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PU-202 / L01 · lecture

Why Classical Physics Fails: Quanta, Spectra, and de Broglie

~8 min read · Tier toggle changes depth (T1 plain · T2 standard · T3 rigour)

BEFORE YOU START

You should be comfortable with basic kinematics, the concept of electric potential energy, and the idea that energy is conserved in isolated systems.

Why this matters

Classical physics cannot explain the discrete spectra of atoms or the emission of electrons by light.

For centuries, the laws of Newton and Maxwell described the macroscopic world with remarkable precision. However, when physicists examined the microscopic realm, particularly the interaction of light with matter, they encountered phenomena that defied classical explanation. The blackbody radiation spectrum, the stability of atoms, and the photoelectric effect all presented paradoxes. This lecture introduces the foundational concepts of quantum mechanics: the quantisation of energy, the particle-like nature of light, and the wave-like nature of matter. These ideas form the bedrock upon which the entire formalism of quantum mechanics is built.

The Breakdown of Classical Intuition

In classical physics, energy is continuous. You can add an infinitesimally small amount of energy to a system, and the system will accept it. Similarly, light is viewed as a continuous electromagnetic wave. If you increase the intensity of a wave, you increase its amplitude, and thus its energy. According to this view, a sufficiently intense light wave should be able to eject electrons from a metal surface, regardless of the light's frequency, provided the light is bright enough. Furthermore, the energy of the ejected electrons should depend on the intensity of the light. However, experiments tell a different story. When light shines on a metal, electrons are only ejected if the light frequency exceeds a specific threshold. Below this threshold, no electrons are emitted, no matter how intense the light is. Moreover, the maximum kinetic energy of the emitted electrons depends linearly on the frequency of the light, not its intensity. This discrete, all-or-nothing behaviour suggests that energy is exchanged in finite packets, or quanta, rather than continuously.

Einstein's Light-Quantum Hypothesis

To resolve the paradox of the photoelectric effect, Albert Einstein proposed (1905) that light consists of discrete particles, later called photons, each carrying energy \(E = hf\), where \(h\) is Planck's constant and \(f\) is the frequency of the light. When a photon strikes an electron in the metal, it transfers all its energy to the electron. Part of this energy is used to overcome the binding force of the metal, known as the work function \(\phi\). The remaining energy appears as the kinetic energy of the emitted electron. This leads to the photoelectric equation, which relates the maximum kinetic energy of the photoelectrons to the frequency of the incident light and the work function of the material.

\[ K_{\max} = hf - \phi \]

This equation explains the experimental observations perfectly. If the frequency \(f\) is too low such that \(hf < \phi\), the photon does not have enough energy to liberate the electron, and no emission occurs. This defines the threshold frequency \(f_0 = \phi/h\). If the frequency is above the threshold, the excess energy \(hf - \phi\) becomes the kinetic energy of the electron. Increasing the intensity of the light increases the number of photons, and thus the number of emitted electrons (the photocurrent), but it does not change the energy of each individual photon, so the maximum kinetic energy remains unchanged. This linear relationship between kinetic energy and frequency is a hallmark of quantum behaviour.

The Wave Nature of Matter

If light, which was traditionally considered a wave, exhibits particle-like properties, it is natural to ask whether matter, which was traditionally considered to consist of particles, exhibits wave-like properties. In 1924, Louis de Broglie proposed that every moving particle is associated with a wave. The wavelength of this matter wave, known as the de Broglie wavelength, is inversely proportional to the momentum of the particle. This hypothesis unifies the particle and wave descriptions of physical entities, suggesting that they are two aspects of the same underlying reality. The de Broglie relation provides a direct link between the mechanical properties of a particle (momentum) and its wave properties (wavelength).

\[ \lambda = \frac{h}{p} \]

For a non-relativistic particle of mass \(m\) and kinetic energy \(E_k\), the momentum is \(p = \sqrt{2mE_k}\). Substituting this into the de Broglie relation gives the wavelength in terms of kinetic energy. If the particle is an electron accelerated from rest through a potential difference \(V\), its kinetic energy is \(eV\), where \(e\) is the elementary charge. The de Broglie wavelength for such an electron is given by the following expression.

\[ \lambda = \frac{h}{\sqrt{2meV}} \]

This result has profound implications. It suggests that electrons, protons, and even larger molecules have wave-like characteristics. The wavelength is typically very small for macroscopic objects due to their large mass, making wave effects negligible. However, for electrons, the wavelength can be comparable to atomic dimensions, making wave effects significant and observable in phenomena such as electron diffraction.

Connecting Quanta and Waves

The concepts of quanta and waves are not separate ideas but are deeply interconnected. The energy of a photon is related to its frequency by \(E = hf\), and its momentum is related to its wavelength by \(p = h/\lambda\). Combining these, we see that the frequency and wavelength of a light wave determine the energy and momentum of its constituent photons. Similarly, for a material particle, the de Broglie relations \(E = \hbar\omega\) and \(p = \hbar k\) connect the particle's energy and momentum to the angular frequency \(\omega\) and wave vector \(\vec{k}\) of its associated wave. This duality is the core of quantum mechanics. It implies that the state of a quantum system is described by a wavefunction, which encodes the probability amplitudes for finding the particle in various states. The transition from classical to quantum physics is thus a transition from deterministic trajectories to probabilistic wave descriptions.

The derivations

◆ DERIVATION EMBED · D-430 · not re-derived here

The Photoelectric Effect and Einstein's Light-Quantum Equation

When light of frequency \(f\) strikes a metal surface with work function \(\phi\), the maximum kinetic energy of the emitted photoelectrons is given by Einstein's light-quantum equation \(K_{\max} = hf - \phi\), where \(h\) is Planck's constant. Electrons are emitted only if \(f\) exceeds the threshold frequency \(f_0 = \phi/h\); the stopping potential \(V_s\) required to halt the fastest electrons satisfies \(eV_s = K_{\max}\), yielding the linear relation \(V_s = \frac{h}{e}f - \frac{\phi}{e}\) with slope \(h/e\) and intercept \(-\phi/e\).

\[ K_{\max} = hf - \phi \qquad \text{or} \qquad eV_s = hf - \phi \]
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◆ DERIVATION EMBED · D-431 · not re-derived here

The de Broglie Wavelength of a Material Particle

A free particle of rest mass \(m\), momentum \(\vec{p}\), and total energy \(E\) is associated with a quantum wave of angular frequency \(\omega\) and wave vector \(\vec{k}\) such that \(E = \hbar\omega\) and \(\vec{p} = \hbar\vec{k}\). The de Broglie wavelength is defined as \(\lambda = 2\pi/|\vec{k}| = h/p\), where \(p = |\vec{p}|\) and \(h = 2\pi\hbar\). For a non-relativistic electron of charge \(e\) accelerated from rest through a potential difference \(V\), this yields \(\lambda = h/\sqrt{2meV}\).

\[ \lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE_k}} = \frac{h}{\sqrt{2meV}} \quad (\text{non-relativistic electron}) \]
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Tier 3 ยท watch out

A common misconception is that the de Broglie wavelength represents a physical wave of matter in the same sense as a water wave. In reality, the wavefunction is a probability amplitude, and its square modulus gives the probability density of finding the particle at a given location. The wave nature is not a mechanical oscillation of mass but a manifestation of the quantum state. C

Worked example

Problem. An electron is accelerated from rest through a potential difference of \(V = 50\ \mathrm{V}\). Calculate the de Broglie wavelength of the electron. Use \(h = 6.626 \times 10^{-34}\ \mathrm{J\,s}\), \(m_e = 9.109 \times 10^{-31}\ \mathrm{kg}\), and \(e = 1.602 \times 10^{-19}\ \mathrm{C}\).

  1. Determine the kinetic energy of the electron using the work done by the electric field: \(E_k = eV\).
  2. Substitute the values into the de Broglie wavelength formula for an electron accelerated through a potential difference: \(\lambda = \frac{h}{\sqrt{2m_e eV}}\).
  3. Calculate the denominator: \(\sqrt{2 \times 9.109 \times 10^{-31}\ \mathrm{kg} \times 1.602 \times 10^{-19}\ \mathrm{C} \times 50\ \mathrm{V}}\).
  4. Compute the final wavelength by dividing Planck's constant by the result from step 3.
\[ \lambda = \frac{6.626 \times 10^{-34}}{\sqrt{2 \times 9.109 \times 10^{-31} \times 1.602 \times 10^{-19} \times 50}} \approx 1.73 \times 10^{-10}\ \mathrm{m} \]

Units check. The numerator is in \(\mathrm{J\,s}\). The denominator is \(\sqrt{\mathrm{kg} \cdot \mathrm{C} \cdot \mathrm{V}}\). Since \(\mathrm{C \cdot V = J}\), the denominator is \(\sqrt{\mathrm{kg \cdot J}} = \sqrt{\mathrm{kg \cdot kg \cdot m^2/s^2}} = \mathrm{kg \cdot m/s}\). Thus, the units are \(\mathrm{J\,s / (kg \cdot m/s)} = (\mathrm{kg \cdot m^2/s^2} \cdot \mathrm{s}) / (\mathrm{kg \cdot m/s}) = \mathrm{m}\). The units are consistent.

CHECK YOURSELF

Three quick tests

  1. What happens to the maximum kinetic energy of photoelectrons if the frequency of incident light is doubled, assuming the initial frequency was above the threshold?
    AnswerThe maximum kinetic energy increases by \(h f\), where \(f\) is the original frequency. It does not double because of the work function term.
  2. Why does the photoelectric effect not occur below the threshold frequency, regardless of light intensity?
    AnswerEach photon has energy \(hf\). If \(hf < \phi\), a single photon cannot provide enough energy to overcome the work function. Intensity only increases the number of photons, not their individual energy.
  3. How does the de Broglie wavelength change if the momentum of a particle is tripled?
    AnswerThe wavelength becomes one-third of its original value, as \(\lambda\) is inversely proportional to \(p\).
PROBLEMS

Problem set for PU-202 →

Each derivation in this unit ends with five graded problems with full solutions.

NOTES

Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.

Derivations cited: D-430 · D-431 · Threads: matter, waves, energy