Wavelength, Frequency & Spectrum

PHYThe EM Spectrum

All EM waves travel at the same speed $c$ in vacuum, so wavelength and frequency are linked by $c=f\lambda$ . These problems convert between $\lambda$ and $f$ and then identify which band of the spectrum the wave belongs to.

Knowing the rough wavelength/frequency ranges of each band (radio to gamma) is essential for the identification mark.

Wave equation in vacuum

c = f\,\lambda

c=3\times10^8\ \text{m s}^{-1}

f

= frequency (Hz),

\lambda

= wavelength (m).

Frequency from wavelength

f = \dfrac{c}{\lambda}

Convert

\lambda

to metres first (e.g. 3 cm

=3\times10^{-2}

m).

Wavelength from frequency

\lambda = \dfrac{c}{f}

Visible light,

f\sim10^{14}

Hz, gives

\lambda

of a few hundred nm (

1\ \text{nm}=10^{-9}

m).

In vacuum every EM wave moves at $c$ ; only $f$ and $\lambda$ change between bands, kept consistent by $c=f\lambda$ .
Approximate boundaries: radio ( $>0.1$ m), microwave (mm-cm), infrared ( $700\ \text{nm}$ to $1\ \text{mm}$ ), visible ( $400$ to $700$ nm), ultraviolet ( $10$ to $400$ nm), X-ray ( $0.01$ to $10$ nm), gamma ( $<0.01$ nm).
Frequency is the invariant when a wave enters a medium; the wavelength changes to $\lambda'=\lambda/n$ while $f$ stays fixed.
Visible light, $f\approx 4$ to $7.5\times10^{14}$ Hz, corresponds to $\lambda\approx 400$ to $700$ nm (violet to red).
A 3 cm microwave has $f=10^{10}$ Hz (10 GHz), squarely in the microwave/radar band.
Higher frequency means shorter wavelength and higher photon energy ( $E=hf$ ), so gamma rays are the most energetic.
Always convert cm, mm, or nm to metres before dividing by or into $c$ .

Where the marks go

Forgetting unit conversion: 3 cm is $3\times10^{-2}$ m, not 3 m — a common factor error in $f=c/\lambda$ .
Misplacing a band boundary, e.g. calling a $10^{14}$ Hz wave a microwave instead of visible/infrared.
Assuming $\lambda$ stays the same in a medium; it is $f$ that is unchanged, with $\lambda$ shrinking by $n$ .
Writing the answer wavelength in metres when the band is naturally quoted in nm, and then misreading the band.

How the board asks it

Numericalc = f lambda, wavelength to frequency
A radio station broadcasts at a wavelength of $30$ m. Calculate the frequency of the broadcast, given $c = 3\times10^{8}\ \text{m s}^{-1}$ , and name the part of the electromagnetic spectrum to which it belongs.
Numericalc = f lambda, frequency to wavelength
An electromagnetic wave has a frequency of $5\times10^{14}\ \text{Hz}$ . Calculate its wavelength and identify the region of the electromagnetic spectrum to which it belongs.
Numericalconvert to metres before applying c = f lambda
A microwave oven produces radiation of wavelength $3$ cm. Calculate its frequency in $\text{Hz}$ and state which band of the electromagnetic spectrum this corresponds to.
Give reasonsfrequency invariant on entering a medium; lambda' = lambda / n
A beam of light of wavelength $600$ nm in air enters glass of refractive index $1.5$ . State, with reasons, what happens to its frequency and to its wavelength inside the glass.
Distinguishhigher frequency means shorter wavelength and higher photon energy
Arrange the following electromagnetic waves in order of increasing frequency: infrared, X-rays, microwaves, visible light. State which of them has the highest photon energy and give a reason.
Assertion–ReasonE = h f, so shorter wavelength means higher energy
Assertion: Gamma rays are more energetic than radio waves. Reason: Gamma rays have a longer wavelength than radio waves. Select the correct option regarding the assertion and the reason.

Practise this topic

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Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.