Notes
Quantum ideas
- Light comes in packets called photons, each with energy $\displaystyle E = hf = \frac{hc}{\lambda}$, where $h = 6.63 \times 10^{-34}$ J·s. Higher frequency (bluer) photons have more energy.
- Photoelectric effect: light shining on metal ejects electrons only if each photon has enough energy. $K_{max} = hf - \phi$, where $\phi$ is the work function. Brighter light (more photons) ejects more electrons but doesn't raise their maximum energy. This is evidence that light is a particle.
- Wave-particle duality: matter also has a wavelength, $\displaystyle \lambda = \frac{h}{p}$ (de Broglie). Electron diffraction shows electrons act like waves.
Atoms and spectra
- Electrons in atoms have discrete energy levels. Dropping from a higher to a lower level emits a photon with $E = E_{high} - E_{low}$; absorbing a photon of exactly that energy raises the electron.
- This gives each element a unique emission or absorption spectrum.
Nuclear physics
- The nucleus has protons and neutrons held by the strong nuclear force. Isotopes have the same protons but different neutrons.
- Alpha decay emits $^4_2\text{He}$ (mass number −4, atomic number −2). Beta-minus decay turns a neutron into a proton and emits an electron (atomic number +1). Gamma decay emits a high-energy photon (no change in particles).
- Nucleon number and charge are conserved in nuclear reactions.
- Half-life: the time for half of a radioactive sample to decay. After $n$ half-lives, $\displaystyle \left(\frac{1}{2}\right)^n$ remains.
- Mass-energy equivalence: $E = mc^2$. In fission and fusion, the products have slightly less mass; the mass defect becomes energy.
Practice
Higher frequency means higher energy.
More photons eject more electrons, but each has the same energy.
$K_{max} = 5.0 - 3.0$.
$-1.5 - (-3.4) = 1.9$ eV.
An alpha particle has 2 protons.
Mass number $238 - 4 = 234$; atomic number $92 - 2 = 90$.
Three half-lives.
Diffraction is a wave behavior.
Free response
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Free-response question
Light of wavelength 300 nm shines on a metal with work function 2.3 eV. Use $hc = 1240$ eV·nm.
(a) Calculate the photon energy in eV.
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$\displaystyle E = \frac{1240}{300} \approx 4.1$ eV.
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(b) Calculate the maximum kinetic energy of ejected electrons.
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$K_{max} = 4.1 - 2.3 = 1.8$ eV.
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(c) The wavelength is changed to 600 nm. Will electrons be ejected? Explain.
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No. $\displaystyle E = \frac{1240}{600} \approx 2.1$ eV, which is less than the 2.3 eV work function, so no electrons are ejected no matter how bright the light is.
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