15.3 Emission and Absorption Spectra

Photon Emission and Absorption

Energy Transfer via Photons

Photons serve as the carriers of electromagnetic energy in atomic processes, facilitating energy transfer when they interact with atoms. This interaction forms the foundation of spectroscopy and quantum mechanics.

• Atoms are modeled as systems consisting of a nucleus surrounded by electrons 🔄
• When a photon is absorbed, its energy is transferred to the atom, causing an electron to move to a higher energy state
• During emission, an atom releases a photon as an electron transitions to a lower energy state, converting internal energy to electromagnetic radiation

Energy Differences Between Atomic States

Atoms can only exist in specific discrete energy states, not in the continuous range of energies we observe in macroscopic objects. This quantization of energy is a fundamental principle of quantum mechanics.

• Energy can only be absorbed or emitted in specific amounts that exactly match the difference between two allowed atomic energy states
• When an atom in a lower energy state encounters a photon with energy equal to the gap between states, it can absorb the photon and jump to the higher energy state
• An excited atom (in a higher energy state) may spontaneously emit a photon and drop to a lower energy state, with the photon carrying away exactly the energy difference
• These energy differences reflect changes in the electron-nucleus interaction:
  • Higher energy states correspond to weaker electron-nucleus interactions (electron is less tightly bound)
  • Lower energy states involve stronger electron-nucleus interactions (electron is more tightly bound)

Atomic Transitions and Photon Energy

The energy of a photon is directly related to its frequency and wavelength through the equation $E = hf = \frac{hc}{\lambda}$, where $h$ is Planck's constant, $f$ is frequency, $c$ is the speed of light, and $\lambda$ is wavelength.

• When an atom transitions between two energy states with energy difference $\Delta E$, it absorbs or emits a photon with energy $E_{photon} = \Delta E = E_2 - E_1$
• The frequency of this photon is given by $f = \frac{\Delta E}{h}$
• The wavelength is given by $\lambda = \frac{hc}{\Delta E}$
• Larger energy transitions produce higher frequency (shorter wavelength) photons
• Smaller energy transitions produce lower frequency (longer wavelength) photons

Unique Atomic Spectra

Each element has a unique set of allowed energy levels due to its specific nuclear and electronic structure. This creates distinctive spectral "fingerprints" that can be used to identify elements.

• Emission spectra show bright lines at specific frequencies where atoms emit photons when transitioning from higher to lower energy states
  • These appear as bright colored lines against a dark background
  • Scientists use emission spectra to identify elements in distant stars and nebulae
  • The specific pattern of lines is unique to each element, like a fingerprint 🌈
• Absorption spectra show dark lines at frequencies where atoms absorb photons
  • These appear as dark lines in an otherwise continuous spectrum
  • When light passes through a gas, the gas atoms absorb specific frequencies
  • The resulting dark lines reveal the composition of the gas
  • Sunlight passing through the Sun's outer layers creates the solar absorption spectrum
• Energy level diagrams provide visual representations of:
  • The allowed energy states of an atom (horizontal lines)
  • The possible transitions between states (vertical arrows)
  • The energy of photons emitted or absorbed during transitions

Binding Energy and Ionization

Binding energy represents the energy that holds an electron to the nucleus. When enough energy is supplied to overcome this binding, ionization occurs, resulting in a free electron and a positive ion.

• The binding energy is the minimum energy required to completely remove an electron from an atom ⚡
• For atoms in the ground state (lowest energy level):
  • The electron is most tightly bound to the nucleus
  • Maximum energy is required for ionization
  • This maximum binding energy is called the ionization energy
• For atoms in excited states:
  • The electron is already partially "lifted" away from the nucleus
  • Less additional energy is needed to completely remove the electron
  • The binding energy decreases as the excitation level increases
• Each element has characteristic binding energies that depend on:
  • The nuclear charge (number of protons)
  • The electron configuration
  • The specific energy state of the electron

Practice Problem 1: Photon Energy in Transitions

Solution

First, we need to find the energy difference between the two states: $\Delta E = E_1 - E_3 = -13.6 \text{ eV} - (-1.51 \text{ eV}) = -12.09 \text{ eV}$

The negative sign indicates energy is being released. The photon energy equals this energy difference: $E_{photon} = |\Delta E| = 12.09 \text{ eV}$

Converting to joules: $E_{photon} = 12.09 \text{ eV} \times 1.602 \times 10^{-19} \text{ J/eV} = 1.94 \times 10^{-18} \text{ J}$

To find the frequency: $f = \frac{E_{photon}}{h} = \frac{1.94 \times 10^{-18} \text{ J}}{6.63 \times 10^{-34} \text{ J\cdot s}} = 2.93 \times 10^{15} \text{ Hz}$

To find the wavelength: $\lambda = \frac{c}{f} = \frac{3.00 \times 10^8 \text{ m/s}}{2.93 \times 10^{15} \text{ Hz}} = 1.02 \times 10^{-7} \text{ m} = 102 \text{ nm}$

This photon is in the ultraviolet region of the electromagnetic spectrum.

Practice Problem 2: Identifying Elements from Spectra

Solution

The first set of wavelengths (410 nm, 434 nm, 486 nm, and 656 nm) matches the Balmer series of hydrogen, which occurs when electrons transition from higher energy levels to the n=2 energy level.

The second set of wavelengths (589 nm, 615 nm, and 498 nm) does not match hydrogen's emission pattern. Since each element has a unique set of energy levels, it produces a unique emission spectrum - like a fingerprint.

The scientist can conclude that:

1. The unknown element is not hydrogen
2. The unknown element must be a different element with its own characteristic energy level structure
3. By comparing these wavelengths to known emission spectra, the scientist could identify the specific element (in this case, the 589 nm line is characteristic of sodium)

This demonstrates how emission spectroscopy allows scientists to identify elements in distant stars and other light sources without physically sampling them.

Practice Problem 3: Binding Energy and Ionization

Solution

For the 12.1 eV photon: Since the binding energy is 13.6 eV and the photon energy (12.1 eV) is less than this value, the electron cannot be completely removed from the atom. Instead, the electron will be excited to a higher energy level.

We can determine which energy level using the energy formula for hydrogen: $E_n = -\frac{13.6 \text{ eV}}{n^2}$

The ground state (n=1) has energy E₁ = -13.6 eV After absorbing 12.1 eV, the new energy is: -13.6 eV + 12.1 eV = -1.5 eV

Finding which energy level has approximately -1.5 eV: $-1.5 \text{ eV} \approx -\frac{13.6 \text{ eV}}{n^2}$ $n^2 \approx \frac{13.6 \text{ eV}}{1.5 \text{ eV}} \approx 9$ $n \approx 3$

So the electron transitions to the n=3 energy level.

For the 15.0 eV photon: Since the photon energy (15.0 eV) exceeds the binding energy (13.6 eV), the electron will be completely removed from the atom, causing ionization. The excess energy (15.0 eV - 13.6 eV = 1.4 eV) becomes kinetic energy of the freed electron.