AP Physics 2 study materials
AP Physics 2 study resources

AP Physics 2: Modern Physics

AP Physics 2 · Work through your exam, one topic at a time.

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Modern Physics

Core concepts

Modern physics covers quantization, the photoelectric effect, atomic energy levels, and nuclear processes. Photon energy is $E=hf=hc/\lambda$, with $h\approx 6.63\times10^{-34},\mathrm{J,s}$ and $hc\approx1240,\mathrm{eV,nm}$. In the photoelectric effect, $K_{max}=hf-\phi$, where $\phi$ is the work function; no emission occurs below the threshold frequency $f_0=\phi/h$. Atomic spectra arise from transitions $\Delta E=hf$ between discrete levels. De Broglie wavelength is $\lambda=h/p$. Nuclear decay types: alpha (emits $^4_2$He, $Z$ drops 2, $A$ drops 4), beta-minus (neutron→proton, $Z$ rises 1), and gamma (photon, no change in $Z$ or $A$). Mass-energy equivalence: $E=mc^2$; binding energy reflects mass defect.

Traps to watch

  • Intensity changes the number of photoelectrons, not their maximum kinetic energy; only frequency changes $K_{max}$.
  • Converting between eV and joules ($1,\mathrm{eV}=1.6\times10^{-19},\mathrm{J}$) is a frequent error source.
  • In beta decay, the atomic number changes but the mass number does not.
  • Emission lines come from electrons dropping to lower levels; absorption lines from rising.
  • Heavier particles have shorter de Broglie wavelengths at the same speed.

Worked example

Light of $\lambda=200,\mathrm{nm}$ strikes a metal with $\phi=2.0,\mathrm{eV}$. Photon energy: $1240/200=6.2,\mathrm{eV}$. Then $K_{max}=6.2-2.0=4.2,\mathrm{eV}$.

Rule of thumb: use $E(\mathrm{eV})=1240/\lambda(\mathrm{nm})$ to skip joule conversions in photon problems.