E1-2. Emission and Absorption Spectra and Photon Energy

1. Continuous vs. Line Spectra

When light from a source passes through a dispersive element (like a diffraction grating or prism), it breaks apart into its component wavelengths, creating a spectrum:

  • Continuous Spectrum: Produced by hot, dense solids, liquids, or high-pressure gases (such as a lightbulb filament or the core of a star). Thermal acceleration of dense particles creates an unbroken rainbow spanning all possible wavelengths.
  • Discrete Line Spectrum: Characterized by sharp, isolated colored lines at very specific wavelengths, separated by wide gaps of dark empty space. These are produced by low-pressure, excited atomic gases. They act as unique chemical fingerprints because each element produces a completely unique array of spectral lines.
Absorption from Ground State E₄ (-1.5 eV) E₃ (-4.2 eV) E₂ (-7.5 eV) E₁ (-12.0 eV) Incoming Photons ΔE = 4.5 eV ΔE = 7.8 eV ΔE = 10.5 eV Not enough energy

2. Physical Mechanism of Gas Spectra

The existence of sharp, discrete lines implies that atomic electrons cannot take on any arbitrary energy value; their energies are strictly quantized into specific bound levels ($E_1, E_2, E_3...$). This leads to two separate phenomena:

I. Emission Spectra (Bright Lines on a Dark Background)

  • When an atomic gas is energized (via electrical discharge or heat), bound electrons absorb kinetic energy and are kicked up to high, unstable orbital states (excitation).
  • To return to a more stable state, the electron falls back down to a lower energy level. In doing so, it sheds its excess energy by emitting a single photon.
  • Because the levels are fixed, the emitted energy difference is exact, producing a single, bright colored line matching that specific wavelength.

II. Absorption Spectra (Dark Lines on a Continuous Background)

  • If a continuous white light beam passes through a cold, low-pressure gas, the gas electrons will selectively absorb incoming photons whose energies match the precise differences between the electron's current state and a higher empty level.
  • Photons with non-matching energies pass straight through unaffected. The absorbed photons are later re-emitted in random, isotropic directions, massively reducing the forward intensity.
  • This leaves distinct dark absorption lines blanking out the continuous background at the exact same wavelengths where emission lines would normally appear.

3. Fundamental Energy-Photon Equations

The energy carried by an emitted or absorbed photon is bound by the Planck-Einstein relation. When an electron transitions between an initial energy level $E_{\text{initial}}$ and a final state $E_{\text{final}}$, the structural energy change is:

$$\Delta E = |E_{\text{initial}} - E_{\text{final}}| = hf = \frac{hc}{\lambda}$$

Where the key physical parameters are defined as:

Symbol Physical Quantity Description Standard SI Units / Equivalent Constants
$\Delta E$ Energy difference of the electronic quantum transition Joules ($\text{J}$) or Electronvolts ($\text{1 eV} = 1.60 \times 10^{-19}\text{ J}$)
$h$ Planck's Constant $6.63 \times 10^{-34}\text{ J s}$ or $4.14 \times 10^{-15}\text{ eV s}$
$f$ Frequency of the wave radiation Hertz ($\text{Hz}$ or $\text{s}^{-1}$)
$c$ Speed of light in a vacuum $3.00 \times 10^8\text{ m s}^{-1}$
$\lambda$ Wavelength of the wave radiation Meters ($\text{m}$) or Nanometers ($\text{nm}$)

4. Examples

Example 1: A Four-Level Quantum System

Problem: A particular atom possesses four discrete bound energy levels: Ground State $E_1 = -12.0\text{ eV}$, $E_2 = -7.5\text{ eV}$, $E_3 = -4.2\text{ eV}$, and $E_4 = -1.5\text{ eV}$.
(a) Sketch an energy level diagram and determine the total number of distinct emission lines possible.
(b) Determine which transition produces the photon with the shortest wavelength, and calculate its value in nanometers.

Energy Level Diagram E₄ -1.5 eV E₃ -4.2 eV E₂ -7.5 eV E₁ -12.0 eV (Ground State) Max ΔE = Min λ 6 Distinct Emission Lines Possible

Solution:

Part (a): The number of distinct emission lines corresponds to all unique downward transitions from higher levels to lower levels. We can compute this combination using: $$\text{Transitions} = \frac{n(n-1)}{2} = \frac{4(3)}{2} = \mathbf{6 \text{ distinct lines}}$$ These transitions are: $(4 \rightarrow 3)$, $(4 \rightarrow 2)$, $(4 \rightarrow 1)$, $(3 \rightarrow 2)$, $(3 \rightarrow 1)$, and $(2 \rightarrow 1)$.

Part (b): The shortest wavelength ($\lambda_{\text{min}}$) corresponds directly to the maximum possible energy change ($\Delta E_{\text{max}}$). Looking at the values, the largest gap is the transition from the highest state to the lowest state ($E_4 \rightarrow E_1$): $$\Delta E_{\text{max}} = E_4 - E_1 = (-1.5\text{ eV}) - (-12.0\text{ eV}) = 10.5\text{ eV}$$ Convert this energy from electronvolts to Joules to use SI units: $$\Delta E = 10.5 \times (1.60 \times 10^{-19}\text{ J}) = 1.68 \times 10^{-18}\text{ J}$$ Now, rearrange the wave equation to solve for wavelength $\lambda$: $$\Delta E = \frac{hc}{\lambda} \implies \lambda = \frac{hc}{\Delta E}$$ $$\begin{aligned} \lambda &= \frac{(6.63 \times 10^{-34}\text{ J s}) \times (3.00 \times 10^8\text{ m s}^{-1})}{1.68 \times 10^{-18}\text{ J}} \\ &= \frac{1.989 \times 10^{-25}}{1.68 \times 10^{-18}} = 1.1839 \times 10^{-7}\text{ m} = \mathbf{118.4\text{ nm}} \end{aligned} $$ Note: This wavelength falls into the ultraviolet (UV) region of the electromagnetic spectrum.

Example 2: Analyzing Absorption Transitions in a Gas Sample

Problem: A low-temperature gas of the same element described in Example 1 is initially in its ground state ($E_1 = -12.0\text{ eV}$). A continuous white light beam containing photons with energies ranging continuously from $2.0\text{ eV}$ up to $9.0\text{ eV}$ is shone through the gas. Identify which photon energies will be missing from the transmission beam (absorbed), and compute the frequency of the lower-frequency absorbed line.

Solution:

Step 1: Check available upward transitions from the ground state ($E_1$): Because the gas is cool, all electrons reside at the ground state $E_1 = -12.0\text{ eV}$. They can only absorb photons whose energies match an upward leap from $E_1$:

  • To level 2: $\Delta E = E_2 - E_1 = -7.5 - (-12.0) = 4.5\text{ eV}$
  • To level 3: $\Delta E = E_3 - E_1 = -4.2 - (-12.0) = 7.8\text{ eV}$
  • To level 4: $\Delta E = E_4 - E_1 = -1.5 - (-12.0) = 10.5\text{ eV}$

Step 2: Match transitions against the source beam energy window ($2.0\text{ eV}$ to $9.0\text{ eV}$): The source beam only contains photons up to $9.0\text{ eV}$. Therefore, the transition to level 4 ($10.5\text{ eV}$) cannot occur because the beam lacks photons with enough energy. The other two transitions fall within the beam's energy range.
Thus, the two photon energies absorbed from the beam are exactly $4.5\text{ eV}$ and $7.8\text{ eV}$.

Step 3: Calculate the frequency of the lower-frequency absorbed line: The lower frequency line corresponds to the smaller energy step ($\Delta E = 4.5\text{ eV}$): $$\Delta E = 4.5 \times (1.60 \times 10^{-19}\text{ J}) = 7.20 \times 10^{-19}\text{ J}$$ Apply the relation $\Delta E = hf$: $$f = \frac{\Delta E}{h} = \frac{7.20 \times 10^{-19}\text{ J}}{6.63 \times 10^{-34}\text{ J s}} = \mathbf{1.09 \times 10^{15}\text{ Hz}}$$