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Study Notes: The Dual Nature of Light

A complete description of the nature of light requires two explanations.

  1.  Light travels in the form of waves.

    The simplest demonstration of the wave nature of light is to throw two stones into a pool of water and watch the waves from each pass through each other and emerge unchanged. Two light beams will pass through each other and emerge unchanged too.

  2. Light travels in the form of bundles or packets of energy called photons.This is sometimes called the particle nature of light. Light will be reflected from a solid mirror surface in the same way that a ball will bounce off a wall.

All of the different behaviours of light can be explained by one (or both) of its wave and particle properties.

Light waves
A wave is a regularly repeating disturbance:

A light wave showing measures for wavelength, amplitude, crest and trough.

The amplitude is the height of a wave.

The wavelength is the distance between two successive crests or troughs in the wave. This can be measured in metres (m), micrometres (microns, µm =10-6 m), nanometres (nm = 10-9 m) or angstroms (Å = 10-10 m).

The reciprocal of wavelength is called wavenumber and is often used in spectroscopy to express a wavelength or a wavelength range. The unit for wavenumber is cm-1 which, in effect, represents the number of waves in a cm.

The frequency is the number of waves that pass by in 1 second. It has the unit Hertz (Hz). A frequency of 1 000 Hz means that 1 thousand complete wave cycles (crest to crest or trough to trough) pass by each second.

The velocity (speed) of a wave can be related to its frequency (nu, called 'Nu') and wavelength (lambda, called 'Lambda') by the rule:

c equals nu times lambda.

where c = the speed of the light wave = 3 x 108 ms-1 (in a vacuum)

Given that c is constant, this equation tells us that nu and lambda must be inversely proportional to each other. That is, as one variable increases, the other decreases.

Sample problem:

Calculate the wavelength of red light in a vacuum if its frequency is 4 X 1014 Hz.

Solution

c equals nu times lambda.

lambda equals c divided by nu, equals 3 times 10 to the 8 divided by 4 times 10 to the 14, which equals 7.5 times 10 to the negative 7 metre.

Photons
A photon is a small packet or bundle of energy.

The energy of a photon is directly proportional to frequency (energy increases with increasing frequency) and is given by the rule:

E equals h times nu.
where h = Planck's constant = 6.63 x 10-34 Js

Combining this rule with the rule for wave speed we obtain:

E equals h times c divided by lambda.

As you would expect, given the relationship between vee and lambda, this equation shows that energy is inversely proportional to wavelength (energy increases as wavelength decreases).

Overall, the energy equations tell us that high frequency/short wavelength light has high energy and low frequency/high wavelength light has low energy. On the electromagnetic spectrum therefore, X-rays have relatively high energy and radio waves have low energy. Within the visible part of the spectrum, violet has the most energy because it is at the high frequency end and red has the least because it is at the low frequency end.

Sample problem:
Calculate the energy of a 2.6 Å X-ray photon (remember that Å = 10-10 m).

Solution

E equals h times c divided by lambda, equals 6.63 times 10 to the negative 34, times 3 times 10 to the 8, all divided by 2.6 times 10 to the negative 10, which equals 7.65 times 10 to the negative 16 joule.

Note that the energy of X-rays is usually expressed in electron volts (eV):

1 J = 6.24 X 1018 eV

Therefore, the energy of the 2.6 Å X-ray photon is equal to:


(7.65 X 10-16) X (6.24 X 1018) = 4 774 eV

 

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