WAVE-PARTICLE DUALITY
- Light waves can behave like particles, i.e. photons, and waves
- This phenomena is called the wave-particle nature of light or wave-particle duality
- Light interacts with matter, such as electrons, as a particle
- The evidence for this is provided by the photoelectric effect
- Light propagates through space as a wave
- The evidence for this comes from the diffraction and interference of light in Young’s Double Slit experiment
Light as a Particle
- Einstein proposed that light can be described as a quanta of energy that behave as particles, called photons
- The photon model of light explains that:
- Electromagnetic waves carry energy in discrete packets called photons
- The energy of the photons are quantised according to the equation $E=h f$
- In the photoelectric effect, each electron can absorb only a single photon – this means only the frequencies of light above the threshold frequency will emit a photo electron
- The wave theory of light does not support a threshold frequency
- The wave theory suggests any frequency of light can give rise to photoelectric emission if the exposure time is long enough
- This is because the wave theory suggests the energy absorbed by each electron will increase gradually with each wave
- Furthermore, the kinetic energy of the emitted electrons should increase with radiation intensity
- However, in the photoelectric effect none of this is observed
- If the frequency is above the threshold and the intensity of the light is increased, more photoelectrons are emitted per second
- Although the wave theory provided good explanations for phenomena such as interference and diffraction, it failed to explain the photoelectric effect
Wave-Particle Duality: Electron Diffraction
- Louis de Broglie discovered that matter, such as electrons, can behave as a wave
- He showed a diffraction pattern is produced when a beam of electrons is directed at a thin graphite film
- Diffraction is a property of waves, and cannot be explained by describing electrons as particles
- In order to observe the diffraction of electrons, they must be focused through a gap similar to their size, such as an atomic lattice
- Graphite film is ideal for this purpose because of its crystalline structure
- The gaps between neighbouring planes of the atoms in the crystals act as slits, allowing the electron waves to spread out and create a diffraction pattern
- The diffraction pattern is observed on the screen as a series of concentric rings
- This phenomenon is similar to the diffraction pattern produced when light passes through a diffraction grating
- If the electrons acted as particles, a pattern would not be observed, instead the particles would be distributed uniformly across the screen
- It is observed that a larger accelerating voltage reduces the diameter of a given ring, while a lower accelerating voltage increases the diameter of the rings
Investigating Electron Diffraction
- Electron diffraction tubes can be used to investigate wave properties of electrons
- The electrons are accelerated in an electron gun to a high potential, such as $5000 \mathrm{~V}$, and are then directed through a thin film of graphite
- The electrons diffract from the gaps between carbon atoms and produce a circular pattern on a fluorescent screen made from phosphor
- Increasing the voltage between the anode and the cathode causes the energy, and hence speed, of the electrons to increase
- The kinetic energy of the electrons is proportional to the voltage across the anode-cathode:
$
E_k=1 / 2 m v^2=e V
$
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 $10 \mathrm{~nm}$
- 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 $(\lambda)$ are related by the equation:
$
\lambda=\frac{h}{p}
$
- Since momentum $p=m v$, the de Broglie wavelength can be related to the speed of a moving particle (v) by the equation:
$
\lambda=\frac{h}{m v}
$
- Since kinetic energy $E=1 / 2 \mathrm{mv}^2$
- Momentum and kinetic energy can be related by:
$
\mathrm{E}=\frac{p^2}{2 m} \text { or } \mathrm{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:
$
\lambda=\frac{h}{\sqrt{2 m E}}
$
- Where:
- $\lambda=$ the de Broglie wavelength $(\mathrm{m})$
- $\mathrm{h}=$ Planck’s constant (J s)
- $p=$ momentum of the particle $\left(\mathrm{kg} \mathrm{m} \mathrm{s}^{-1}\right)$
- $E=$ kinetic energy of the particle (J)
- $\mathrm{m}=$ mass of the particle $(\mathrm{kg})$
- $v=$ speed of the particle $\left(\mathrm{m} \mathrm{s}^{-1}\right)$
Worked example: de Broglie wavelength
A proton and an electron are each accelerated from rest through the same potential difference. Determine the ratio:
$
\frac{\text { de Broglie wavelength of the proton }}{\text { de Broglie wavelength of the electron }}
$
Mass of a proton $=1.67 \times 10^{-27} \mathrm{~kg}$
Mass of an electron $=9.11 \times 10^{-31} \mathrm{~kg}$
Answer/Explanation
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 which relates the de Broglie wavelength of a particle to its kinetic energy:
$
\lambda=\frac{h}{p}=\frac{h}{\sqrt{2 m E}}
$
$
\lambda \propto \frac{1}{\sqrt{m}}
$
Step 3:
Calculate the ratio:
$
\begin{aligned}
& \frac{\text { de Broglie wavelength of the proton }}{\text { de Broglie wavelength of the electron }}=\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}
\end{aligned}
$
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