Wave-particle duality

In a nutshell

Electromagnetic waves can exhibit both wave-like properties, such as being diffracted, and particle-like properties, such as the photoelectric effect. All particles with mass can be described in terms of their wavelength using the de Broglie equation.

Equations

DESCRIPTION	EQUATION
de Broglie equation	$\lambda = \dfrac{h}{p} = \dfrac{h}{mv}$
Accelerated particles	$\lambda = \dfrac{h}{\sqrt{2E_km}}$

Constants

CONSTANT	SYMBOL	VALUE
$Planck's \ constant$	$h$	$6.63 \times 10^{-34} \,Js$
$mass\ of \ an\ electron$	$m_e$	$9.11 \times 10^{-31}\,kg$
$charge \ of \ an \ electron$	$e$	$1.6 \times 10^{-19} \ J$

Variables

QUANTITY NAME	SYMBOL	DERIVED UNIT	SI UNIT
$de \ Broglie \ wavelength$	$\lambda$	$m$	$m$
$velocity$	$v$	$ms^{-1}$	$ms^{-1}$
$momentum$	$p$	$kgms^{-1}$	$kgms^{-1}$
$kinetic\ energy$	$E_k$	$J$	$kgm^2s^{-2}$
$mass$	$m$	$kg$	$kg$
$voltage$	$V$	$V$	$kg m^2 s^{-3} A^{ -1}$

Diffraction of particles

Louis de Broglie proposed that all matter has both particle and wave-like properties. All particles have a wavelength and this can be shown by their ability to diffract.

Electromagnetic waves, such as visible light, also diffract due to their wave-like properties. However, the photoelectric effect relies on light consisting of photons. This is because electromagnetic waves have a dual nature, a wave-particle duality.

Electron diffraction

Electrons can be made to diffract under certain specific conditions. The electrons are accelerated in an electron gun, then fired at a thin film of graphite. The electrons will diffract through the gaps between carbon atoms in the film, which are of the distance as the wavelength of the electrons (~ $10^{10}\,m$ ). This causes a circular diffraction pattern on a fluorescent screen.

Physics; Quantum physics; KS5 Year 12; Wave-particle duality

This confirms wave-particle duality, as particles are not able to diffract and this is a property of waves only.

The de Broglie equation

De Broglie realised the wavelength associated with a particle is directly proportional to its momentum, which can be described by the equation:

$\lambda = \dfrac{h}{p}\\[0.1in]$

Substituting the equation for momentum of the particle, $p$ , into the first equation:

$p = mv\\[0.1in] \lambda = \dfrac{h}{mv}$

This equation applies to all particles with mass, however the heavier the object the shorter their wavelength and therefore harder it is to observe wave-like properties.

Example

Calculate the de Broglie wavelength of a proton travelling at a speed of $1.5 \times 10^3 \, ms^{-1}$

State variables:

$v = 1.5 \times 10^3 \, ms^{-1}\\ m_p= 1.67 \times 10^{-27}\,kg$

State equation:

$\lambda = \dfrac{h}{mv}$

Sub in and solve:

$\lambda = \dfrac{6.67 \times 10^{-34}}{1.5 \times 10^3 \times 1.67 \times 10^{-27}}\\[0.2in] \lambda = 2.7 \times 10^{-10}\,m$

The de Broglie wavelength of this proton is $\underline{\lambda = 2.7 \times 10^{-10}\,m}$ .

Accelerated particles

An object's kinetic energy can be related to its de Broglie wavelength, using the general equation for kinetic energy and de Broglie wavelength shown above. When a particle is accelerated through a potential difference:

$E_k = \dfrac{1}{2}mv^2\\[0.1in] 2E_k = mv^2\\[0.1in] 2E_k\, m = m^2v^2\\[0.1in] \sqrt{2E_k\,m} = mv$

Substituting this into the de Broglie equation:

$\lambda = \dfrac{h}{mv} = \dfrac{h}{\sqrt{2E_km}}$

Investigating electron diffraction

Electron diffraction tubes can be used to investigate the wave properties of an electron. Particles are accelerated in an electron gun, then directed at a film of graphite, and diffract between the carbon atoms. Increasing the voltage between the anode and cathode causes the energy of the electrons to increase.

The kinetic energy of electrons is proportional to the voltage across the anode-cathode:

$E_k= \dfrac{1}{2}mv^2 =eV$

Example

An electron is accelerated in an electron gun through a potential difference of $200\,V$ , calculate its wavelength, using its kinetic energy.

State variables:

$V= 200\,V$

State equation:

$E_k = eV$

$\lambda = \dfrac{h}{\sqrt{2E_km}}$

$\lambda = \dfrac{h}{\sqrt{2eVm}}$

Sub in and solve:

$\lambda = \dfrac{6.63 \times 10^{-34}}{\sqrt{2 \times 1.6 \times 10^{-19} \times 200 \times 9.11 \times 10^{-31}}}$

$\lambda = 8.7 \times 10^{-11}\,m$

The wavelength is $\underline{8.7 \times 10^{-11}\,m}$