Calculate the wavelength, frequency and wavenumber of a light wave whose period is 2.0 × 10⁻¹⁰s.

Exercises · 2.7

2.0 \times 10^{-10}\ s

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Concept understanding — Electromagnetic Wave Wavelength

What is Wavelength? The Intuition First

Imagine dropping a pebble into a still pond. Ripples spread outward — evenly spaced circles. The distance between two consecutive crests (the highest points) is what we call wavelength. It's the "repeat length" of the wave.

Now, electromagnetic waves are not water waves. They don't need a medium. But the same idea holds: an EM wave is a travelling disturbance in electric and magnetic fields. As it moves, the fields oscillate — they go up, down, up, down. The wavelength ( $\lambda$ , Greek letter lambda) is the distance over which the wave's shape repeats.

Note

In a vacuum, all EM waves travel at the same speed: $c = 3 \times 10^8 \ \text{m/s}$ . What changes from one wave to another is the wavelength (and its partner, frequency).

The Precise Definition

For a sinusoidal electromagnetic wave travelling in one direction, the electric field at a fixed instant of time looks like a sine curve in space. The wavelength $\lambda$ is the spatial distance between two successive points that are in phase — for example, from one crest to the next crest, or from one trough to the next trough.

Mathematically, if the electric field at position $x$ and time $t$ is given by

E(x,t) = E_0 \sin(kx - \omega t)

then the wave number $k$ is related to wavelength by

k = \frac{2\pi}{\lambda}

So $\lambda$ is the distance needed for the argument $kx$ to change by $2\pi$ — one full cycle of the sine wave.

The Fundamental Relationship

The wavelength, frequency $f$ , and speed $c$ are tied together by a simple equation:

c = f \lambda

This is the wave equation for EM waves in vacuum. It means:

If the wavelength is long, the frequency is low.
If the wavelength is short, the frequency is high.
The product is always $c$ , a constant.

Tip

To remember: think of a marching band. If soldiers take long strides (large $\lambda$ ), they take fewer steps per second (low $f$ ). If they take short, quick steps (small $\lambda$ ), they take many steps per second (high $f$ ). The speed of the band is fixed — stride length × steps per second.

The Electromagnetic Spectrum

Wavelength is what separates different kinds of EM radiation. Here's the spectrum from longest to shortest wavelength:

Type of EM Wave	Approximate Wavelength Range
Radio waves	$> 0.1 \ \text{m}$ (up to km)
Microwaves	$1 \ \text{mm}$ to $0.1 \ \text{m}$
Infrared	$700 \ \text{nm}$ to $1 \ \text{mm}$
Visible light	$400 \ \text{nm}$ to $700 \ \text{nm}$
Ultraviolet	$10 \ \text{nm}$ to $400 \ \text{nm}$
X-rays	$0.01 \ \text{nm}$ to $10 \ \text{nm}$
Gamma rays	$< 0.01 \ \text{nm}$

Watch out

A common mistake: thinking that wavelength is the "size" of the wave. It's not. It's the repeat distance. A radio wave can have a wavelength of 1 km, but its amplitude (the strength of the field) might be tiny. Wavelength and amplitude are independent properties.

Why Wavelength Matters …

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The key idea is that wavelength, frequency, and wavenumber are all linked through the speed of light and the period. Given the period $T = 2.0 \times 10^{-10}\ \text{s}$ , the frequency is $f = 5.0 \times 10^{9}\ \text{Hz}$ , the wavelength is $\lambda = 0.06\ \text{m}$ (or $6.0\ \text{cm}$ ), and the wavenumber is $\bar{\nu} \approx 16.7\ \text{m}^{-1}$ .

Concept and Intuition

When we talk about a light wave, its period $T$ is the time it takes for one complete oscillation at a fixed point. The frequency $f$ is simply how many such oscillations happen per second — they are reciprocals: $f = 1/T$ .

Once we know the frequency, the wavelength $\lambda$ tells us the spatial distance between successive crests. For any electromagnetic wave in vacuum, the product of wavelength and frequency equals the speed of light $c$ : $\lambda f = c$ .

The wavenumber $\bar{\nu}$ (often denoted by $\tilde{\nu}$ or $k$ in different contexts) is the number of wavelengths per unit distance. In spectroscopy, it's usually defined as $\bar{\nu} = 1/\lambda$ , giving units of $\text{m}^{-1}$ (or $\text{cm}^{-1}$ ).

So the path is: period → frequency → wavelength → wavenumber. Each step uses a simple relation, but the key is to keep track of units and not confuse angular wavenumber ( $k = 2\pi/\lambda$ ) with the spectroscopic wavenumber ( $1/\lambda$ ). The problem asks for "wavenumber" in the simplest sense — so we'll use $1/\lambda$ .

Step-by-Step Solution

1. Find the frequency from the period.

The period $T$ is given as $2.0 \times 10^{-10}\ \text{s}$ . Frequency is the reciprocal:

f = \frac{1}{T} = \frac{1}{2.0 \times 10^{-10}} = 5.0 \times 10^{9}\ \text{s}^{-1}

Since $1\ \text{s}^{-1} = 1\ \text{Hz}$ , we have $f = 5.0 \times 10^{9}\ \text{Hz}$ .

Tip

A quick check: $10^{-10}$ s is 0.1 nanosecond. The reciprocal gives $10^{10}$ Hz, but here it's $2.0 \times 10^{-10}$ , so half that — $5 \times 10^{9}$ Hz. That's in the microwave region of the EM spectrum.

2. Calculate the wavelength using $c = \lambda f$ .

The speed of light in vacuum is $c = 3.0 \times 10^{8}\ \text{m/s}$ . Rearranging:

\lambda = \frac{c}{f} = \frac{3.0 \times 10^{8}}{5.0 \times 10^{9}} = 0.06\ \text{m}

That's $6.0 \times 10^{-2}\ \text{m}$ , or $6.0\ \text{cm}$ . …

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Q.Calculate the wavelength, frequency and wavenumber of a light wave whose period is 2.0×10−10 s2.0 \times 10^{-10}\ s2.0×10−10 s.

Concept and Intuition

Step-by-Step Solution

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

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