Apply Quantum Physics — Apply E.2 Quantum physics in problem-solving and explanations - IB Physics

SECTION 1

Historical Context & Motivation

At the turn of the twentieth century, classical physics could explain the motion of planets and the behavior of magnets, but it stumbled badly when it tried to describe the very small. Experiments involving heated objects, light striking metals, and glowing gases produced results that simply did not match the predictions of Newtonian mechanics or Maxwell's electromagnetic theory. These failures were not minor—they shook the foundations of physics and forced scientists to rethink the nature of energy and matter at the atomic scale.

The resolution came in stages over about three decades, with each breakthrough introducing a new quantum idea—that energy is not continuous but comes in discrete packets. The timeline below traces the key milestones that built the quantum framework you will apply in IB Physics E.2.

1900 • Planck's Quantum Hypothesis

Max Planck proposed that energy is emitted in discrete packets called quanta, each with energy E = hf, to solve the ultraviolet catastrophe in blackbody radiation.

1905 • Einstein and the Photoelectric Effect

Albert Einstein explained the photoelectric effect by treating light as a stream of particle-like photons, each carrying energy E = hf. This earned him the Nobel Prize in 1921.

1913 • Bohr's Atomic Model

Niels Bohr applied quantization to the hydrogen atom, proposing that electrons occupy only certain allowed energy levels and emit or absorb photons when transitioning between them.

1924 • de Broglie's Matter Waves

Louis de Broglie suggested that particles such as electrons also exhibit wave-like behavior, with wavelength λ = h/p. This was confirmed experimentally by electron diffraction.

1926 • Schrödinger's Wave Equation

Erwin Schrödinger formulated a mathematical wave equation that describes how quantum states evolve, unifying the wave and particle descriptions into modern quantum mechanics.

The central question that E.2 asks you to answer is this: how do we use the quantum equations—for photon energy, the photoelectric effect, matter waves, and energy-level transitions—to make quantitative predictions and provide coherent physical explanations? The rest of this lesson equips you to do exactly that.

SECTION 2

Core Principles & Definitions

Before you can solve quantum problems, you need to internalize four foundational ideas. These concepts appear repeatedly in IB exam questions, and understanding them deeply will keep you from making common errors. Each principle below connects an observable phenomenon to a mathematical relationship.

Photon Energy

Light consists of photons—discrete packets of electromagnetic energy. Each photon carries energy E = hf, where h is Planck's constant (6.63 × 10⁻³⁴ J·s) and f is the frequency of the light. Higher-frequency light means higher-energy photons.

Photoelectric Effect

When a photon strikes a metal surface, it can eject an electron if the photon energy exceeds the metal's work function (φ). The leftover energy becomes kinetic energy of the ejected electron: E_k(max) = hf − φ.

Matter Waves (de Broglie)

Every moving particle has a de Broglie wavelength given by λ = h/p, where p is the particle's momentum. For electrons accelerated through a potential difference, this wavelength is small enough to produce diffraction patterns.

Atomic Energy Levels

Electrons in atoms occupy discrete energy levels. A photon is emitted or absorbed when an electron transitions between levels. The photon's energy equals the energy difference: ΔE = hf. This explains the line spectra of elements.

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KEY TAKEAWAY

Think of photon energy like coins of specific denominations. You cannot pay with half a coin—either you have enough coins of the right value to "buy" an electron's freedom from the metal, or you don't. No amount of small-value coins (low-frequency photons) will work if each individual coin is too small, no matter how many you pile up. This is why intensity alone cannot trigger the photoelectric effect below the threshold frequency—each photon must individually carry enough energy.

SECTION 3

Visual Explanation — Energy Levels & Photon Emission

The diagram below illustrates how quantized energy levels in the hydrogen atom produce the visible Balmer series of spectral lines. Each downward arrow represents an electron transition from a higher level (n = 3, 4, 5, or 6) to the n = 2 level, releasing a photon whose energy corresponds to a specific color of visible light.

Each horizontal line represents a quantized energy level of hydrogen (values in eV on the left axis). The downward arrows show electron transitions to the n = 2 level, each producing a photon with a characteristic wavelength in the visible Balmer series. Notice that transitions from higher starting levels produce shorter wavelengths (higher-energy photons).

When solving IB problems involving spectral lines, always start by identifying the two energy levels involved in the transition. The energy of the emitted (or absorbed) photon equals the absolute difference between those levels: ΔE = |E_upper − E_lower|. From there, you can find frequency using f = ΔE/h and wavelength using λ = c/f. Many students lose marks by forgetting that the energy values are negative, so taking the absolute difference is essential.

SECTION 4

Mathematical Framework

Quantum physics in E.2 revolves around a small set of powerful equations. Each one links an observable quantity—like a wavelength or kinetic energy—to quantum properties. You will use these equations in virtually every calculation, so understanding what each variable means is just as important as memorizing the formula.

PHOTON ENERGY

E = hf = hc / λ

E = photon energy (J or eV), h = Planck's constant (6.63 × 10⁻³⁴ J·s), f = frequency (Hz), c = speed of light (3.00 × 10⁸ m/s), λ = wavelength (m). Use the second form when wavelength is given instead of frequency.

PHOTOELECTRIC EQUATION

E_k(max) = hf − φ

E_k(max) = maximum kinetic energy of ejected photoelectrons (J or eV), φ = work function of the metal (J or eV). No electrons are emitted when hf < φ. The threshold frequency is f₀ = φ/h.

DE BROGLIE WAVELENGTH

λ = h / p = h / (mv)

λ = de Broglie wavelength (m), p = momentum (kg·m/s), m = particle mass (kg), v = speed (m/s). For an electron accelerated through potential difference V, use p = √(2m_eeV) where e = 1.60 × 10⁻¹⁹ C.

ENERGY-LEVEL TRANSITIONS

ΔE = hf = E_upper − E_lower

ΔE = energy of the photon emitted or absorbed during a transition. For hydrogen, E_n = −13.6 / n² eV, where n is the principal quantum number. Use absolute values when calculating photon energy.

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UNIT CHECK TIP

IB exams frequently give energies in electron-volts (eV) and require answers in joules, or vice versa. Always remember: 1 eV = 1.60 × 10⁻¹⁹ J. Convert before substituting into equations that use SI units. Many marks are lost to unit errors, not conceptual mistakes.

SECTION 5

The Photoelectric Effect — Detailed Breakdown

The photoelectric effect is arguably the most-tested quantum topic in IB Physics. Understanding its graph—a plot of maximum kinetic energy of photoelectrons against the frequency of incident light—is essential for both paper 1 multiple-choice and paper 2 structured questions. The diagram below shows what this graph looks like and how to extract key information from it.

The graph of maximum kinetic energy versus frequency is a straight line with gradient equal to Planck's constant h. The x-intercept gives the threshold frequency f₀, and extrapolating the line to the y-axis gives −φ (the negative work function). The shaded region to the left of f₀ indicates frequencies where no photoelectrons are emitted.

There are several observations that classical wave theory cannot explain but the photon model handles perfectly. First, no electrons are emitted below the threshold frequency, regardless of light intensity—classically, brighter light should always deliver enough energy eventually. Second, photoelectrons appear almost instantaneously (within ~10⁻⁹ s) even at low intensity—classical theory predicts a significant time delay. Third, increasing intensity increases the number of photoelectrons but not their maximum kinetic energy, while increasing frequency increases E_k(max). These facts only make sense when light is treated as a stream of individual photons.

Observation	Classical Prediction	Quantum Explanation
Threshold frequency exists	Any frequency should work if intensity is high enough	Each photon must individually have E ≥ φ; frequency determines single-photon energy
Instantaneous emission	Energy accumulates slowly; expect a delay	A single photon delivers all its energy to one electron in a single interaction
Intensity increases current, not E_k(max)	Higher intensity should increase electron energy	More photons eject more electrons; each photon still carries the same energy hf

Comparison of classical and quantum predictions for the photoelectric effect

SECTION 6

Worked Example — Photoelectric Effect & de Broglie Wavelength

The following example combines the photoelectric equation with the de Broglie wavelength—a common style for IB exam questions that require you to connect two quantum ideas in a single problem.

Photoelectric Emission & de Broglie Wavelength of Ejected Electrons

1
Step 1 — Read the Problem

Ultraviolet light of wavelength 200 nm strikes a potassium surface whose work function is φ = 2.30 eV. Calculate (a) the maximum kinetic energy of the ejected photoelectrons in eV, and (b) the de Broglie wavelength of the fastest ejected electrons.
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Step 2 — Find the Photon Energy

Use E = hc/λ. Converting the wavelength: λ = 200 nm = 200 × 10⁻⁹ m = 2.00 × 10⁻⁷ m. Then E = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (2.00 × 10⁻⁷) = 9.945 × 10⁻¹⁹ J. Converting to eV: E = 9.945 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ = 6.22 eV.

E_photon = 6.22 eV
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Step 3 — Apply the Photoelectric Equation

E_k(max) = hf − φ = 6.22 − 2.30 = 3.92 eV. This is part (a) of the answer.

E_k(max) = 3.92 eV
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Step 4 — Convert Kinetic Energy to SI for Part (b)

We need kinetic energy in joules to find momentum. E_k = 3.92 × 1.60 × 10⁻¹⁹ = 6.272 × 10⁻¹⁹ J.

E_k = 6.27 × 10⁻¹⁹ J
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Step 5 — Find Momentum, Then de Broglie Wavelength

From E_k = p²/(2m_e), we get p = √(2m_eE_k) = √(2 × 9.11 × 10⁻³¹ × 6.272 × 10⁻¹⁹) = √(1.143 × 10⁻⁴⁸) = 1.069 × 10⁻²⁴ kg·m/s. Then λ = h/p = 6.63 × 10⁻³⁴ / 1.069 × 10⁻²⁴ = 6.20 × 10⁻¹⁰ m ≈ 0.620 nm.

λ_{de Broglie} ≈ 0.620 nm (6.20 × 10⁻¹⁰ m)

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EXAM STRATEGY

In IB structured questions, always show your unit conversions explicitly. If you convert eV to joules, write the conversion factor. If you convert nm to m, write it out. Examiners award process marks even when the final numerical answer is slightly off.

SECTION 7

Classical Physics vs Quantum Physics — Strengths & Limitations

Quantum physics did not replace classical physics—it extended it into domains where classical models fail. It is important to understand where each framework works well and where it breaks down. IB examiners frequently ask you to explain why a classical explanation is inadequate for a given phenomenon, so the table below is worth studying carefully.

Feature	Classical Physics	Quantum Physics
Energy	Continuous — any value allowed	Quantized — comes in discrete packets (E = hf)
Light	Purely a wave (EM theory)	Wave-particle duality (photon model + interference)
Electrons	Point particles with definite paths	Exhibit wave-like diffraction; described by probability
Atomic structure	Electrons spiral inward, radiating energy continuously (unstable!)	Electrons occupy stable, quantized energy levels
Scale of validity	Macroscopic objects (planets, baseballs)	Atomic and subatomic scales (essential when λ_dB ≈ object size)

Classical vs quantum physics: where each framework applies

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KEY TAKEAWAY

Think of classical physics as a road map that works perfectly at the scale of cities and highways. Quantum physics is like a GPS that also works inside buildings and narrow alleyways. You don't throw away the road map—it's still excellent for highway driving. But when you need to navigate a complex building interior (the atomic scale), you need the GPS. The quantum equations of E.2 are your GPS for the subatomic world: they give precise, testable predictions where classical physics gives wrong or nonsensical answers.

SECTION 8

Connection to Advanced Quantum Theory

The E.2 quantum physics you learn in IB is the foundation for far more sophisticated theories studied in university-level physics. The Bohr model, the photoelectric effect, and the de Broglie hypothesis were all stepping stones to the full framework of quantum mechanics developed in the 1920s. The table below shows how each E.2 concept connects to its more advanced counterpart.

IB E.2 Concept	Advanced Extension	What Changes
Bohr model energy levels	Schrödinger equation solutions	Circular orbits replaced by 3D probability clouds (orbitals); quantum numbers l and m added
de Broglie wavelength λ = h/p	Wave functions ψ(x, t)	A full mathematical description of particle behavior; \|ψ\|² gives probability density
Photon energy E = hf	Quantum electrodynamics (QED)	Photons are excitations of the electromagnetic field; interactions described by Feynman diagrams
Photoelectric threshold	Band theory of solids	Work function explained via energy bands in metals, semiconductors, and insulators

How IB E.2 topics extend into university-level quantum mechanics

You do not need to know these advanced topics for the IB exam, but recognizing where the E.2 models are heading helps you appreciate why they were so revolutionary. Every quantum technology you use daily—from LEDs and laser pointers to MRI scanners and semiconductor chips—relies on the principles you are mastering in E.2.

SECTION 9

Practice Problems

Work through these five problems in order. They increase in difficulty and cover the full range of E.2 quantum physics topics. Show all working, and pay careful attention to unit conversions.

Question 1: CONCEPTUAL

A metal surface is illuminated with red light of low intensity, and no photoelectrons are emitted. A student suggests that increasing the intensity of the red light will eventually cause emission. Explain, using the photon model, whether the student is correct.

Show answer

The student is incorrect. In the photon model, each photon carries energy E = hf. Red light has a fixed (low) frequency, so each photon carries a fixed amount of energy. If this energy is less than the work function φ of the metal, no single photon can eject an electron, regardless of how many photons arrive per second. Increasing intensity only increases the number of photons (hence the rate of photon-surface interactions), but it does not change the energy of each individual photon. Therefore, no photoelectrons will be emitted at any intensity if the frequency is below the threshold frequency f₀ = φ/h.

Question 2: BASIC CALCULATION

Calculate the energy of a photon of green light with wavelength 530 nm. Express your answer in both joules and electron-volts. (h = 6.63 × 10⁻³⁴ J·s, c = 3.00 × 10⁸ m/s, 1 eV = 1.60 × 10⁻¹⁹ J)

Show answer

Using E = hc/λ: λ = 530 × 10⁻⁹ m = 5.30 × 10⁻⁷ m. E = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (5.30 × 10⁻⁷) = 1.989 × 10⁻²⁵ / 5.30 × 10⁻⁷ = 3.75 × 10⁻¹⁹ J. Converting to eV: E = 3.75 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ = 2.34 eV. The photon energy is approximately 3.75 × 10⁻¹⁹ J or 2.34 eV, consistent with the visible portion of the electromagnetic spectrum.

Question 3: INTERMEDIATE

Light of frequency 1.20 × 10¹⁵ Hz strikes a sodium surface (φ = 2.28 eV). (a) Calculate the maximum kinetic energy of the emitted photoelectrons in eV. (b) Calculate the stopping voltage required to reduce the photocurrent to zero.

Show answer

(a) First find the photon energy: E = hf = 6.63 × 10⁻³⁴ × 1.20 × 10¹⁵ = 7.956 × 10⁻¹⁹ J = 7.956 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ = 4.97 eV. Then

E_k

(max) = hf − φ = 4.97 − 2.28 = 2.69 eV. (b) The stopping voltage

V_s

is defined by eV_s =

E_k

(max), so

V_s

E_k

(max) / e = 2.69 eV / e = 2.69 V. The stopping potential equals the maximum kinetic energy expressed in eV divided by the elementary charge, which numerically gives

V_s

= 2.69 V.

Question 4: APPLIED

An electron microscope accelerates electrons through a potential difference of 5.00 kV. (a) Calculate the de Broglie wavelength of these electrons. (b) Explain why electron microscopes can resolve much finer detail than optical microscopes. (

m_e

= 9.11 × 10⁻³¹ kg, e = 1.60 × 10⁻¹⁹ C, h = 6.63 × 10⁻³⁴ J·s)

Show answer

(a) The kinetic energy gained is

E_k

= eV = 1.60 × 10⁻¹⁹ × 5000 = 8.00 × 10⁻¹⁶ J. The momentum is p = √(2m_e

E_k

) = √(2 × 9.11 × 10⁻³¹ × 8.00 × 10⁻¹⁶) = √(1.458 × 10⁻⁴⁵) = 3.818 × 10⁻²³ kg·m/s. The de Broglie wavelength is λ = h/p = 6.63 × 10⁻³⁴ / 3.818 × 10⁻²³ = 1.74 × 10⁻¹¹ m = 0.0174 nm. (b) Resolution in any microscope is limited by the wavelength used. Visible light has wavelengths around 400–700 nm, while these electrons have a wavelength of about 0.017 nm—roughly 25,000 times shorter. Since diffraction limits the smallest resolvable feature to approximately the wavelength being used, electron microscopes can image individual atoms and molecular structures that are completely invisible to optical microscopes.

Question 5: CRITICAL THINKING

A hydrogen atom transitions from the n = 4 energy level to the n = 2 energy level, emitting a photon. (a) Using

E_n

= −13.6/n² eV, calculate the wavelength of the emitted photon. (b) The same photon is then directed at a cesium surface with work function φ = 2.10 eV. Determine whether this photon can eject a photoelectron, and if so, find the maximum kinetic energy of the ejected electron.

Show answer

(a) E₄ = −13.6/4² = −13.6/16 = −0.850 eV. E₂ = −13.6/2² = −13.6/4 = −3.40 eV. ΔE = E₄ − E₂ = −0.850 − (−3.40) = 2.55 eV. Converting to joules: ΔE = 2.55 × 1.60 × 10⁻¹⁹ = 4.08 × 10⁻¹⁹ J. Wavelength: λ = hc/ΔE = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (4.08 × 10⁻¹⁹) = 1.989 × 10⁻²⁵ / 4.08 × 10⁻¹⁹ = 4.88 × 10⁻⁷ m = 488 nm. This is in the blue-cyan part of the visible spectrum, corresponding to the Hβ Balmer line. (b) The photon energy is 2.55 eV, and the cesium work function is 2.10 eV. Since 2.55 eV > 2.10 eV, the photon can eject a photoelectron.

E_k

(max) = hf − φ = 2.55 − 2.10 = 0.45 eV. This problem demonstrates how an atomic transition in one system (hydrogen) can produce a photon that drives a quantum process (photoelectric effect) in another system (cesium), linking two core E.2 topics in a single physical scenario.

SUMMARY

Lesson Summary

Summary

IB Physics E.2 centres on applying quantum principles quantitatively. Photon energy is calculated using E = hf = hc/λ, linking every photon to a specific frequency and wavelength. The photoelectric effect demonstrates that light behaves as individual photons: only photons with energy above the metal's work function φ can eject electrons, and the maximum kinetic energy of ejected electrons is E_k(max) = hf − φ. The graph of E_k(max) versus frequency is a straight line with gradient equal to Planck's constant h and x-intercept equal to the threshold frequency f₀.

The de Broglie wavelength λ = h/p reveals the wave nature of particles and explains why electron diffraction is observable at atomic scales. For atoms, electrons occupy discrete energy levels, and transitions between levels produce or absorb photons of specific energies (ΔE = hf), which accounts for the line spectra of elements. In every E.2 problem, the strategy is the same: identify the relevant quantum equation, convert units carefully (especially between eV and joules), substitute, solve, and check that your answer has sensible magnitude and units.