Wave Nature of Electromagnetic Radiation and the Electromagnetic Spectrum

Why an Atom Chapter Suddenly Talks About Light

Rutherford's nuclear atom carries a fatal flaw. An electron circling a nucleus is a charged particle under constant acceleration, and nineteenth-century electromagnetic theory says an accelerating charge must radiate energy. A radiating electron spirals inward and crashes into the nucleus in about $10^{-8}$ s. Atoms are stable, so the classical picture was wrong. It also said nothing about where the electrons are or what energies they have.

Niels Bohr fixed this in 1913 using two lines of evidence on how radiation and matter interact:

1. The dual character of electromagnetic radiation — radiation behaves like a wave in some experiments and like a stream of particles in others.
2. Experimental results on atomic spectra — atoms emit and absorb light only at certain sharply defined wavelengths.

Key Point: Bohr's model rests on two pillars: (i) the wave-particle duality of radiation, and (ii) the line spectra of atoms. This section builds the wave picture of radiation; the particle picture comes in Section 4, and atomic spectra in Section 5.

Corpuscles or waves: a three-century argument

People argued about what light is long before anyone could measure it properly.

Newton's authority kept the corpuscular view alive for over a century, but by 1900 the wave nature of light looked settled. Interference, diffraction (the bending of a wave around an obstacle) and polarisation all had clean wave explanations and no particle explanation.

The wave picture then ran into three walls of its own — black-body radiation, the photoelectric effect and the heat capacities of solids (Section 4). Radiation needs both descriptions, which is what "dual character" means.

Where thermal radiation fits in

Mid-nineteenth-century physicists studied the absorption and emission of radiation by heated objects — thermal radiation. A hot iron rod glows red, then orange, then white as it heats; a warm object you cannot see still radiates in the infrared. Thermal radiation is a mixture of electromagnetic waves of many frequencies, and the way its intensity is spread over those frequencies is the black-body problem that broke classical physics. Study of thermal-radiation laws began in the 1850s; the theory of the waves arrived with Maxwell in the early 1870s.

[Board] "Name the two developments that led to Bohr's model" is a standard two-mark question. The answer is the numbered list above: dual nature of electromagnetic radiation, and atomic spectra.

Maxwell's Electromagnetic Waves

When an electrically charged particle moves with acceleration, it produces alternating electric and magnetic fields, and these fields are transmitted outward in the form of waves. Those waves are electromagnetic waves or electromagnetic radiation. This was James Clerk Maxwell's insight (1870). It also showed that light itself is one of these waves, with an oscillating electric character and an oscillating magnetic character travelling together.

Shake a charge back and forth and you create a ripple in the electric field around it. A changing electric field creates a magnetic field and a changing magnetic field creates an electric field, so the ripple regenerates itself as it travels.

The four properties you must know

(i) The two fields are mutually perpendicular, and both are perpendicular to the direction of travel. In the figure the electric field ($\vec{E}$) oscillates up and down, the magnetic field ($\vec{B}$) in and out of the page, and the wave travels to the right. The two components share the same wavelength, frequency, speed and amplitude.

(ii) Electromagnetic waves need no medium. Sound needs air and a water wave needs water, but an electromagnetic wave carries its own fields through a perfect vacuum. That is how sunlight crosses 150 million kilometres of empty space.

(iii) There are many kinds of electromagnetic radiation, and they differ only in wavelength (or frequency). Together they make up the electromagnetic spectrum — radio, microwave, infrared, visible, ultraviolet, X-ray, gamma. The physics is the same throughout.

(iv) Different kinds of units are used to describe the radiation. Frequency in hertz, wavelength in metres, nanometres or ångströms, wavenumber in cm$^{-1}$.

Key Point: An electromagnetic wave is a pair of oscillating electric and magnetic fields, perpendicular to each other and to the direction of propagation, that needs no medium and travels through vacuum at the speed of light. All electromagnetic radiations differ from one another only in wavelength or frequency.

A wave is transverse

The fields oscillate across the direction of travel, so an electromagnetic wave is a transverse wave, like a wave on a rope; sound is longitudinal. Being transverse is why light can be polarised and sound cannot.

[JEE Main] It is the oscillating (accelerating) charge that radiates. A charge at rest produces a static electric field, and a charge in uniform motion produces a steady magnetic field; neither radiates.

The speed of light

In vacuum every electromagnetic radiation, whatever its wavelength, travels at the same speed:

$$c = 3.0 \times 10^{8}\ \mathrm{m\ s^{-1}} \qquad (2.997925 \times 10^{8}\ \mathrm{m\ s^{-1}}\text{, to be precise})$$

This is the speed of light, symbol $c$. Radio waves, gamma rays and yellow light all move at exactly this speed in vacuum. In glass or water the speed is lower, and different wavelengths slow by different amounts, which is why a prism spreads white light into colours.

Use $c = 3.0 \times 10^{8}$ m s$^{-1}$ unless a question supplies a different value; the precise value changes answers only in the fourth significant figure.

Describing a Wave: Wavelength, Frequency, Amplitude and Speed

Any wave is described by a handful of quantities. Every problem in this section is these definitions plus one equation.

The vocabulary

Key Point (Definition): The SI unit of frequency is the hertz (Hz, equal to s$^{-1}$), named after Heinrich Hertz. One hertz means one complete wave passing a given point per second.

Amplitude controls the intensity (brightness) of the radiation, not its colour: a brighter red lamp has larger-amplitude waves of the same wavelength. This matters for the photoelectric effect later.

Time period is the reciprocal of frequency. If 5 waves pass per second ($\nu = 5$ Hz), each takes $1/5 = 0.2$ s to pass.

The one equation: $c = \nu\lambda$

In one second, $\nu$ waves pass a point and each wave is $\lambda$ long, so the distance the wave front moves in one second is $\nu$ times $\lambda$:

$$c = \nu \lambda$$

Rearranged in the two forms you will actually use:

$$\nu = \frac{c}{\lambda} \qquad\text{and}\qquad \lambda = \frac{c}{\nu}$$

Since $c$ is fixed in vacuum, frequency and wavelength are inversely proportional: double the wavelength and the frequency halves. That settles every "arrange in increasing order" question.

Key Point: $c = \nu\lambda$ with $c = 3.0 \times 10^{8}$ m s$^{-1}$. Put $\lambda$ in metres and you get $\nu$ in hertz — no other unit pair works without conversion.

Smaller units of wavelength

Electromagnetic wavelengths run from kilometres (radio) to less than a picometre (gamma rays), so smaller units are used constantly.

The conversions between them:

$$1\ \text{\AA} = 10^{-10}\ \mathrm{m} = 10^{-8}\ \mathrm{cm} = 0.1\ \mathrm{nm} = 100\ \mathrm{pm}$$

$$1\ \mathrm{nm} = 10^{-9}\ \mathrm{m} = 10^{-7}\ \mathrm{cm} = 10\ \text{\AA} = 1000\ \mathrm{pm}$$

Yellow sodium light can be written as 580 nm, 5800 Å, $5.8 \times 10^{-7}$ m or $5.8 \times 10^{-5}$ cm — all the same wavelength.

Units of frequency

Frequency uses the ordinary SI prefixes: 1 kHz $= 10^{3}$ Hz, 1 MHz $= 10^{6}$ Hz, 1 GHz $= 10^{9}$ Hz. AM radio broadcasts in kHz, FM in MHz, ovens and phones in GHz. Given "1368 kHz", write $1.368 \times 10^{6}$ s$^{-1}$ first.

[Exam Tip] Before touching $c = \nu\lambda$, convert everything to metres and seconds. Most wrong answers here come from leaving $\lambda$ in nm or $\nu$ in MHz, or from taking Å as $10^{-9}$ m instead of $10^{-10}$ m.

Wavenumber: The Spectroscopist's Favourite

The wavenumber, written $\bar{\nu}$ (read "nu-bar"), is the number of wavelengths per unit length — how many complete waves fit into one metre, or one centimetre, of the beam.

Key Point (Definition): $\bar{\nu} = \dfrac{1}{\lambda}$. Wavenumber has units that are the reciprocal of the wavelength unit: the SI unit is m$^{-1}$, but the unit used in practice is cm$^{-1}$ (not an SI unit).

Why chemists use it

Frequency decides the energy of a photon (Section 4), but light frequencies are around $10^{14}$ or $10^{15}$ Hz, awkward to write. Wavenumber is directly proportional to frequency ($\bar{\nu} = \nu/c$) and gives friendlier numbers: visible light lies between about 13,000 and 25,000 cm$^{-1}$, and the whole hydrogen spectrum runs on one constant, 109,677 cm$^{-1}$. Infrared spectroscopy reports every peak in cm$^{-1}$.

The three quantities together

Because $c = \nu\lambda$, frequency, wavelength and wavenumber are locked together:

$$\bar{\nu} = \frac{1}{\lambda} = \frac{\nu}{c} \qquad\Longleftrightarrow\qquad \nu = c\,\bar{\nu}$$

In an ordering question, "increasing wavenumber" means the same as "increasing frequency" and the opposite of "increasing wavelength".

Converting m$^{-1}$ to cm$^{-1}$

Since 1 m $= 100$ cm, a wavenumber in m$^{-1}$ is 100 times larger than the same wavenumber in cm$^{-1}$:

$$1\ \mathrm{cm^{-1}} = 100\ \mathrm{m^{-1}} \qquad\text{so}\qquad \bar{\nu}\,(\mathrm{cm^{-1}}) = \frac{\bar{\nu}\,(\mathrm{m^{-1}})}{100}$$

Students routinely multiply instead of divide. A centimetre is shorter than a metre, so fewer waves fit into it and the cm$^{-1}$ number must be smaller.

Yellow light of wavelength 5800 Å:

$$\lambda = 5800 \times 10^{-10}\ \mathrm{m} = 5.8 \times 10^{-7}\ \mathrm{m} = 5.8 \times 10^{-5}\ \mathrm{cm}$$

$$\bar{\nu} = \frac{1}{5.8 \times 10^{-7}\ \mathrm{m}} = 1.724 \times 10^{6}\ \mathrm{m^{-1}} = \frac{1}{5.8 \times 10^{-5}\ \mathrm{cm}} = 1.724 \times 10^{4}\ \mathrm{cm^{-1}}$$

Same light, two units, a factor of 100 between them.

[JEE Main] For visible and nearby light, $\bar{\nu}\,(\mathrm{cm^{-1}}) = \dfrac{10^{7}}{\lambda\,(\mathrm{nm})}$. Check with 580 nm: $10^{7}/580 = 17{,}241$ cm$^{-1}$, which is $1.724 \times 10^{4}$ cm$^{-1}$.

A note on the symbol

Do not confuse the three "nu"s: $\nu$ is frequency, $\bar{\nu}$ is wavenumber, and $v$ (Latin vee) is the velocity of a particle, from Section 4 onward. In handwriting, put a clear bar over the wavenumber and keep the Greek $\nu$ curly.

The Electromagnetic Spectrum

All electromagnetic radiations travel at $c$ and obey $c = \nu\lambda$; they differ only in wavelength and frequency. Laid out from longest wavelength to shortest, they form the electromagnetic spectrum. The boundaries between regions are conventions, but the order is fixed and you must know it in both directions.

The regions, with representative numbers

Cosmic rays from outer space are more energetic still and sit beyond gamma rays at the high-frequency end.

Memorise the powers of ten in the frequency column: radio $10^{6}$, microwave $10^{10}$, infrared $10^{13}$, visible $10^{15}$, ultraviolet $10^{16}$.

Key Point: In order of increasing frequency (equivalently increasing wavenumber, and — as Section 4 shows — increasing energy per photon): $$\text{radio} < \text{microwave} < \text{infrared} < \text{visible} < \text{ultraviolet} < \text{X-rays} < \gamma\text{-rays}$$ Wavelength runs the opposite way: radio waves are the longest, gamma rays the shortest.

The visible window

The portion around $10^{15}$ Hz is visible light. Everything else needs an instrument: an antenna for radio, a thermal sensor for infrared, a detector for X-rays.

The visible range runs from violet at 400 nm to red at 750 nm. Converting with $c = \nu\lambda$:

$$\nu_{\text{violet}} = \frac{3.0 \times 10^{8}}{400 \times 10^{-9}} = 7.5 \times 10^{14}\ \mathrm{Hz} \qquad \nu_{\text{red}} = \frac{3.0 \times 10^{8}}{750 \times 10^{-9}} = 4.0 \times 10^{14}\ \mathrm{Hz}$$

Visible light spans $4.0 \times 10^{14}$ to $7.5 \times 10^{14}$ Hz. Violet has the shortest wavelength and highest frequency; red the longest wavelength and lowest frequency. Inside the window the colours run in order:

VIBGYOR lists them from violet to red, short to long wavelength. Yellow sodium light at 580 nm and the neon line at 616 nm sit inside this window; a hydrogen line at 1285 nm is infrared and invisible.

Just beyond the window

Just shorter than violet is the ultraviolet, part of the Sun's radiation and the cause of sunburn. Just longer than red is the infrared, felt as heat from a fire. The visible band is a sliver in a spectrum spanning more than twenty powers of ten.

[NEET] Ordering questions usually mix one item from each region. Slot each into its region and read the order off the spectrum. Amber is visible (orange-yellow, roughly 590 nm), so it sits between infrared and ultraviolet.

[JEE Main] When wavelengths come in mixed units, convert them all to one convenient unit (nm for visible and UV, m for radio) rather than to metres, then compare. Long wavelength always means low frequency.

Working the Numbers: A Method That Never Fails

Every calculation here has one shape: you are given one of ($\lambda$, $\nu$, $\bar{\nu}$, $T$) and asked for the others.

The four-step routine

1. Convert what is given to SI. Wavelength to metres (nm $\times 10^{-9}$, Å $\times 10^{-10}$, cm $\times 10^{-2}$); frequency to hertz (kHz $\times 10^{3}$, MHz $\times 10^{6}$, GHz $\times 10^{9}$); period in seconds.
2. Pick the right form of $c = \nu\lambda$, or $\nu = 1/T$ if a period is given.
3. Do the powers of ten separately from the digits. For $\nu = 3.0 \times 10^{8} / (5.8 \times 10^{-7})$: $3.0/5.8 = 0.517$, $10^{8}/10^{-7} = 10^{15}$, so $5.17 \times 10^{14}$ Hz.
4. Convert to the unit asked for and check the order of magnitude. A visible frequency must land around $10^{14}$ to $10^{15}$ Hz; a radio wavelength in metres.

Order-of-magnitude anchors

Distance and time from the speed of light

Since $c$ is a speed, two more problem types appear:

• Distance travelled in time $t$: $d = ct$. Light travelling for 30 s covers $3.0 \times 10^{8} \times 30 = 9.0 \times 10^{9}$ m. Wavelength is irrelevant; every electromagnetic wave covers the same distance in the same time.
• Time to cover distance $d$: $t = d/c$. The Sun is $1.5 \times 10^{11}$ m away, so sunlight takes $1.5 \times 10^{11} / (3.0 \times 10^{8}) = 500$ s, about 8.3 minutes.

The mistakes that cost marks

Key Point: Convert to SI, apply $c = \nu\lambda$, separate digits from powers of ten, and check the answer against the spectrum.

[Exam Tip] With $c = 3.0 \times 10^{8}$ (two significant figures) answers are good to two or three figures. Quote $5.17 \times 10^{14}$ Hz, not $5.172413 \times 10^{14}$.

Solved Examples

Question 1: The wavelength of a radio station

The Vividh Bharati station of All India Radio, Delhi, broadcasts on a frequency of 1368 kHz. Calculate the wavelength of the electromagnetic radiation emitted by the transmitter. Which part of the electromagnetic spectrum does it belong to?

Answer:

First the frequency in SI units: $\nu = 1368\ \mathrm{kHz} = 1.368 \times 10^{6}\ \mathrm{s^{-1}}$. I want wavelength, so I use $\lambda = c/\nu$:

$$\lambda = \frac{3.0 \times 10^{8}\ \mathrm{m\ s^{-1}}}{1.368 \times 10^{6}\ \mathrm{s^{-1}}}$$

Digits and powers separately: $3.0/1.368 = 2.193$ and $10^{8}/10^{6} = 10^{2}$, so $\lambda = 2.193 \times 10^{2}\ \mathrm{m} = 219.3$ m. A few hundred metres, or a frequency near $10^{6}$ Hz, is the radio-wave region.

Ans: $\lambda = 219.3$ m; radio waves. Watch out: The s$^{-1}$ must cancel to leave metres. If it does not, a unit conversion was skipped.

Question 2: Frequencies at the two ends of the visible spectrum

The wavelength range of the visible spectrum extends from violet (400 nm) to red (750 nm). Express these wavelengths in frequencies (Hz). (1 nm $= 10^{-9}$ m)

Answer:

Both wavelengths into metres: violet $\lambda = 4.00 \times 10^{-7}$ m, red $\lambda = 7.50 \times 10^{-7}$ m. For violet:

$$\nu = \frac{c}{\lambda} = \frac{3.00 \times 10^{8}\ \mathrm{m\ s^{-1}}}{4.00 \times 10^{-7}\ \mathrm{m}} = 7.50 \times 10^{14}\ \mathrm{Hz}$$

For red:

$$\nu = \frac{3.00 \times 10^{8}\ \mathrm{m\ s^{-1}}}{7.50 \times 10^{-7}\ \mathrm{m}} = 4.00 \times 10^{14}\ \mathrm{Hz}$$

The visible band therefore runs from $4.0 \times 10^{14}$ Hz (red) to $7.5 \times 10^{14}$ Hz (violet).

Ans: Violet: $7.50 \times 10^{14}$ Hz; red: $4.00 \times 10^{14}$ Hz; visible range $4.0 \times 10^{14}$ to $7.5 \times 10^{14}$ Hz. Watch out: The longer wavelength (red) gives the lower frequency, not the higher one.

Question 3: Wavenumber and frequency of yellow light in ångström

Calculate (a) the wavenumber and (b) the frequency of yellow radiation having wavelength 5800 Å.

Answer:

The wavelength in both units I need: $\lambda = 5800\ \text{\AA} = 5.8 \times 10^{-7}$ m $= 5.8 \times 10^{-5}$ cm.

(a) Wavenumber in m$^{-1}$:

$$\bar{\nu} = \frac{1}{\lambda} = \frac{1}{5.8 \times 10^{-7}\ \mathrm{m}} = 1.724 \times 10^{6}\ \mathrm{m^{-1}}$$

In cm$^{-1}$:

$$\bar{\nu} = \frac{1}{5.8 \times 10^{-5}\ \mathrm{cm}} = 1.724 \times 10^{4}\ \mathrm{cm^{-1}}$$

Dividing the m$^{-1}$ value by 100 gives the same number.

(b) Frequency:

$$\nu = \frac{c}{\lambda} = \frac{3.0 \times 10^{8}\ \mathrm{m\ s^{-1}}}{5.8 \times 10^{-7}\ \mathrm{m}} = 5.172 \times 10^{14}\ \mathrm{s^{-1}}$$

The other route agrees: $\nu = c\,\bar{\nu} = 3.0 \times 10^{8} \times 1.724 \times 10^{6} = 5.172 \times 10^{14}$ Hz.

Ans: (a) $\bar{\nu} = 1.724 \times 10^{6}$ m$^{-1}$ $= 1.724 \times 10^{4}$ cm$^{-1}$; (b) $\nu = 5.172 \times 10^{14}$ Hz. Watch out: The wavenumber unit is the reciprocal of whatever unit the wavelength is in. Put $\lambda$ in cm if the answer is wanted in cm$^{-1}$; the two values differ by exactly a factor of 100.

Question 4: The sodium lamp in nanometres

Yellow light emitted from a sodium lamp has a wavelength of 580 nm. Calculate the frequency and the wavenumber of the yellow light.

Answer:

In metres, $\lambda = 580\ \mathrm{nm} = 580 \times 10^{-9}\ \mathrm{m} = 5.80 \times 10^{-7}$ m.

$$\nu = \frac{c}{\lambda} = \frac{3.0 \times 10^{8}}{5.80 \times 10^{-7}} = 5.17 \times 10^{14}\ \mathrm{s^{-1}}$$

$$\bar{\nu} = \frac{1}{\lambda} = \frac{1}{5.80 \times 10^{-7}\ \mathrm{m}} = 1.72 \times 10^{6}\ \mathrm{m^{-1}}$$

Dividing by 100 gives $1.72 \times 10^{4}$ cm$^{-1}$, and the shortcut $10^{7}/580 = 17{,}241$ cm$^{-1}$ matches. 580 nm and 5800 Å are the same wavelength, so the answers match Question 3.

Ans: $\nu = 5.17 \times 10^{14}$ Hz; $\bar{\nu} = 1.72 \times 10^{6}$ m$^{-1}$ ($1.72 \times 10^{4}$ cm$^{-1}$).

Question 5: From time period to everything else

Calculate the wavelength, frequency and wavenumber of a light wave whose period is $2.0 \times 10^{-10}$ s.

Answer:

The period gives the frequency:

$$\nu = \frac{1}{T} = \frac{1}{2.0 \times 10^{-10}\ \mathrm{s}} = 5.0 \times 10^{9}\ \mathrm{s^{-1}}$$

Then the wavelength:

$$\lambda = \frac{c}{\nu} = \frac{3.0 \times 10^{8}\ \mathrm{m\ s^{-1}}}{5.0 \times 10^{9}\ \mathrm{s^{-1}}} = 6.0 \times 10^{-2}\ \mathrm{m} = 6.0\ \mathrm{cm}$$

And the wavenumber:

$$\bar{\nu} = \frac{1}{\lambda} = \frac{1}{6.0 \times 10^{-2}\ \mathrm{m}} = 16.67\ \mathrm{m^{-1}} = 0.1667\ \mathrm{cm^{-1}}$$

5 GHz with a wavelength of 6 cm sits in the microwave region.

Ans: $\nu = 5.0 \times 10^{9}$ Hz; $\lambda = 6.0 \times 10^{-2}$ m (6.0 cm); $\bar{\nu} = 16.67$ m$^{-1}$ ($0.167$ cm$^{-1}$). Watch out: "Light wave" in a question just means electromagnetic wave. This one is not visible light, so check the region yourself before naming it.

Question 6: Arranging radiations by frequency

Arrange the following types of radiation in increasing order of frequency: (a) radiation from a microwave oven, (b) amber light from a traffic signal, (c) radiation from an FM radio, (d) cosmic rays from outer space, (e) X-rays.

Answer:

I place each item in its region first. FM radio: radio waves, around $10^{8}$ Hz. Microwave oven: microwaves, around $2.45 \times 10^{9}$ Hz. Amber traffic light: visible (orange-yellow, roughly 590 nm), around $5 \times 10^{14}$ Hz. X-rays: around $10^{18}$ Hz. Cosmic rays: beyond gamma rays, above $10^{20}$ Hz.

Frequency increases from radio towards gamma, so the order is FM radio $<$ microwave oven $<$ amber light $<$ X-rays $<$ cosmic rays. The same list in increasing wavelength is exactly reversed.

Ans: (c) < (a) < (b) < (e) < (d). Watch out: Decide first whether the question asks for increasing frequency or increasing wavelength — the two orders are opposites.

Question 7: A neon sign — frequency and distance travelled

Neon gas is generally used in sign boards. If it emits strongly at 616 nm, calculate (a) the frequency of the emission, and (b) the distance travelled by this radiation in 30 s.

Answer:

In metres, $\lambda = 616\ \mathrm{nm} = 6.16 \times 10^{-7}$ m.

(a) $\nu = \dfrac{c}{\lambda} = \dfrac{3.0 \times 10^{8}}{6.16 \times 10^{-7}}$. Digits: $3.0/6.16 = 0.487$; powers: $10^{8}/10^{-7} = 10^{15}$. So $\nu = 4.87 \times 10^{14}\ \mathrm{Hz}$. That lies between $4.0$ and $7.5 \times 10^{14}$ Hz, so it is visible, orange-red as a neon sign should be.

(b) Light travels at $c$ whatever its wavelength, so

$$d = c\,t = 3.0 \times 10^{8}\ \mathrm{m\ s^{-1}} \times 30\ \mathrm{s} = 9.0 \times 10^{9}\ \mathrm{m}$$

Ans: (a) $\nu = 4.87 \times 10^{14}$ Hz; (b) $d = 9.0 \times 10^{9}$ m. Watch out: Part (b) needs only $c$ and $t$. The wavelength does not enter it at all.

Question 8: Wavelength inside a microwave oven

A domestic microwave oven operates at a frequency of 2450 MHz. Calculate the wavelength of the radiation in (a) metres and (b) centimetres, and state its wavenumber in m$^{-1}$.

Answer:

In hertz, $\nu = 2450\ \mathrm{MHz} = 2.45 \times 10^{9}\ \mathrm{s^{-1}}$.

(a) $\lambda = \dfrac{c}{\nu} = \dfrac{3.0 \times 10^{8}}{2.45 \times 10^{9}} = 1.224 \times 10^{-1}\ \mathrm{m} = 0.122$ m.

(b) In centimetres, $0.122 \times 100 = 12.2$ cm. Wavenumber:

$$\bar{\nu} = \frac{1}{\lambda} = \frac{1}{0.1224\ \mathrm{m}} = 8.17\ \mathrm{m^{-1}} \approx 0.082\ \mathrm{cm^{-1}}$$

12 cm at $2.45 \times 10^{9}$ Hz is squarely microwave.

Ans: (a) $0.122$ m; (b) 12.2 cm; $\bar{\nu} \approx 8.2$ m$^{-1}$. Watch out: Microwave wavelengths are centimetre-sized. The oven door mesh has holes a few millimetres across, far smaller than 12 cm, so the radiation cannot escape while visible light (500 nm) passes through.

Question 9: An FM station and a mobile-phone band

(i) An FM radio station broadcasts at 100 MHz. What is the wavelength? (ii) A mobile-phone signal has a wavelength of 15 cm. What is its frequency in GHz? (iii) Which of the two has the greater wavenumber?

Answer:

(i) $\nu = 1.0 \times 10^{8}$ s$^{-1}$, so $\lambda = \dfrac{3.0 \times 10^{8}}{1.0 \times 10^{8}} = 3.0$ m.

(ii) 15 cm is 0.15 m, so $\nu = \dfrac{c}{\lambda} = \dfrac{3.0 \times 10^{8}}{0.15} = 2.0 \times 10^{9}\ \mathrm{s^{-1}} = 2.0$ GHz.

(iii) FM: $\bar{\nu} = 1/3.0 = 0.33$ m$^{-1}$. Phone: $\bar{\nu} = 1/0.15 = 6.7$ m$^{-1}$. The phone signal has the shorter wavelength, so the greater wavenumber.

Ans: (i) 3.0 m; (ii) 2.0 GHz; (iii) the mobile-phone signal. Watch out: Wavenumber always ranks the same way as frequency, never the same way as wavelength. A useful anchor: 300 MHz is exactly 1 m, so 100 MHz is 3 m and 3 GHz is 10 cm.

Question 10: Same wavelength, four units

The strong red line of hydrogen (the first Balmer line) has a wavelength of 656.3 nm. Express this wavelength in (a) metres, (b) ångström, (c) centimetres and (d) picometres, and then find its wavenumber in cm$^{-1}$.

Answer:

(a) $656.3\ \mathrm{nm} = 6.563 \times 10^{-7}$ m.

(b) 1 nm $= 10\ \text{\AA}$, so $656.3\ \mathrm{nm} = 6563\ \text{\AA}$.

(c) $6.563 \times 10^{-7}\ \mathrm{m} \times 100 = 6.563 \times 10^{-5}$ cm.

(d) 1 nm $= 1000$ pm, so $656.3\ \mathrm{nm} = 6.563 \times 10^{5}$ pm.

Wavenumber, from the centimetre value:

$$\bar{\nu} = \frac{1}{6.563 \times 10^{-5}\ \mathrm{cm}} = 1.524 \times 10^{4}\ \mathrm{cm^{-1}}$$

The shortcut agrees: $10^{7}/656.3 = 15{,}237$ cm$^{-1}$.

Ans: (a) $6.563 \times 10^{-7}$ m; (b) 6563 Å; (c) $6.563 \times 10^{-5}$ cm; (d) $6.563 \times 10^{5}$ pm; $\bar{\nu} = 1.524 \times 10^{4}$ cm$^{-1}$. Watch out: Moving between nm, Å and pm shifts the decimal point by one place and by three places. This line reappears in Section 5 as the first line of the Balmer series.

Question 11: How long does sunlight take to reach us?

The average distance between the Sun and the Earth is $1.5 \times 10^{11}$ m. (i) How long does light from the Sun take to reach the Earth? (ii) The Moon is $3.84 \times 10^{8}$ m away; how long does moonlight take? (iii) Would ultraviolet radiation from the Sun arrive sooner or later than visible light?

Answer:

(i) Time is distance over speed:

$$t = \frac{d}{c} = \frac{1.5 \times 10^{11}\ \mathrm{m}}{3.0 \times 10^{8}\ \mathrm{m\ s^{-1}}} = 5.0 \times 10^{2}\ \mathrm{s} = 500\ \mathrm{s}$$

That is $500/60 = 8.33$ min, about 8 minutes 20 seconds.

(ii) For the Moon, $t = \dfrac{3.84 \times 10^{8}}{3.0 \times 10^{8}} = 1.28$ s.

(iii) Same time. In vacuum all electromagnetic radiations travel at $c$ whatever their wavelength, so ultraviolet and visible light arrive together after 500 s.

Ans: (i) 500 s (about 8.3 min); (ii) 1.28 s; (iii) neither — they arrive together. Watch out: Speed in vacuum is independent of wavelength. Only in a medium do different wavelengths travel at slightly different speeds.

Question 12: An X-ray and a gamma ray by the numbers

An X-ray used in a hospital has a wavelength of 0.50 Å and a gamma ray from cobalt-60 has a frequency of $2.8 \times 10^{20}$ Hz. Find (i) the frequency of the X-ray, (ii) the wavelength of the gamma ray in picometres, and (iii) the ratio of their wavenumbers.

Answer:

(i) $\lambda = 0.50\ \text{\AA} = 5.0 \times 10^{-11}$ m, so

$$\nu = \frac{3.0 \times 10^{8}}{5.0 \times 10^{-11}} = 6.0 \times 10^{18}\ \mathrm{Hz}$$

(ii) $\lambda = \dfrac{c}{\nu} = \dfrac{3.0 \times 10^{8}}{2.8 \times 10^{20}} = 1.07 \times 10^{-12}\ \mathrm{m} = 1.07$ pm.

(iii) X-ray $\bar{\nu} = 1/(5.0 \times 10^{-11}) = 2.0 \times 10^{10}$ m$^{-1}$; gamma $\bar{\nu} = 1/(1.07 \times 10^{-12}) = 9.3 \times 10^{11}$ m$^{-1}$. The ratio gamma : X-ray is $9.3 \times 10^{11} / 2.0 \times 10^{10} \approx 47$. Since $\bar{\nu} \propto \nu$, the frequencies give it directly: $2.8 \times 10^{20} / 6.0 \times 10^{18} = 47$.

Ans: (i) $6.0 \times 10^{18}$ Hz; (ii) 1.07 pm; (iii) about 47 : 1 (gamma : X-ray). Watch out: Ratios of wavenumbers equal ratios of frequencies, so skip the wavelengths whenever only a ratio is wanted.