HL only Nuclear and Quantum Physics BETA E.2

Quantum Physics

This topic explores the transition from classical physics to quantum mechanics, demonstrating that electromagnetic radiation exhibits particle-like properties and matter exhibits wave-like properties. Through the phenomena of the photoelectric effect, de Broglie matter waves, and Compton scattering, the standard classical view of nature is replaced by a quantum framework of wave-particle duality.

Part of IB Physics (2025-2030 syllabus) — Standard and Higher Level.

Key points

The photoelectric effect provides key evidence for the particle nature of light because classical wave theory cannot explain the threshold frequency or the instantaneous emission of electrons under low intensity.
Einstein explained the photoelectric effect by proposing that light is composed of localized packets of energy called photons, where the energy of a single photon is directly proportional to its frequency, expressed as $E = hf$ .
At the threshold frequency the photon energy exactly matches the work function, $\Phi = hf_0$ , so electrons are released with zero kinetic energy; below $f_0$ no emission occurs at any intensity.
Matter waves suggest that every moving particle has an associated wave character, with a de Broglie wavelength $\lambda$ that is inversely proportional to its momentum $p$ .
Wave-particle duality describes how physical entities, such as photons and electrons, exhibit both wave-like behaviors (like interference and diffraction) and particle-like behaviors (like localized collisions and ionization) depending on the experimental setup.
Compton scattering demonstrates that photons carry momentum $p = \frac{h}{\lambda}$ because when high-energy photons collide with stationary electrons, they scatter with a longer wavelength, reflecting a loss of energy and momentum in a particle-like collision.

Subtopic by subtopic

The photoelectric effect

The photoelectric effect is the emission of electrons from a metal surface when electromagnetic radiation of sufficiently high frequency shines on it. Three experimental facts define the effect:

For every metal there is a threshold frequency $f_0$ : radiation below this frequency releases no electrons, no matter how intense it is.
Emission is effectively instantaneous, even when the light is extremely faint.
Increasing the intensity at a fixed frequency increases the number of electrons emitted per second (the photocurrent) but has no effect on their maximum kinetic energy; only raising the frequency does that.

Classical wave theory fails on all three counts. A wave delivers energy continuously across the whole surface, so:

any frequency should eventually eject electrons
a measurable time delay should occur at low intensity
brighter light should mean faster electrons

A classic demonstration shows the conflict: a negatively charged zinc plate discharges when ultraviolet light falls on it, but stays charged under even the brightest visible light.

You should be able to describe the photocell experiment, in which light strikes an emitting electrode and the resulting current is measured, and explain how a reverse potential difference can be used to probe the kinetic energy of the emitted electrons. Most importantly, you must state these observations and explain clearly why each one contradicts the predictions of the classical wave model of light.

Einstein's photoelectric equation and the work function

Einstein explained the photoelectric effect by treating light as a stream of photons, each carrying energy $E = hf$ . A single photon interacts with a single electron in an all-or-nothing event. Part of the photon's energy, the work function $\Phi$ , is spent freeing the electron from the metal; whatever remains appears as kinetic energy.

The fastest electrons come from the surface itself and obey Einstein's equation:

$E_{max} = hf - \Phi$

This one equation explains every observation. Emission requires $hf \geq \Phi$ , which defines the threshold frequency through $\Phi = hf_0$ . Because each electron is released by a single photon, emission is immediate, and intensity (photons arriving per second) controls only the current.

Sodium, with $\Phi \approx 2.3\ \text{eV}$ , emits electrons under blue light but not under red light, whose photons each carry too little energy.

Experimentally, $E_{max}$ is measured with a stopping voltage: the reverse potential difference $V_s$ that just halts the fastest electrons, so $E_{max} = eV_s$ . Combining the two results gives $eV_s = hf - \Phi$ , so a graph of $V_s$ against $f$ is a straight line with:

gradient $\frac{h}{e}$
an $x$ -intercept at $f_0$
a $y$ -intercept of $-\frac{\Phi}{e}$

You must be able to convert fluently between electronvolts and joules, rearrange the equation for any quantity, and extract $h$ , $\Phi$ and $f_0$ from experimental graphs.

Matter waves and the de Broglie wavelength

De Broglie proposed that wave-particle duality works in both directions: if light waves can behave as particles, then particles of matter should behave as waves. Every particle with momentum $p$ has an associated de Broglie wavelength, given by:

$\lambda = \frac{h}{p}$

The smaller the momentum, the longer the wavelength, so wave behaviour is only noticeable for very light particles such as electrons.

The hypothesis was confirmed by electron diffraction. Electrons accelerated through a potential difference of around $100\ \text{V}$ have wavelengths near $0.1\ \text{nm}$ , comparable to the spacing between atoms in a crystal, so a thin crystal acts as a diffraction grating and produces a pattern of bright rings on a screen. Diffraction is a uniquely wave property, so this is direct evidence that moving matter has wave character.

By contrast, a $0.1\ \text{kg}$ ball thrown at $10\ \text{m s}^{-1}$ has $\lambda \approx 6.6 \times 10^{-34}\ \text{m}$ , far too small for any conceivable aperture to diffract, which is why everyday objects show no observable wave behaviour.

For calculations, you must handle charged particles accelerated from rest through a potential difference $V$ :

the kinetic energy is $E_k = qV$
the momentum follows from $p = \sqrt{2mE_k}$
the wavelength is then $\lambda = \frac{h}{p}$

Be ready to compare electron wavelengths with atomic spacings and to explain why diffraction confirms the matter-wave model.

Wave-particle duality

Wave-particle duality is the recognition that neither the wave model nor the particle model alone describes light or matter completely. Photons and electrons show wave-like behaviour, such as interference and diffraction, in some experiments, and particle-like behaviour, such as localized collisions and ionization, in others; the experimental setup determines which aspect is revealed.

The double-slit experiment with matter demonstrates this duality at its most fundamental level. When electrons are fired one at a time through a pair of narrow slits, each electron registers as a single, localized impact on the detector screen, exactly as a particle should. Yet as thousands of impacts accumulate, a clear interference pattern of bright and dark fringes builds up, exactly as a wave should.

The detection event is localized, but the probability of an electron arriving at any given point is governed by a wave: the wave nature is physical, described by the de Broglie wavelength $\lambda = \frac{h}{p}$ , and it determines the statistical distribution of where the particles land.

A useful way to organize the evidence:

the photoelectric effect and Compton scattering reveal the particle nature of electromagnetic radiation
interference and diffraction reveal its wave nature
electron diffraction reveals the wave nature of matter
discrete impacts and collision tracks reveal its particle nature

You should be able to name the evidence on each side, describe the single-electron double-slit experiment, and explain why both models are needed.

Compton scattering

Compton scattering is the increase in wavelength of a high-energy photon, typically an X-ray or gamma ray, when it scatters off an electron that is effectively free and at rest. In classical electromagnetism, a wave scattering off a charge should keep its frequency, and electromagnetic waves carry energy but no localized momentum like discrete particles.

The observed shift resolved this: the interaction behaves exactly like a billiard-ball collision in which the photon, carrying momentum $p = \frac{h}{\lambda}$ , transfers some momentum and energy to the electron. Because the photon loses energy, the scattered photon has a lower frequency and therefore a longer wavelength ( $\lambda_f > \lambda_i$ ).

Applying conservation of energy and momentum to the collision gives the wavelength shift:

$\Delta \lambda = \lambda_f - \lambda_i = \frac{h}{m_e c}(1 - \cos\theta)$

where $\theta$ is the angle through which the photon scatters.

The constant $\frac{h}{m_e c} = 2.43 \times 10^{-12}\ \text{m}$ , the Compton wavelength of the electron, sets the scale of the effect: the shift is zero at $\theta = 0$ and reaches its maximum of twice this value for back-scattering at $\theta = 180^\circ$ . The angular dependence of the shift matches experimental results perfectly, providing decisive proof of the photon's localized, particle-like momentum.

You must be able to:

calculate $\Delta \lambda$ and $\lambda_f$ for a given angle
explain why the wavelength increases
contrast Compton scattering (photon scattered with reduced energy) with the photoelectric effect (photon completely absorbed)

Formulae

E = hf = \frac{hc}{\lambda}

To calculate the energy of a single photon from its frequency $f$ or its wavelength $\lambda$ .

E_{max} = hf - \Phi

To calculate the maximum kinetic energy of emitted photoelectrons when electromagnetic radiation of frequency $f$ is incident on a metal with work function $\Phi$ .

\lambda = \frac{h}{p}

To determine the de Broglie wavelength of any moving particle with known momentum $p$ .

\lambda_f - \lambda_i = \Delta\lambda = \frac{h}{m_e c}(1 - \cos\theta)

To calculate the change in wavelength of an X-ray or gamma-ray photon after scattering off an electron at a scattering angle of $\theta$ .

p = \frac{h}{\lambda}

To determine the momentum of a photon given its wavelength $\lambda$ .

Definitions

Photon: A discrete quantum of electromagnetic radiation that carries energy $E = hf$ and momentum $p = \frac{h}{\lambda}$ and interacts with matter in all-or-nothing events.
Work function ( $\Phi$ ): The minimum energy required to liberate a conduction electron from the surface of a metal.
Threshold frequency ( $f_0$ ): The minimum frequency of incident electromagnetic radiation required to cause the emission of photoelectrons from a given metal surface.
Stopping voltage ( $V_s$ ): The negative potential difference applied to an anode relative to the photocathode that is just sufficient to stop the most energetic photoelectrons from reaching the anode, such that $E_{max} = eV_s$ .
De Broglie wavelength ( $\lambda$ ): The wavelength associated with a moving particle of matter, determined by dividing Planck's constant by the particle's momentum.

Worked examples

Monochromatic light of wavelength $250\ \text{nm}$ shines on a sodium surface which has a work function of $2.28\ \text{eV}$ . Calculate the maximum kinetic energy of the emitted photoelectrons in Joules, and determine the stopping voltage required to reduce the photoelectric current to zero.

1
Step 1: Convert the work function from electronvolts to Joules: $\Phi = 2.28 \times 1.60 \times 10^{-19}\ \text{J} = 3.65 \times 10^{-19}\ \text{J}$ .
2
Step 2: Calculate the energy of the incident photon: $E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34}\ \text{J s} \times 3.00 \times 10^8\ \text{m s}^{-1}}{250 \times 10^{-9}\ \text{m}} = 7.96 \times 10^{-19}\ \text{J}$ .
3
Step 3: Calculate the maximum kinetic energy: $E_{max} = E - \Phi = 7.96 \times 10^{-19}\ \text{J} - 3.65 \times 10^{-19}\ \text{J} = 4.31 \times 10^{-19}\ \text{J}$ .
4
Step 4: Calculate the stopping voltage using $E_{max} = eV_s$ : $V_s = \frac{E_{max}}{e} = \frac{4.31 \times 10^{-19}\ \text{J}}{1.60 \times 10^{-19}\ \text{C}} \approx 2.69\ \text{V}$ .

Answer: Maximum kinetic energy = $4.31 \times 10^{-19}\ \text{J}$ , Stopping voltage = $2.69\ \text{V}$

An electron is accelerated from rest through a potential difference of $150\ \text{V}$ . Calculate the de Broglie wavelength of the electron, and comment on why electrons at this energy are suitable for diffraction by a crystal.

1
Step 1: Calculate the kinetic energy gained by the electron: $E_k = eV = 1.60 \times 10^{-19}\ \text{C} \times 150\ \text{V} = 2.40 \times 10^{-17}\ \text{J}$ .
2
Step 2: Evaluate the quantity $2m_e E_k = 2 \times 9.11 \times 10^{-31}\ \text{kg} \times 2.40 \times 10^{-17}\ \text{J} = 4.37 \times 10^{-47}\ \text{kg}^2\ \text{m}^2\ \text{s}^{-2}$ .
3
Step 3: Calculate the momentum: $p = \sqrt{2m_e E_k} = \sqrt{4.37 \times 10^{-47}} = 6.61 \times 10^{-24}\ \text{kg m s}^{-1}$ .
4
Step 4: Calculate the de Broglie wavelength: $\lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34}\ \text{J s}}{6.61 \times 10^{-24}\ \text{kg m s}^{-1}} = 1.00 \times 10^{-10}\ \text{m}$ .
5
Step 5: This wavelength of $0.100\ \text{nm}$ is comparable to the spacing between atoms in a crystal lattice, so the crystal acts as a diffraction grating and produces an observable diffraction pattern.

Answer: $\lambda = 1.00 \times 10^{-10}\ \text{m}$ ( $0.100\ \text{nm}$ ), comparable to atomic spacing in crystals

An X-ray photon of wavelength $\lambda_i = 0.0240\ \text{nm}$ undergoes Compton scattering off a stationary electron. The photon is scattered at an angle of $90.0^\circ$ relative to its incident direction. Calculate the wavelength of the scattered photon.

1
Step 1: Write down the Compton scattering formula: $\Delta \lambda = \lambda_f - \lambda_i = \frac{h}{m_e c}(1 - \cos\theta)$ .
2
Step 2: Substitute $\theta = 90.0^\circ$ . Since $\cos(90.0^\circ) = 0$ , the term $(1 - \cos\theta)$ simplifies to $1$ .
3
Step 3: Calculate the shift: $\Delta \lambda = \frac{6.63 \times 10^{-34}\ \text{J s}}{9.11 \times 10^{-31}\ \text{kg} \times 3.00 \times 10^8\ \text{m s}^{-1}} = 2.43 \times 10^{-12}\ \text{m} = 0.00243\ \text{nm}$ .
4
Step 4: Calculate the final wavelength: $\lambda_f = \lambda_i + \Delta \lambda = 0.0240\ \text{nm} + 0.00243\ \text{nm} = 0.02643\ \text{nm}$ .

Answer: $0.0264\ \text{nm}$ (or $2.64 \times 10^{-11}\ \text{m}$ )

Common mistakes

×Confusing light intensity with photon energy: Students often wrongly think that increasing the intensity of incident light increases the maximum kinetic energy of photoelectrons. Remember that intensity only increases the *rate* of photon emission (and thus the current), whereas photon energy is solely dependent on frequency ( $E = hf$ ).
×Unit conversion errors between eV and Joules: Working in the wrong unit system is highly common. Always convert work functions or energies given in $\text{eV}$ into Joules ( $1\ \text{eV} = 1.60 \times 10^{-19}\ \text{J}$ ) before utilizing Planck's constant in SI units ( $6.63 \times 10^{-34}\ \text{J s}$ ).
×Confusing the Compton shift with the final wavelength: Students sometimes calculate the shift $\Delta \lambda$ and write it down as the final wavelength of the photon. Ensure you add $\Delta \lambda$ to the initial wavelength $\lambda_i$ to obtain the final wavelength $\lambda_f$ .

Exam tips

✓To earn full marks on questions asking to *describe* the photoelectric effect, contrast the wave theory predictions with the experimental realities. State that the immediate emission of photoelectrons at low light intensity contradicts classical expectations of time delays.
✓When asked to *sketch* a graph of stopping voltage $V_s$ against frequency $f$ , draw a straight line with a negative $y$ -intercept. State clearly that the gradient of this line is $\frac{h}{e}$ and the $x$ -intercept corresponds to the threshold frequency $f_0$ .
✓When you *calculate* the de Broglie wavelength of an accelerated charged particle, first determine its kinetic energy using $E_k = qV$ , then its momentum using $p = \sqrt{2mE_k}$ , and finally use $\lambda = \frac{h}{p}$ .

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