Chemistry: Atomic Structure - Electromagnetic Spectrum & Quantum Theory

The electromagnetic spectrum encompasses a vast range of radiant energy.
Visible light constitutes only a tiny fraction of this spectrum.
The spectrum ranges from ultrashort-wavelength, high-frequency gamma ( $\gamma$ ) rays to long-wavelength, low-frequency radio waves.
Frequency increases from right to left across the typical spectrum representation, while wavelength increases from left to right.

According to Maxwell's theory, electromagnetic radiation propagates through space as waves.
These waves consist of two mutually perpendicular components: an oscillating electric field and an oscillating magnetic field.
Maxwell's derived equations, based on this oscillating-wave model, accurately describe nearly all observed properties of radiant energy.

Every wave of electromagnetic radiation possesses characteristic properties:
- Wavelength ( $\lambda$ ): The distance from one wave crest to the next. For example, Wave A having twice the wavelength of Wave B indicates a longer distance between its crests.
- Frequency ( $\nu$ ): The number of crests that pass a stationary point in space every second.
Units of Frequency:
- Waves per second
- Cycles per second ( $\text{s}^{-1}$ )
- Hertz ( $\text{Hz}$ ), where $1 \text{ Hz} = 1 \text{ s}^{-1}$ .
The speed of light ( $c$ ), frequency ( $\nu$ ), and wavelength ( $\lambda$ ) are related by the equation: $\nu = c/\lambda$

Problem: What is the frequency of infrared light that the James Webb Space Telescope (JWST) collects, given a wavelength of $28.5 \; \mu\text{m}$ ?
Collect and Organize:
- Given: Wavelength ( $\lambda$ ) = $28.5 \; \mu\text{m}$
- To find: Frequency ( $\nu$ )
- Relationship: $\nu = c/\lambda$
- Constant: Speed of light ( $c$ ) = $2.998 \times 10^8 \; \text{m/s}$
Analyze:
- The speed of light ( $c$ ) is in meters per second, but the given wavelength is in micrometers ( $\mu\text{m}$ ).
- Conversion factor: $1 \; \mu\text{m} = 10^{-6} \; \text{m}$ must be used to convert the wavelength to meters.
Solve:
- Convert wavelength: $28.5 \; \mu\text{m} \times \frac{10^{-6} \; \text{m}}{1 \; \mu\text{m}} = 28.5 \times 10^{-6} \; \text{m}$
- Substitute into the equation:
  $\nu = \frac{c}{\lambda} = \frac{2.998 \times 10^8 \; \text{m/s}}{28.5 \times 10^{-6} \; \text{m}} = 1.05 \times 10^{13} \; \text{s}^{-1}$
- Result: The frequency is $1.05 \times 10^{13} \; \text{s}^{-1}$ .

At very high temperatures, the atoms of each unique element emit a characteristic spectrum.
These emission spectra (e.g., hydrogen, helium, neon shown in Figure 3.13) consist of distinct lines of color, rather than a continuous rainbow.
This observation was key to understanding the quantized nature of atomic energy levels.

Kirchhoff's group initially studied the continuous emission spectra produced by hot, incandescent objects, such as heated metals (Figure 3.13).
Max Planck (1900), commissioned to improve incandescent lightbulb efficiency, made a revolutionary assumption about radiant energy:
- Radiant energy is never truly continuous.
- Instead, objects emit electromagnetic radiation only in integral multiples of an elementary unit of energy, known as a quantum.
Planck's equation defining the energy of a quantum: $E = h\nu$
- $E$ = energy (in Joules, J)
- $h$ = Planck's constant
- $\nu$ = frequency (in Hz or s $^{-1}$ )

Quantization can be visualized by comparing a set of stairs to a ramp (Figure 3.16).
- Stairs exemplify quantization: each step rises by a discrete, fixed height to the next step. You can only be at specific heights.
- A ramp exemplifies continuous variation: heights are not quantized; one can be at any height along the ramp, with infinitesimal changes possible.

The energy of a photon (light particle) can be calculated using two related formulas:
1. Based on frequency: $E = h\nu$
2. Based on wavelength: $E = \frac{hc}{\lambda}$
Constants:
- Planck's constant ( $h$ ) = $6.626 \times 10^{-34} \; \text{J}\cdot\text{s}$
- Speed of light ( $c$ ) = $3.0 \times 10^8 \; \text{m/s}$ (approximately, more precise $2.998 \times 10^8 \; \text{m/s}$ used in calculations)

Problem: Figure 3.18 shows emission intensity vs. wavelength for three quantum dots of different sizes.
- a. What is the energy of one photon of light from the largest quantum dot (wavelength $640 \; \text{nm}$ )? What is the energy of 1 mole of these photons?
- b. Is that energy greater than or less than the energies of photons produced by smaller quantum dots?
Solve (a):
- Convert wavelength: $640 \; \text{nm} \times \frac{10^{-9} \; \text{m}}{1 \; \text{nm}} = 6.40 \times 10^{-7} \; \text{m}$
- Energy of one photon:
  $E = \frac{hc}{\lambda} = \frac{(6.626 \times 10^{-34} \; \text{J}\cdot\text{s})(2.998 \times 10^8 \; \text{m/s})}{6.40 \times 10^{-7} \; \text{m}} = 3.10 \times 10^{-19} \; \text{J/photon}$
- Energy of 1 mole of photons (using Avogadro's number, $N<em>A = 6.022 \times 10^{23} \; \text{photons/mol}$ ): $E</em>{\text{mol}} = (3.10 \times 10^{-19} \; \text{J/photon}) \times (6.022 \times 10^{23} \; \text{photons/mol}) = 1.87 \times 10^5 \; \text{J/mol}$ (approximately $1.9 \times 10^5 \; \text{J/mol}$ )
Solve (b):
- Figure 3.18 indicates that smaller quantum dots produce photons with shorter wavelengths.
- According to the equation $E = hc/\lambda$ , energy ( $E$ ) is inversely proportional to wavelength ( $\lambda$ ).
- Therefore, these shorter-wavelength photons from smaller quantum dots possess more energy than the photon emitted at $640 \; \text{nm}$ .

While Planck's quantum model effectively explained emission spectra of hot objects, experimental evidence for the existence of quanta was lacking in 1900.
In 1905, Albert Einstein provided this crucial experimental evidence through his studies of the photoelectric effect.
The photoelectric effect is the phenomenon where electrons are emitted from metals when they are illuminated by and absorb electromagnetic radiation.
The emitted electrons are called photoelectrons (from Greek photo-, meaning