19} J$ ✓, and $8 \times 1 0^{- 19} /1.6 \times 1 0^{- 19} = 5 eV$ , right in UV territory. So the data is consistent provided $λ$ is in nanometres. Trap avoided: had you plugged $λ = 250 m$ , you'd get $E \sim 1 0^{- 27} J$ (radio!) — absurd for a "UV lamp."

Active recall

Recall Which cell does each trick belong to?

Given $ν$ , want $λ$ — which formula? ::: $λ = c / ν$ (rearranged $c = ν λ$ ). Given $E$ in eV, want $λ$ — first move? ::: Convert eV → joules (× $1.6 \times 1 0^{- 19}$ ) before using $h c / λ$ . Fastest way to compare two photon energies without a calculator? ::: Ratio $E_{1} / E_{2} = λ_{2} / λ_{1}$ — constants cancel. What is the always-available sanity check on any $(ν, λ)$ pair? ::: Their product must equal $c = 3 \times 1 0^{8}$ m/s. As $ν \to 0$ , what happens to $λ$ and $E$ ? ::: $λ \to \infty$ , $E \to 0$ — the empty (non-photon) limit. An exam gives $λ = 250$ with no unit for a UV lamp — what unit do you supply? ::: Nanometres (nm), because that yields a UV-scale energy of ~5 eV.