Modern Physics (Phys 2) — AP Physics 2 Phys 2 Study Guide

For: AP Physics 2 candidates sitting AP Physics 2.

Covers: The photoelectric effect, wave-particle duality, atomic spectra and the Bohr model, nuclear decay modes, and mass-energy equivalence as outlined in the AP Physics 2 Course and Exam Description.

You should already know: AP Physics 1 or equivalent.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics 2 style for educational use. They are not reproductions of past College Board papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official College Board mark schemes for grading conventions.

1. What Is Modern Physics (Phys 2)?

Modern physics is a set of 20th-century physical frameworks that describe phenomena at subatomic, atomic, and high-energy scales that classical Newtonian mechanics and electromagnetic theory cannot explain. It accounts for counterintuitive experimental observations that defied 19th-century physical models, and makes up 10-15% of your total AP Physics 2 exam score per the official CED. Common related terms include quantum physics and nuclear physics, which are the two core subsets of modern physics tested on this exam.

2. Photoelectric effect — Einstein's equation

The photoelectric effect describes the emission of electrons from a metal surface when light of sufficient frequency shines on it. Classical wave theory incorrectly predicted that higher light intensity (brighter light) would eject electrons at any frequency, but experiments showed no electrons are emitted below a material-specific threshold frequency, regardless of intensity. Einstein resolved this contradiction by proposing that light behaves as discrete packets of energy called photons, where each photon has energy proportional to its frequency: $E=hf$, where $h$ is Planck's constant ($4.14\times 10^{-15}\text{ eVcdotps}$ for AP Physics 2 calculations). His famous photoelectric equation relates the maximum kinetic energy of ejected electrons to photon energy and the material's work function $\Phi$, the minimum energy required to free an electron from the metal surface: $$K_{\text{max}}=hf-\Phi$$ The maximum kinetic energy can also be measured using stopping potential $V_{\text{stop}}$, the voltage required to stop all ejected electrons: $K_{\text{max}}=e{V}_{\text{stop}}$, where $e$ is the charge of an electron.

Worked Example: Cesium has a work function of 2.14 eV. If 550 nm yellow light shines on a cesium sample, what is the maximum kinetic energy of ejected electrons, and the corresponding stopping potential?

1. Calculate photon energy using the shortcut $hc=1240\text{ eVcdotpnm}$: $E=\frac{hc}{\lambda }=\frac{1240}{550}\approx 2.25\text{ eV}$
2. Apply Einstein's equation: $K_{\text{max}}=2.25\text{ eV}-2.14\text{ eV}=0.11\text{ eV}$
3. Stopping potential equals the maximum kinetic energy in eV divided by electron charge, so $V_{\text{stop}}=0.11\text{ V}$ Exam tip: Examiners frequently test that increasing intensity only increases the number of ejected electrons if the frequency is above threshold; it does not increase the maximum kinetic energy of individual electrons.

3. Wave-particle duality

If light (long thought to be a wave) exhibits particle-like behavior in the photoelectric effect, French physicist Louis de Broglie hypothesized that matter (long thought to be made of particles) exhibits wave-like behavior. The de Broglie wavelength of a moving particle is inversely proportional to its momentum: $$\lambda =\frac{h}{p}$$ where $p=mv$ for non-relativistic speeds (the only case tested on AP Physics 2). Experimental proof of this hypothesis came from electron diffraction experiments, where electrons fired through a crystal lattice produced interference patterns, a property exclusive to waves. Wave properties of matter are only observable when the de Broglie wavelength is comparable to the size of the obstacles or slits the particle interacts with. For macroscopic objects (e.g., a 0.1 kg baseball moving at 30 m/s), the de Broglie wavelength is ~$2\times 10^{-34}$ m, far smaller than atomic nuclei, so wave behavior is undetectable. For electrons, however, wavelengths are comparable to atomic spacing in crystals, making diffraction easily observable.

Worked Example: What is the de Broglie wavelength of an electron (mass $9.11\times 10^{-31}\text{ kg}$) accelerated through a 50 V potential difference?

1. Calculate electron kinetic energy: $K=qV=1.6\times 10^{-19}\text{ C}\times 50\text{ V}=8\times 10^{-18}\text{ J}$
2. Relate kinetic energy to momentum: $K=\frac{p^2}{2m}⟹p=\sqrt{2mK}=\sqrt{2\times 9.11\times 10^{-31}\times 8\times 10^{-18}}\approx 3.82\times 10^{-24}\text{ kgcdotpm/s}$
3. Calculate de Broglie wavelength: $\lambda =\frac{6.626\times 10^{-34}}{3.82\times 10^{-24}}\approx 1.73\times 10^{-10}\text{ m}$, which matches typical atomic spacing in crystals, so diffraction will occur.

4. Atomic spectra and the Bohr model

When low-pressure gas is heated, it emits light only at discrete, element-specific wavelengths called an emission spectrum, rather than the continuous spectrum classical physics predicted. Classical models also predicted orbiting electrons would radiate energy and crash into the nucleus, making stable atoms impossible. Niels Bohr resolved these contradictions with a quantized model of the hydrogen atom (the only atom the model accurately describes):

1. Electrons orbit the nucleus in discrete stationary states with fixed energy levels, and do not radiate energy while in these states.
2. Electrons transition between levels by absorbing or emitting a photon with energy equal to the difference between the two levels: $hf=|E_{\text{high}}-E_{\text{low}}|$
3. Electron angular momentum is quantized: $L=n\frac{h}{2\pi }$, where $n$ is the principal quantum number ($n=1,2,\mathrm{3...}$) The energy levels of hydrogen are given by: $$E_n=\frac{-13.6\text{ eV}}{n^2}$$ The negative sign indicates the electron is bound to the nucleus; $n=1$ is the ground state (lowest energy, $E=-13.6$ eV), and $n=\infty$ corresponds to a free electron with 0 energy. Transitions to $n=1$ emit ultraviolet photons (Lyman series), transitions to $n=2$ emit visible photons (Balmer series), and transitions to $n=3$ emit infrared photons (Paschen series).

Worked Example: What wavelength of light is required to excite a hydrogen electron from the ground state to the $n=3$ excited state?

1. Calculate the energy difference: $\Delta E=E_3-E_1=\frac{-13.6}{9}-(-13.6)\approx -1.51+13.6=12.09\text{ eV}$
2. Calculate wavelength using the $hc$ shortcut: $\lambda =\frac{1240\text{ eVcdotpnm}}{12.09\text{ eV}}\approx 103\text{ nm}$, which falls in the ultraviolet range.

5. Nuclear decay — alpha, beta, gamma

The atomic nucleus consists of protons (atomic number $Z$, positive charge) and neutrons (neutron number $N$, neutral charge), with total mass number $A=Z+N$. Isotopes are atoms of the same element with equal $Z$ but different $N$. Unstable nuclei undergo spontaneous radioactive decay to reach a more stable state, emitting three common types of radiation:

1. Alpha decay: The nucleus emits an alpha particle (a helium nucleus, ${}_2^4\text{He}$), losing 2 protons and 2 neutrons. $Z$ decreases by 2, $A$ decreases by 4. Example: ${}_{92}^{238}\text{U}{\to }_{90}^{234}\text{Th}{+}_2^4\text{He}$. Alpha particles have low penetration, stopped by paper or skin.
2. Beta decay: Two subtypes:

• Beta-minus (${\beta }^{-}$): A neutron decays into a proton, emitting an electron (${}_{-1}^{0}e$) and antineutrino. $Z$ increases by 1, $A$ stays constant. Example: ${}_6^{14}\text{C}{\to }_7^{14}\text{N}{+}_{-1}^{0}e+{\bar{\nu }}_e$
• Beta-plus (${\beta }^{+}$): A proton decays into a neutron, emitting a positron (${}_{+1}^{0}e$) and neutrino. $Z$ decreases by 1, $A$ stays constant. Example: ${}_{11}^{22}\text{Na}{\to }_{10}^{22}\text{Ne}{+}_{+1}^{0}e+{\nu }_e$ Beta particles have medium penetration, stopped by aluminum foil.

3. Gamma decay: An excited nucleus drops to a lower energy state, emitting a high-energy gamma photon. No change to $Z$ or $A$, only energy is released. Gamma rays have high penetration, requiring lead or thick concrete to stop. Radioactive decay follows a half-life rule: $T_{1/2}$ is the time required for half of a sample of unstable nuclei to decay. The remaining number of nuclei at time $t$ is $N(t)=N_0{\left(\frac{1}{2}\right)}^{t/T_{1/2}}$.

Worked Example: Iodine-131, used in medical imaging, has a half-life of 8 days. If a patient receives a 100 MBq dose, how much radioactive iodine remains in their body after 32 days?

1. Calculate number of half-lives passed: $\frac{32}{8}=4$
2. Apply half-life formula: $N=100\times {\left(\frac{1}{2}\right)}^4=100/16=6.25\text{ MBq}$

6. Mass-energy equivalence $E=m{c}^2$

Einstein's special relativity showed that mass is a concentrated form of energy, related by the famous equation: $$E=m{c}^2$$ The mass defect $\Delta m$ of a nucleus is the difference between the total mass of its individual protons and neutrons, and the mass of the bound nucleus. This "missing" mass is converted into nuclear binding energy, the energy required to split the nucleus into its constituent nucleons: $E_b=\Delta m{c}^2$. For AP Physics 2, use the convenient conversion factor: $1\text{ atomic mass unit (u)}=931.5\text{ MeV}/c^2$, so binding energy can be calculated directly as $E_b(\text{MeV})=\Delta m(\text{u})\times 931.5\text{ MeV/u}$. Higher binding energy per nucleon corresponds to a more stable nucleus; fission of heavy nuclei and fusion of light nuclei both release energy because they produce products with higher binding energy per nucleon.

Worked Example: The mass of an iron-56 nucleus is 55.9207 u. Proton mass = 1.00728 u, neutron mass = 1.00866 u. Calculate the binding energy per nucleon for iron-56.

1. Iron-56 has 26 protons and 30 neutrons: total mass of free nucleons = $(26\times 1.00728)+(30\times 1.00866)=26.1893+30.2598=56.4491\text{ u}$
2. Mass defect: $\Delta m=56.4491-55.9207=0.5284\text{ u}$
3. Total binding energy: $0.5284\times 931.5\approx 492.2\text{ MeV}$
4. Binding energy per nucleon: $\frac{492.2}{56}\approx 8.79\text{ MeV/nucleon}$, one of the highest values for any element, making iron-56 extremely stable.

7. Common Pitfalls (and how to avoid them)

• Pitfall 1: Assuming higher light intensity will eject electrons below the threshold frequency for the photoelectric effect. Why it happens: Students confuse classical wave intensity (total energy per area) with quantum per-photon energy. Correct move: Intensity only controls the number of photons, not the energy per photon; only frequency determines if photons have enough energy to overcome the work function.
• Pitfall 2: Using $E=hf$ to calculate the energy of a moving electron. Why it happens: Students mix up photon energy rules with matter particle rules. Correct move: For massive particles, use $K=\frac{p^2}{2m}$ for kinetic energy and $\lambda =h/p$ for de Broglie wavelength; reserve $E=hf$ exclusively for photons.
• Pitfall 3: Forgetting hydrogen energy levels are negative, leading to incorrect transition energy calculations. Why it happens: Students treat $E_n$ as a positive value instead of a bound-state energy relative to a free electron. Correct move: Always use $\Delta E=|E_{\text{final}}-E_{\text{initial}}|$ for photon energy; negative $\Delta E$ means a photon is emitted, positive means a photon is absorbed.
• Pitfall 4: Mixing up the change in atomic number for beta-plus and beta-minus decay. Why it happens: Students can't remember which particle is emitted in each decay mode. Correct move: Beta-minus emits a negatively charged electron, so the nucleus gains +1 charge ($Z$ increases by 1); beta-plus emits a positively charged positron, so the nucleus loses +1 charge ($Z$ decreases by 1).
• Pitfall 5: Converting mass defect to energy using SI units, leading to arithmetic errors. Why it happens: Students forget the AP-provided conversion factor for atomic mass units. Correct move: Use $1\text{ u}=931.5{\text{ MeV/c}}^2$ to directly convert mass defect in u to binding energy in MeV without converting to kilograms or using $c=3\times 10^8\text{ m/s}$.

8. Practice Questions (AP Physics 2 Style)

Question 1 (Multiple Choice)

A metal has a work function of 2.9 eV. Which of the following light sources will eject electrons from the metal? A) 350 nm ultraviolet light B) 450 nm blue light C) 550 nm green light D) 650 nm red light

Solution: First calculate the maximum wavelength of light that can eject electrons: ${\lambda }_{\text{max}}=\frac{hc}{\Phi }=\frac{1240\text{ eVcdotpnm}}{2.9\text{ eV}}\approx 427\text{ nm}$. Only light with wavelength shorter than 427 nm has enough energy. Option A (350 nm) is the only valid choice. Answer: A

Question 2 (Free Response)

A hydrogen electron transitions from the $n=4$ excited state to the $n=2$ state. a) Calculate the energy of the emitted photon in eV. b) Calculate the wavelength of the photon in nm, and state what region of the electromagnetic spectrum it falls into.

Solution: a) $E_4=\frac{-13.6}{16}=-0.85\text{ eV}$, $E_2=\frac{-13.6}{4}=-3.4\text{ eV}$. $\Delta E=|E_2-E_4|=|-3.4-(-0.85)|=2.55\text{ eV}$ b) $\lambda =\frac{1240}{2.55}\approx 486\text{ nm}$, which falls in the visible (blue-green) range.

Question 3 (Free Response)

The half-life of carbon-14 is 5730 years. An archaeological sample has 1/32 of the original carbon-14 content of living tissue. a) How many half-lives have passed since the organism died? b) Calculate the age of the sample.

Solution: a) $\frac{1}{32}={\left(\frac{1}{2}\right)}^5$, so 5 half-lives have passed. b) Age = $5\times 5730=28650$ years.

9. Quick Reference Cheatsheet

10. What's Next

This modern physics content connects directly to cross-topic questions you will encounter on the AP Physics 2 exam, particularly with optics and thermodynamics. Wave interference principles from physical optics are used to explain electron diffraction, and binding energy calculations are often paired with thermodynamics questions about energy release from fission and fusion reactions. If you pursue higher level physics, this content is the foundation for AP Physics C quantum mechanics topics and university-level modern physics courses.

If you struggle with any of the concepts in this guide, from balancing decay equations to calculating photoelectric stopping potential, you can get personalized, 24/7 help from Ollie our AI tutor right on the homepage. You can also access full-length AP Physics 2 mock exams, targeted topic drills, and additional worked examples to reinforce your understanding and maximize your exam score.