Quantum and Nuclear Physics (HL) — IB Physics HL HL Study Guide

For: IB Physics HL candidates sitting IB Physics HL.

Covers: Photoelectric effect and Einstein's equation, wave-particle duality and de Broglie wavelength, atomic spectra and energy levels, nuclear fission and fusion reactions, and radioactive half-life and decay law.

You should already know: IGCSE Physics, basic algebra and calculus.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the IB Physics HL style for educational use. They are not reproductions of past IBO papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official IBO mark schemes for grading conventions.

1. What Is Quantum and Nuclear Physics?

Quantum and Nuclear Physics is the branch of physics that describes the behavior of matter and energy at the subatomic (atomic, nuclear, and particle) scale, where classical Newtonian physics no longer provides accurate predictions. It is a core HL-only topic, accounting for 12-15% of marks across Paper 2 and Paper 3, with questions ranging from 1-mark multiple choice to 8-mark extended response problems focused on real-world applications like solar cells and nuclear power.

2. Photoelectric effect and Einstein's equation

The photoelectric effect is the emission of electrons (called photoelectrons) from a clean metal surface when electromagnetic radiation of sufficiently high frequency is incident on it. Classical wave physics predicted that electron emission would depend on radiation intensity, but experimental results showed emission only occurs above a material-specific threshold frequency (f_0), regardless of intensity.

Key definitions:

• Work function (\Phi): The minimum energy required to eject an electron from the metal surface, equal to (hf_0), where (h = 6.63 \times 10^{-34}\ \text{J s}) is Planck's constant.
• Stopping potential (V_s): The reverse voltage required to reduce the photocurrent to zero, so the maximum kinetic energy of emitted electrons is (KE_{max} = eV_s), where (e) is the electron charge.

Einstein explained the effect by proposing that light behaves as discrete packets of energy called photons, each with energy (E = hf). When a photon hits an electron, it transfers all its energy to the electron, leading to his famous photoelectric equation: $$hf=\Phi +K{E}_{max}$$

Worked example

A cesium metal sample has a work function of 2.1 eV. If it is illuminated by UV light of frequency (1.1 \times 10^{15}\ \text{Hz}), calculate the maximum kinetic energy of emitted photoelectrons in eV.

1. Calculate photon energy in eV using (h = 4.14 \times 10^{-15}\ \text{eV s}): $$hf=4.14\times 10^{-15}\times 1.1\times 10^{15}=4.55\text{eV}$$
2. Rearrange Einstein's equation to solve for (KE_{max}): $$K{E}_{max}=4.55-2.1=2.45\text{eV}$$

Examiners frequently test that increasing radiation intensity only increases the number of emitted electrons, not their maximum kinetic energy, so be prepared to explain this distinction for 2-3 mark response questions.

3. Wave-particle duality, de Broglie wavelength

Wave-particle duality states that all matter and radiation exhibit both wave and particle properties, depending on the experiment performed. Light shows particle behavior in the photoelectric effect, and wave behavior in diffraction and interference experiments. In 1924, Louis de Broglie extended this duality to all moving particles, proposing that every particle has an associated matter wave with a wavelength given by: $$\lambda =\frac{h}{p}$$ where (p = mv) is the particle's momentum (for non-relativistic speeds, i.e., (v \ll 3 \times 10^8\ \text{m s}^{-1})). This hypothesis was confirmed experimentally by the diffraction of electrons off crystalline solids, which produce wave-like interference patterns.

Worked example

Calculate the de Broglie wavelength of an electron accelerated through a potential difference of 150 V. The mass of an electron is (9.11 \times 10^{-31}\ \text{kg}).

1. Calculate the kinetic energy of the electron: $$KE=eV=1.6\times 10^{-19}\times 150=2.4\times 10^{-17}\text{J}$$
2. Relate KE to momentum: (KE = \frac{p^2}{2m}) so (p = \sqrt{2mKE}): $$p=\sqrt{2\times 9.11\times 10^{-31}\times 2.4\times 10^{-17}}=6.61\times 10^{-24}{\text{kg m s}}^{-1}$$
3. Solve for de Broglie wavelength: $$\lambda =\frac{6.63\times 10^{-34}}{6.61\times 10^{-24}}=1.00\times 10^{-10}\text{m}$$ This wavelength matches the spacing between atoms in a crystal lattice, explaining why electrons diffract off solid surfaces.

4. Atomic spectra and energy levels

Atomic spectra are unique line patterns produced when electrons in atoms transition between discrete, quantized energy levels. There are two types of spectra:

• Emission spectra: Produced when electrons fall from higher to lower energy levels, emitting photons of energy equal to the difference between the two levels.
• Absorption spectra: Produced when electrons absorb photons to jump from lower to higher energy levels, so those specific photon frequencies are missing from a continuous incident light spectrum.

Energy levels in atoms are negative because they represent bound states: the ground state (lowest energy level, (n=1)) is the most negative, and the ionization energy is the energy required to move an electron from the ground state to (E=0) (free, unbound electron). The energy of a photon emitted or absorbed during a transition is: $$hf=|E_{upper}-E_{lower}|$$

Worked example

The energy levels of hydrogen are (E_1 = -13.6\ \text{eV}), (E_2 = -3.4\ \text{eV}), (E_3 = -1.51\ \text{eV}), (E_4 = -0.85\ \text{eV}). Calculate the wavelength of the photon emitted when an electron falls from (n=4) to (n=2).

1. Calculate the magnitude of the energy difference: $$\Delta E=|-0.85-(-3.4)|=2.55\text{eV}=4.08\times 10^{-19}\text{J}$$
2. Solve for frequency: (f = \frac{\Delta E}{h} = \frac{4.08 \times 10^{-19}}{6.63 \times 10^{-34}} = 6.15 \times 10^{14}\ \text{Hz})
3. Solve for wavelength: (\lambda = \frac{c}{f} = \frac{3 \times 10^8}{6.15 \times 10^{14}} = 4.88 \times 10^{-7}\ \text{m} = 488\ \text{nm}), which is visible blue-green light.

5. Nuclear reactions, fission and fusion

All nuclei are composed of protons (atomic number (Z)) and neutrons (neutron number (N)), with mass number (A = Z + N). Isotopes are nuclei with the same (Z) but different (N). All nuclear reactions conserve charge, nucleon number, mass-energy, and momentum.

Key definitions:

• Mass defect (\Delta m): The difference between the total mass of individual nucleons in a nucleus and the mass of the intact nucleus: (\Delta m = Zm_p + Nm_n - m_{nucleus})
• Binding energy (E_b): The energy equivalent of the mass defect, given by (E_b = \Delta m c^2). Higher binding energy per nucleon means a more stable nucleus, with iron-56 being the most stable nucleus.

There are two energy-releasing nuclear reaction types:

1. Fission: A heavy, unstable nucleus (e.g., U-235) splits into two lighter nuclei when hit by a neutron, releasing energy and extra neutrons that can cause a chain reaction. Fission is used in nuclear power plants. A typical fission reaction: $${}_{92}^{235}U{+}_0^{1}n{\to }_{56}^{141}Ba{+}_{36}^{92}Kr+3_0^{1}n$$
2. Fusion: Two light nuclei combine to form a heavier nucleus, releasing energy. Fusion powers all stars, including our Sun, and requires extremely high temperatures and pressures to overcome electrostatic repulsion between positively charged nuclei. A typical fusion reaction: $${}_1^{2}H{+}_1^{3}H{\to }_2^{4}He{+}_0^{1}n+17.6\text{MeV}$$

Worked example

Calculate the energy released in the fusion reaction above, given masses: (^2_1H = 2.01410\ \text{u}), (^3_1H = 3.01605\ \text{u}), (^4_2He = 4.00260\ \text{u}), (^1_0n = 1.00867\ \text{u}), and (1\ \text{u} = 931.5\ \text{MeV}/c^2).

1. Calculate total mass before reaction: (2.01410 + 3.01605 = 5.03015\ \text{u})
2. Calculate total mass after reaction: (4.00260 + 1.00867 = 5.01127\ \text{u})
3. Mass defect: (\Delta m = 5.03015 - 5.01127 = 0.01888\ \text{u})
4. Energy released: (0.01888 \times 931.5 = 17.6\ \text{MeV})

6. Half-life and decay law

Radioactive decay is a spontaneous, random process: it is impossible to predict when an individual nucleus will decay, but we can model the decay of a large sample statistically.

Key definitions:

• Decay constant (\lambda): The probability that a single nucleus will decay per unit time, with units of (\text{s}^{-1}) or equivalent time units.
• Activity (A): The number of decays per second in a sample, measured in becquerels (Bq), equal to (A = \lambda N) where (N) is the number of undecayed nuclei.

The radioactive decay law describes the number of undecayed nuclei or activity remaining in a sample after time (t): $$N(t)=N_0{e}^{-\lambda t},A(t)=A_0{e}^{-\lambda t}$$ where (N_0) and (A_0) are the initial number of nuclei and initial activity, respectively. The half-life (T_{1/2}) is the time taken for half the nuclei in a sample to decay, or for the activity to drop to half its initial value. The relationship between half-life and decay constant is: $$T_{1/2}=\frac{\ln 2}{\lambda }$$

Worked example

A sample of carbon-14 has an initial activity of 120 Bq, and a half-life of 5730 years. Calculate the activity remaining after 17190 years.

1. Calculate the number of half-lives elapsed: (\frac{17190}{5730} = 3)
2. Activity after (n) half-lives: (A = A_0 \times \left(\frac{1}{2}\right)^n = 120 \times \left(\frac{1}{2}\right)^3 = 15\ \text{Bq}) For non-integer numbers of half-lives, use the exponential decay formula directly.

7. Common Pitfalls (and how to avoid them)

• Wrong move: Using light intensity to calculate the maximum kinetic energy of photoelectrons. Why students do it: Confuse classical wave energy proportional to intensity with the photon model. Correct move: Only incident light frequency affects maximum kinetic energy; intensity only affects the number of emitted electrons.
• Wrong move: Using the negative sign of energy level differences to calculate photon energy. Why students do it: Forget energy levels are negative bound states, so transitions involve the magnitude of the energy gap. Correct move: Always take (|\Delta E|) when calculating photon frequency or wavelength.
• Wrong move: Forgetting that 1 u = 931.5 MeV/c², so multiplying mass defect in u by (c^2) unnecessarily. Why students do it: Mix up SI and atomic mass unit conventions. Correct move: If mass defect is in u, multiply directly by 931.5 MeV/u to get energy in MeV without (c^2).
• Wrong move: Assuming all nuclear reactions release energy. Why students do it: Confuse fission and fusion direction relative to iron-56. Correct move: Reactions moving towards iron-56 (fission of heavier nuclei, fusion of lighter nuclei) release energy; the reverse absorbs energy.
• Wrong move: Calculating half-life as (1/\lambda) instead of (\ln 2/\lambda). Why students do it: Mix up decay constant inverse with half-life. Correct move: Memorize the relationship, and double check units match before calculating.

8. Practice Questions (IB Physics HL Style)

Question 1

A sodium metal surface has a work function of 2.28 eV. It is illuminated with monochromatic light of wavelength 500 nm. (a) Calculate the maximum kinetic energy of emitted photoelectrons, in eV. (b) Calculate the stopping potential required to reduce the photocurrent to zero. (c) State and explain the effect on the stopping potential if the intensity of the incident light is tripled.

Solution

(a) Photon energy: (E = \frac{hc}{\lambda} = \frac{1240\ \text{eV nm}}{500\ \text{nm}} = 2.48\ \text{eV}). (KE_{max} = E - \Phi = 2.48 - 2.28 = 0.20\ \text{eV}) (b) Stopping potential (V_s = \frac{KE_{max}}{e} = 0.20\ \text{V}) (c) No change. Tripling intensity increases the number of incident photons per second, but each photon still has the same energy, so the maximum kinetic energy of photoelectrons and thus stopping potential remains the same.

Question 2

Calculate the de Broglie wavelength of an alpha particle (mass = (6.64 \times 10^{-27}\ \text{kg})) traveling at 5% of the speed of light. Give your answer in meters.

Solution

Speed of alpha particle: (v = 0.05 \times 3 \times 10^8 = 1.5 \times 10^7\ \text{m s}^{-1}) Momentum: (p = mv = 6.64 \times 10^{-27} \times 1.5 \times 10^7 = 9.96 \times 10^{-20}\ \text{kg m s}^{-1}) de Broglie wavelength: (\lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34}}{9.96 \times 10^{-20}} = 6.66 \times 10^{-15}\ \text{m})

Question 3

Cobalt-60, used in radiotherapy, has a half-life of 5.27 years. A hospital sample has an initial activity of 800 MBq. (a) Calculate the decay constant of cobalt-60, in (\text{s}^{-1}). (b) Calculate the activity remaining after 10.54 years. (c) Calculate the number of undecayed cobalt-60 nuclei in the initial sample.

Solution

(a) Convert half-life to seconds: (5.27\ \text{years} = 5.27 \times 365 \times 24 \times 3600 = 1.66 \times 10^8\ \text{s}). (\lambda = \frac{\ln 2}{T_{1/2}} = \frac{0.693}{1.66 \times 10^8} = 4.17 \times 10^{-9}\ \text{s}^{-1}) (b) Number of half-lives: (10.54 / 5.27 = 2). Activity: (800 \times (1/2)^2 = 200\ \text{MBq}) (c) Initial number of nuclei: (N_0 = \frac{A_0}{\lambda} = \frac{800 \times 10^6}{4.17 \times 10^{-9}} = 1.92 \times 10^{17}\ \text{nuclei})

9. Quick Reference Cheatsheet

10. What's Next

This topic forms the foundation for multiple other HL Physics topics: it connects directly to Particle Physics, where you will extend nuclear reaction conservation rules to fundamental particles and interactions, and Astrophysics, where fusion reactions and radioactive decay are used to model stellar evolution, stellar lifetimes, and the age of celestial objects. If you are studying an optional topic, concepts from this guide appear in Medical Physics (radiotherapy, X-ray production) and Engineering Physics (quantum tunneling, semiconductor physics) as well. To reinforce your mastery of this topic, practice more extended response questions that require both numerical calculation and conceptual explanation, as these make up the majority of marks for this topic in IB exams. If you get stuck on any concept, or need additional practice problems tailored to your skill level, you can ask Ollie, our AI tutor, for personalized support at any time. You can find more IB Physics HL study resources and past paper practice on the homepage.