Wave-particle duality

1. Overview

Wave-particle duality is the cornerstone of quantum mechanics, asserting that every fundamental entity exhibits both wave-like and particle-like properties depending on the experimental context. Electromagnetic (EM) radiation, traditionally viewed as a continuous wave, demonstrates a particulate nature through the photoelectric effect, where it interacts with matter as discrete packets of energy called photons. Conversely, matter—specifically particles with mass such as electrons—demonstrates a wave nature through phenomena like diffraction and interference. This duality implies that the classical distinction between "waves" and "particles" is an approximation; the true nature of the universe is described by the de Broglie relationship, which links the momentum of a particle to its associated wavelength.

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Key Definitions

• Photon: A discrete quantum (or packet) of electromagnetic radiation. The energy of a photon is proportional to the frequency of the radiation ($E = hf$).
• Photoelectric Effect: The process whereby electrons are emitted from the surface of a metal when electromagnetic radiation of sufficiently high frequency is incident upon it.
• Threshold Frequency ($f_0$): The minimum frequency of incident electromagnetic radiation required to liberate an electron from the surface of a particular metal.
• Work Function ($\Phi$): The minimum energy required to liberate an electron from the surface of a metal.
• Diffraction: The spreading of a wave as it passes through a gap or around an obstacle, which occurs significantly when the gap width is comparable to the wavelength.
• Interference: The superposition of two or more coherent waves to produce a resultant displacement, resulting in patterns of maxima and minima.
• de Broglie Wavelength ($\lambda$): The wavelength associated with a moving particle, defined by the ratio of the Planck constant to the particle's momentum.

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Content

3.1 Evidence for the Dual Nature of Electromagnetic Radiation

Historically, light was modeled as a longitudinal wave (Huygens) and later a transverse electromagnetic wave (Maxwell). While the wave model is highly successful for propagation, it fails to explain how light exchanges energy with matter.

Evidence for the Wave Nature:

• Interference: Experiments such as Young’s Double Slit show that light produces regions of reinforcement and cancellation. This is a property unique to waves, where displacements can sum to zero (destructive interference).
• Diffraction: Light spreads when passing through narrow apertures. The degree of spreading depends on the wavelength, a behavior that cannot be explained by a stream of classical particles traveling in straight lines.

Evidence for the Particulate Nature (The Photoelectric Effect): The wave theory of light makes three predictions that are contradicted by the photoelectric effect:

1. Intensity vs. Emission: Wave theory predicts that any frequency of light should eventually emit electrons if the intensity is high enough (energy would "build up"). Observation: No electrons are emitted if the frequency is below the threshold frequency, regardless of intensity.
2. Time Delay: Wave theory predicts a time lag at low intensities while the electron absorbs enough energy to escape. Observation: Emission is instantaneous (less than $10^{-9}$ s).
3. Kinetic Energy: Wave theory predicts that higher intensity (greater amplitude) should give electrons more kinetic energy. Observation: The maximum kinetic energy ($E_{k(\text{max})}$) depends only on the frequency of the light, not the intensity.

Conclusion: These observations prove that light arrives in discrete "packets" (photons). One photon interacts with exactly one electron in a 1-to-1 interaction. If the photon energy ($hf$) is greater than the work function ($\Phi$), the electron is emitted.

3.2 Evidence for the Wave Nature of Particles (Electron Diffraction)

In 1924, Louis de Broglie hypothesized that if light (a wave) could behave like a particle, then matter (particles) should behave like waves.

The Electron Diffraction Experiment:

1. Setup: An electron gun fires a beam of electrons in a vacuum. The electrons are accelerated by a high potential difference ($V$).
2. Target: The beam passes through a thin film of polycrystalline graphite.
3. Detection: The electrons strike a fluorescent screen, creating a visible pattern.

Observations and Interpretation:

• The Pattern: Instead of a single bright spot, the screen shows a series of concentric bright and dark rings.
• Interpretation: This is a diffraction pattern. Diffraction is an exclusively wave-like property. The gaps between the carbon atoms in the graphite act as "slits." Because the spacing between atoms is roughly $10^{-10}$ m (similar to the wavelength of accelerated electrons), the electrons undergo diffraction.
• Conclusion: Since electrons (which have mass and charge) produce a diffraction pattern, they must possess wave-like properties.

Factors Affecting the Pattern:

• If the accelerating voltage ($V$) is increased:
  • The kinetic energy ($E_k = eV$) increases.
  • The momentum ($p = \sqrt{2mE_k}$) increases.
  • The de Broglie wavelength ($\lambda = h/p$) decreases.
  • A smaller wavelength results in less diffraction (the rings move closer to the center, reducing their radius).

3.3 The de Broglie Wavelength

The de Broglie hypothesis states that a particle of momentum $p$ has an associated wavelength $\lambda$.

The de Broglie Equation: $$\lambda = \frac{h}{p}$$ (This equation is on the Data Sheet)

Since momentum $p = mv$: $$\lambda = \frac{h}{mv}$$

Where:

• $\lambda$ = de Broglie wavelength (m)
• $h$ = Planck constant ($6.63 \times 10^{-34}$ J s)
• $m$ = mass of the particle (kg)
• $v$ = velocity of the particle (m s⁻¹)

Derivation: Relating $\lambda$ to Accelerating Voltage ($V$): In many exam questions, you are given the potential difference $V$ used to accelerate an electron from rest.

1. Energy Conservation: The electrical work done on the electron equals its gain in kinetic energy: $$eV = \frac{1}{2}mv^2$$
2. Relate $E_k$ to Momentum: $$E_k = \frac{p^2}{2m} \implies p = \sqrt{2mE_k}$$
3. Substitute $E_k = eV$: $$p = \sqrt{2meV}$$
4. Substitute into de Broglie Equation: $$\lambda = \frac{h}{\sqrt{2meV}}$$

3.4 Worked Examples

Worked example 1 — Calculating Electron Wavelength

An electron is accelerated from rest through a potential difference of 5.0 kV. Calculate its de Broglie wavelength. (Mass of electron $m_e = 9.11 \times 10^{-31}$ kg; elementary charge $e = 1.60 \times 10^{-19}$ C)

Step 1: Calculate the Kinetic Energy ($E_k$) in Joules $$E_k = eV = (1.60 \times 10^{-19} \text{ C}) \times (5000 \text{ V}) = 8.0 \times 10^{-16} \text{ J}$$

Step 2: Calculate the Momentum ($p$) $$p = \sqrt{2mE_k} = \sqrt{2 \times (9.11 \times 10^{-31} \text{ kg}) \times (8.0 \times 10^{-16} \text{ J})}$$ $$p = \sqrt{1.4576 \times 10^{-45}} = 3.818 \times 10^{-23} \text{ kg m s}^{-1}$$

Step 3: Calculate the Wavelength ($\lambda$) $$\lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34} \text{ J s}}{3.818 \times 10^{-23} \text{ kg m s}^{-1}}$$ Answer: $\lambda = 1.7 \times 10^{-11} \text{ m}$ (or 0.017 nm)

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Worked example 2 — Comparing Particles

A proton and an alpha particle are moving with the same velocity. Determine the ratio: $$\frac{\text{de Broglie wavelength of proton}}{\text{de Broglie wavelength of alpha particle}}$$ (Mass of alpha particle $\approx 4 \times$ mass of proton)

Step 1: State the relationship $$\lambda = \frac{h}{mv}$$ Since $h$ and $v$ are constant for both particles: $$\lambda \propto \frac{1}{m}$$

Step 2: Set up the ratio $$\frac{\lambda_p}{\lambda_\alpha} = \frac{m_\alpha}{m_p}$$

Step 3: Substitute the mass relationship $$\frac{\lambda_p}{\lambda_\alpha} = \frac{4m_p}{m_p} = 4$$ Answer: The ratio is 4:1. (The proton has 4 times the wavelength of the alpha particle).

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Worked example 3 — Determining Velocity from Wavelength

A neutron has a de Broglie wavelength of $2.0 \times 10^{-10}$ m. Calculate the velocity of this neutron. (Mass of neutron $m_n = 1.67 \times 10^{-27}$ kg)

Step 1: Rearrange the de Broglie equation for $v$ $$\lambda = \frac{h}{mv} \implies v = \frac{h}{m\lambda}$$

Step 2: Substitute the values $$v = \frac{6.63 \times 10^{-34}}{(1.67 \times 10^{-27}) \times (2.0 \times 10^{-10})}$$ $$v = \frac{6.63 \times 10^{-34}}{3.34 \times 10^{-37}}$$ Answer: $v = 1.99 \times 10^3 \text{ m s}^{-1}$

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Key Equations

Equation	Description	Data Sheet?
$\lambda = \frac{h}{p}$	de Broglie wavelength	Yes
$p = mv$	Momentum of a particle	No (AS Level)
$E_k = \frac{p^2}{2m}$	Kinetic energy in terms of momentum	No
$E_k = eV$	Energy gained by a charge in a PD	No
$E = hf = \frac{hc}{\lambda}$	Energy of a photon	Yes

Standard Constants:

• Planck constant ($h$): $6.63 \times 10^{-34}$ J s
• Mass of electron ($m_e$): $9.11 \times 10^{-31}$ kg
• Mass of proton ($m_p$): $1.67 \times 10^{-27}$ kg
• Elementary charge ($e$): $1.60 \times 10^{-19}$ C

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Common Mistakes to Avoid

• ❌ Wrong: Using $v = c$ (speed of light) for particles like electrons or protons. ✓ Right: Only photons travel at $c$. Particles with mass travel at speeds $v < c$. You must calculate $v$ from kinetic energy or momentum.
• ❌ Wrong: Confusing the photon energy equation ($E = hc/\lambda$) with the de Broglie equation ($\lambda = h/p$) when dealing with electrons. ✓ Right: For an electron, $E_k = \frac{1}{2}mv^2$. The equation $E = hf$ applies only to photons.
• ❌ Wrong: Forgetting to convert units, especially keV to Joules or nm to meters. ✓ Right: Always work in SI units. $1 \text{ eV} = 1.60 \times 10^{-19} \text{ J}$. $1 \text{ nm} = 10^{-9} \text{ m}$.
• ❌ Wrong: Thinking that diffraction occurs because electrons "collide" with atoms. ✓ Right: Diffraction is a wave phenomenon occurring because the electron's wavefront passes through the gaps in the crystal lattice and interferes with itself.
• ❌ Wrong: Stating that the photoelectric effect proves light is a wave. ✓ Right: The photoelectric effect is the primary evidence for the particulate (photon) nature of light.

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Exam Tips

1. The "Logic Chain" for Diffraction: If asked how the diffraction pattern changes when accelerating voltage $V$ increases, use this exact sequence:
   • $V$ increases, so Kinetic Energy ($E_k = eV$) increases.
   • Momentum ($p = \sqrt{2mE_k}$) increases.
   • de Broglie wavelength ($\lambda = h/p$) decreases.
   • The angle of diffraction $\theta$ decreases (since $\sin \theta \propto \lambda$).
   • The rings on the screen become smaller in radius and closer together.
2. Describing the Evidence:
   • If the question asks for evidence of the wave nature of matter, the answer is electron diffraction.
   • If the question asks for evidence of the particulate nature of radiation, the answer is the photoelectric effect.
3. Why Graphite?: You may be asked why graphite is used. Graphite has a regular crystalline structure where the spacing between atomic layers is approximately $10^{-10}$ m. This is the same order of magnitude as the de Broglie wavelength of electrons accelerated by a few kilovolts, which is the necessary condition for observable diffraction.
4. Significant Figures: Always provide your final answer to the same number of significant figures as the least precise data value given in the question (usually 2 or 3 s.f.).
5. Interpreting "Qualitatively": Syllabus objective 22.3.2 asks for a qualitative interpretation. This means you should be able to describe the appearance of the pattern (concentric rings) and explain what it implies (wave nature) without necessarily performing complex 3D geometry calculations.