The de Broglie Wavelength

De Broglie proposed that electrons travel through space as a wave
- This would explain why they can exhibit behaviour such as diffraction
He therefore suggested that electrons must also hold wave properties, such as wavelength
- This became known as the de Broglie wavelength
However, he realised all particles can show wave-like properties, not just electrons
So, the de Broglie wavelength can be defined as:

The wavelength associated with a moving particle

The majority of the time, and for everyday objects travelling at normal speeds, the de Broglie wavelength is far too small for any quantum effects to be observed
A typical electron in a metal has a de Broglie wavelength of about 10 nm
Therefore, quantum mechanical effects will only be observable when the width of the sample is around that value
The electron diffraction tube can be used to investigate how the wavelength of electrons depends on their speed
- The smaller the radius of the rings, the smaller the de Broglie wavelength of the electrons
As the voltage is increased:
- The energy of the electrons increases
- The radius of the diffraction pattern decreases
This shows as the speed of the electrons increases, the de Broglie wavelength of the electrons decreases

Calculating de Broglie Wavelength

Using ideas based upon the quantum theory and Einstein’s theory of relativity, de Broglie suggested that the momentum (p) of a particle and its associated wavelength (λ) are related by the equation:

$λ = \frac{h}{p}$

Since momentum p = mv, the de Broglie wavelength can be related to the speed of a moving particle (v) by the equation:

$λ = \frac{h}{m v}$

Kinetic energy $E = \frac{1}{2} m v^{2}$
Therefore, momentum and kinetic energy can be related by:

$E = \frac{p^{2}}{2 m}$ or $p = \sqrt{2 m E}$

Combining this with the de Broglie equation gives a form which relates the de Broglie wavelength of a particle to its kinetic energy:

$λ = \frac{h}{\sqrt{2 m E}}$

Where:
- λ = the de Broglie wavelength (m)
- h = Planck’s constant (J s)
- p = momentum of the particle (kg m s^-1)
- E = kinetic energy of the particle (J)
- m = mass of the particle (kg)
- v = speed of the particle (m s^-1)

Worked example

A proton and an electron are each accelerated from rest through the same potential difference.

Determine the ratio: $\frac{d e B r o g l i e w a v e l e n g t h o f t h e p r o t o n}{d e B r o g l i e w a v e l e n g t h o f t h e e l e c t r o n}$

Mass of a proton = 1.67 × 10^–27 kg
Mass of an electron = 9.11 × 10^–31 kg

Answer:

Step 1: Consider how the proton and electron can be related via their masses

The proton and electron are accelerated through the same p.d., therefore, they both have the same kinetic energy

Step 2: Write the equation relating the de Broglie wavelength of a particle to its kinetic energy

$λ = \frac{h}{p} = \frac{h}{\sqrt{2 m E}}$

$λ \propto \frac{1}{\sqrt{m}}$

Step 3: Calculate the ratio

$\frac{d e B r o g l i e w a v e l e n g t h o f t h e p r o t o n}{d e B r o g l i e w a v e l e n g t h o f t h e e l e c t r o n} = \frac{1}{\sqrt{m_{p}}} \div \frac{1}{\sqrt{m_{e}}}$

$\sqrt{\frac{m_{e}}{m_{p}}} = \sqrt{\frac{9.11 \times 10^{- 31}}{1.67 \times 10^{- 27}}} = 2.3 \times 10^{- 2}$

This means the de Broglie wavelength of the proton is 0.023 times smaller than that of the electron OR the de Broglie wavelength of the electron is about 40 times larger than that of the proton

Exam Tip

Although you don't need to know the derivation of the de Broglie wavelength in terms of energy, it is useful to know this version of the equation in case you are given the kinetic energy instead of momentum.

The de Broglie Wavelength (CIE A Level Physics)

Revision Note