Bucknell Physics 212 Supplement

Section 4.5 Absorption, Emission, and Spectroscopy

Okay, we are now ready to address some questions that you might have asked when you were a kid: “Why is a tomato red?” or “How do glow-in-the-dark shirts work?” or “Why is it that black-light illumination (which looks kinda deep violet in color) cause certain pigments to glow orange” or “Why do certain laundry detergents make my clothes look so unnaturally bright white?” These questions all fall under the category of the topic of photophysics or photochemistry, i.e., the physics and chemistry of how light interacts with matter. In addition to providing answers to these questions, this subject also relates to the important topic of spectroscopy which is a critical diagnostic technique that spans all fields of science and engineering.

There are a few different ways in which light (and any electromagnetic radiation) can interact with matter, but two important processes are absorption and emission of individual photons of light. Figure 4.14 shows these two processes. If a photon strikes a material, it can be absorbed by an electron, resulting in an increase in the energy of that electron from an initial (lower) value \(E_1\) to a higher value \(E_2\text{,}\) as shown in Figure 4.14a. This process must conserve energy, which means that the energy gained by the electron must be the same as the energy of the absorbed photon:

Note that the upper energy level state must initially be unoccupied for this process to work. As we'll see in Chapter 6, two electrons cannot occupy the same quantum state.

The process of emission is the same thing in reverse (Figure 4.14b). An electron in a higher energy state can drop to a lower-energy state (if initially unoccupied), releasing a single photon with an energy given by (4.20).

These simple quantum principles now enable us to resolve the third of the great failures of classical physics — the observation in the early 1900s that when energy is pumped into atoms, they radiate EM waves only at certain distinct wavelengths; i.e., the problem of discrete atomic spectral lines discussed in Chapter 2. The explanation is actually quite simple: each element or molecule has its own discrete, quantized energy level structure (e.g., Figure 4.15). An electron in an atom can make transitions between the energy levels. Any transition from a higher to a lower energy level will result in the emission of a photon. But since there are only certain, well-defined, quantized energy levels for the electron, then there are only certain well-defined differences \(\Delta E\) allowed for the transitions between these levels. And since the photons emitted by these transitions have energy \(E_\text{ ph } = |\Delta E|\text{,}\) that means that the emitted photons have only certain, well-defined energies. And since the energy of the photon is related to the frequency and wavelength of the electromagnetic radiation via \(E_\text{ ph } = hf = hc/\lambda\text{,}\) that means that the light emitted by atoms has only certain, well-defined, discrete wavelengths.

For a molecule with the four-level system in Figure 4.15, assume that \(E_0 = -11.5\Xunits{eV}\text{,}\) \(E_1=-6.2\Xunits{eV}\text{,}\) \(E_2 = -5.4\Xunits{eV}\text{,}\) and \(E_3 = -2.5\Xunits{eV}\text{.}\) (a) Determine how many different wavelengths of EM radiation can be emitted from this molecule, and calculate (b) the largest and (c) the smallest wavelengths of EM radiation that can be emitted from this molecule.

Solution.

(a) There are several possible transitions that an electron can make that will emit a photon in this system. An electron can drop from energy level 3 down to level 0, 1 or 2; it can drop from energy level 2 down to level 0 or 1; and it can drop from energy level 1 down to level 0. Altogether, there are 6 possible downward energy transitions, so there are 6 possible wavelengths of light that can be emitted.

For parts (b) and (c), for any emitted photon, the energy \(E_text{ph} = |\Delta E|\) and since \(E_\text{ ph } = hc/\lambda\text{,}\) it follows that \(hc/\lambda = |\Delta E|\) and \(\lambda = hc/\Delta E\text{.}\) So, the largest and smallest wavelengths correspond to the smallest and largest energy differences \(\Delta E\text{.}\) (b) The smallest energy difference \(\Delta E\) given the allowed electron energies is the energy difference between levels 2 and 1. So, the largest emitted wavelength is:

\begin{equation} \lambda = \frac{hc}{\Delta E} = \frac{1240 \Xunits{eV \cdot nm}}{|-5.4 \Xunits{eV} - (-6.2 \Xunits{eV})|} = \frac{1240 \Xunits{eV \cdot nm}}{0.8 \Xunits{eV}} = 1550 \Xunits{nm}\text{,}\tag{4.21} \end{equation}

where we have used the result \(hc = 1240\Xunits{eV \cdot nm}\text{,}\) which is convenient for spectra problems where the energy levels and differences are expressed in electron volts and the wavelengths of emitted and absorbed light are expressed in nanometers. We could have converted everything into SI units and used typical values for \(h\) and \(c\text{,}\) but that is more time-consuming.

Note that the longest wavelength emitted is a value beyond the visible spectrum (this is in the infrared spectrum), so the human eye wouldn't be able to see this emitted light. (c) For the shortest wavelength for emitted EM radiation, we want the largest \(\Delta E\) possible for these four energy levels, which corresponds to a transition from energy level 3 down to energy level 0.

\begin{equation} \lambda = \frac{hc}{\Delta E} = \frac{1240 \Xunits{eV \cdot nm}}{|-2.5 \Xunits{eV} - (-11.5 \Xunits{eV})|} = \frac{1240 \Xunits{eV \cdot nm}}{9.0 \Xunits{eV}} = 138 \Xunits{nm}\tag{4.22} \end{equation}

This wavelength is also outside the visible spectrum (it's ultraviolet, in this case).

Everything in the previous example works the other way for absorption. But instead of an electron dropping to a lower energy level and emitting a photon, a photon is absorbed and the electron jumps up to a higher energy level. But this only works if the incoming photon has the right energy. From a practical perspective, if you shine white light through a gas of a particle element or molecule, all the light will pass through except for light with particular wavelengths and energies corresponding to the energy differences \(\Delta E\) for the sample.

Since each element and each molecule has its own distinctive energy-level structure, the spectrum of EM radiation emitted is unique to each, as is the spectrum of radiation absorbed. This “fingerprint” is often used to identify materials in systems as diverse as a microscopic sample of DNA¹ up to astronomical-scale objects such as stars and planets; large, nebular clouds that are collapsing to form new stars and planetary systems; and accretion disks of gas spiraling into supermassive black holes at the centers of “active” galaxies.²

So, why is a tomato red? Answer: because it contains a pigment called lycopene that has electronic energy levels with an energy difference corresponding to the energy of red photons. When an electron in an excited state in lycopene drops to a lower level, it emits red photons. (Lycopene is actually a fluorescent pigment; fluorescence is discussed in the next section.)

Finally, absorption and emission also work for solids with band structures similar to that in Figure 4.13. There are many, closely-spaced energy levels within the bands, allowing for EM radiation with very large wavelengths to be emitted and absorbed. But it is also possible for electrons to make transitions between the bands, resulting in larger energy differences and smaller wavelengths for emitted or absorbed light.