3.11 Spectroscopy and the Electromagnetic Spectrum · 3.12 Properties of Photons

3.11 Spectroscopy and the Electromagnetic Spectrum · 3.12 Properties of Photons

A molecule can rotate, vibrate, or rearrange its electrons only by changing its energy. Spectroscopy studies how matter absorbs or emits electromagnetic radiation to reveal those energy changes. The sharp question is: Why does microwave radiation make a molecule rotate, while ultraviolet radiation can move an electron to a different energy level?

Learning Objective 3.11.A: Explain the relationship between a region of the electromagnetic spectrum and the types of molecular or electronic transitions associated with that region.

The electromagnetic spectrum is the continuous range of electromagnetic radiation, organized by wavelength, frequency, and energy. A spectral region is a portion of that range whose photons have similar energies and therefore interact with matter in characteristic ways.

The key relationship is:

$$ c = \lambda \nu $$

where $c$ is the speed of light, $\lambda$ is wavelength, and $\nu$ is frequency. Because the speed of light is constant in a vacuum, wavelength and frequency are inversely related: shorter-wavelength radiation has higher frequency.

Different spectral regions correspond to different internal motions because the energy gaps between those motions differ.

Spectral region	Matter’s response	Transition described
Microwave	Molecule changes its rotational state	Molecular rotational levels
Infrared	Bonds change their vibrational state	Molecular vibrational levels
Ultraviolet/visible	Electrons change energy levels	Electronic energy levels

Photons: energy delivered in discrete packets

A photon is a discrete packet, or quantum, of electromagnetic energy. Matter does not absorb an arbitrary amount of light; an atom or molecule absorbs a photon when the photon’s energy exactly matches an allowed energy difference.

Essential Knowledge 3.11.A.1: Differences in absorption or emission of photons in different spectral regions are related to different types of molecular motion or electronic transition.

For a transition between two energy states,

$$ \Delta E_{\text{matter}} = E_{\text{photon}} $$

The sign depends on the process. During absorption, matter gains energy, so $\Delta E_{\text{matter}} > 0$. During emission, matter loses energy and releases a photon.

Photon energy is related to frequency by

$$ E = h\nu $$

and to wavelength by

$$ E = \frac{hc}{\lambda} $$

where $h$ is Planck’s constant, $6.626 \times 10^{-34}\ \mathrm{J \cdot s}$, and $c$ is approximately $3.00 \times 10^8\ \mathrm{m \cdot s^{-1}}$.

These equations produce two exam-ready conclusions:

• Higher-frequency photons have greater energy.
• Shorter-wavelength photons have greater energy.

A common misconception is that higher frequency means longer wavelength because both describe “more intense” light. Frequency and wavelength are not directly proportional: when $\nu$ increases, $\lambda$ decreases because $c = \lambda \nu$.

Worked example: calculating a photon’s energy

A photon has wavelength $500\ \mathrm{nm}$. Calculate its energy in joules.

First convert nanometers to meters:

$$ 500\ \mathrm{nm} \left(\frac{1\ \mathrm{m}}{10^9\ \mathrm{nm}}\right) = 5.00 \times 10^{-7}\ \mathrm{m} $$

Then substitute into the wavelength equation:

$$ E = \frac{(6.626 \times 10^{-34}\ \mathrm{J \cdot s})(3.00 \times 10^8\ \mathrm{m \cdot s^{-1}})} {5.00 \times 10^{-7}\ \mathrm{m}} $$

$$ E = 3.98 \times 10^{-19}\ \mathrm{J\ photon^{-1}} $$

The units demonstrate why dimensional analysis matters: meters cancel, seconds cancel, and the result is joules per photon. Because the wavelength has three significant figures, the final answer should be reported as $3.98 \times 10^{-19}\ \mathrm{J\ photon^{-1}}$.

Connecting spectra to molecular behavior

Microwave absorption changes a molecule’s rotational level. Infrared absorption changes the vibrational motion of bonds—such as stretching or bending. Ultraviolet or visible absorption changes an electron’s energy, often moving it from a lower-energy orbital or electronic state to a higher-energy state.

When an excited electron returns to a lower allowed energy level, the atom or molecule emits a photon. The emitted photon’s energy equals the energy difference between the two states:

$$ E_{\text{photon}} = E_{\text{higher}} - E_{\text{lower}} $$

A larger energy gap produces a higher-frequency, shorter-wavelength photon. This is why electronic transitions generally involve ultraviolet or visible radiation, while rotational and vibrational transitions involve lower-energy regions.

Misconception check: absorption does not mean that a molecule is destroyed or ionized automatically. It means that the molecule takes in a photon whose energy matches an allowed transition. Ionization occurs only if the absorbed energy is sufficient to remove an electron completely.

AP skills in action

This topic explicitly develops Science Practice 4, Model Analysis, especially 4.A: Predict and/or explain chemical properties or phenomena using given chemical theories, models, and representations. A spectrum, energy-level diagram, or molecular-motion model can be used to predict which type of transition occurred.

It also develops Science Practice 5, Mathematical Routines, specifically 5.F: Calculate, estimate, or predict an unknown quantity from known quantities by selecting and following a logical computational pathway and attending to precision. A strong solution identifies the known quantity, converts units before substitution, chooses $E=h\nu$ or $E=hc/\lambda$, and reports an appropriately precise result.

Retrieval check: A molecule absorbs infrared radiation and later emits a photon with a shorter wavelength than the absorbed photon. Is the emitted photon more energetic or less energetic? The shorter wavelength means higher frequency and therefore greater photon energy; the molecule’s emitted transition has the larger energy difference.