AcademyMatter Waves
Academy
Atomic Spectra
Level 1 - Physics topic page in Matter Waves.
Principle
Atomic line spectra come from photons emitted or absorbed between discrete energy levels.
Notation
\(E_i,E_f\)
initial and final atomic energies
\(\mathrm{J}\)
\(\Delta E\)
energy difference
\(\mathrm{J}\)
\(f\)
photon frequency
\(\mathrm{Hz}\)
\(\lambda\)
photon wavelength
\(\mathrm{m}\)
\(h\)
Planck constant
\(\mathrm{J\,s}\)
\(c\)
speed of light
\(\mathrm{m\,s^{-1}}\)
Method
Derivation 1: Connect energy gaps to photons
An atom emits a photon when it moves from a higher energy level to a lower one.
Emission energy
\[hf=E_i-E_f\]
Photon wavelength
\[\lambda=\frac{c}{f}\]
Energy wavelength
\[E_i-E_f=\frac{hc}{\lambda}\]
Derivation 2: Absorb only allowed energies
Absorption occurs when the photon energy matches an available upward transition.
Absorption condition
\[hf=E_f-E_i\]
Derivation 3: Use spectra as fingerprints
Different atoms have different level spacings, so their line spectra identify the element.
Rules
Photon energy
\[E_\gamma=hf=\frac{hc}{\lambda}\]
Emission
\[E_\gamma=E_i-E_f\]
Absorption
\[E_\gamma=E_f-E_i\]
Electronvolt
\[1\,\mathrm{eV}=1.60\times10^{-19}\,\mathrm J\]
Examples
Question
A transition releases
\[3.00\,\mathrm{eV}\]
Find the photon wavelength using \[hc=1240\,\mathrm{eV\,nm}\]
Answer
\[\lambda=\frac{hc}{E}=\frac{1240}{3.00}=413\,\mathrm{nm}\]
Checks
- Emission: atom loses energy, photon carries the gap.
- Absorption: atom gains energy, photon must match an allowed gap.
- Shorter wavelength means larger photon energy.
- Line spectra imply discrete atomic energies.