Light & Atomic Energy: Calculating Wavelength, Frequency, and Photon Energy

Understanding light is the key to unlocking the structure of the atom. Whether you are analyzing the crimson red of a strontium firework or calculating the electron energy levels in a hydrogen atom, the math remains the same. At the quantum level, light behaves as both a wave and a particle, and mastering the relationship between wavelength, frequency, and photon energy is essential for chemistry exams.

This guide breaks down the core equations—Planck’s Law, the Rydberg formula, and de Broglie wavelength—with step-by-step solutions to common exam problems.

Wave Properties of Light: Wavelength, Frequency, and Speed

Before diving into energy, we must define the relationship between the physical dimensions of a light wave. Electromagnetic radiation travels at a constant speed in a vacuum.

The fundamental equation connecting these variables is:

Practice Problem: Converting Wavelength to Frequency

Scenario: UV radiation from the sun is intense in the 320–400 nm range. If a specific UV ray has a wavelength of 360 nm, what is its frequency?

Step-by-Step Solution:

1. Convert units to meters. The speed of light is in meters per second, so wavelength must match.
   $360 \text{ nm} = 360 \times 10^{-9} \text{ m} = 3.60 \times 10^{-7} \text{ m}$.
2. Rearrange the equation for frequency ($\nu$).
   $\nu = \frac{c}{\lambda}$
3. Calculate.
   $$ \nu = \frac{3.00 \times 10^8 \text{ m/s}}{3.60 \times 10^{-7} \text{ m}} = 8.33 \times 10^{14} \text{ s}^{-1} $$

The frequency is $8.33 \times 10^{14} \text{ Hz}$.

The Photoelectric Effect: Planck’s Equation and Photon Energy

Classical physics assumed energy was continuous, but Max Planck discovered that energy is quantized. Light travels in discrete packets called photons. The energy of a single photon is directly proportional to its frequency.

The equation for photon energy is:

Where $h$ is Planck’s Constant: $6.626 \times 10^{-34} \text{ J}\cdot\text{s}$.

Illustrative Example: From Energy to Frequency

Problem: When sodium is heated, it emits a distinct yellow light (the "sodium doublet"). If the energy of a single photon in this light is $3.37 \times 10^{-19} \text{ J}$, what is its frequency?

Solution:

1. Identify knowns: $E = 3.37 \times 10^{-19} \text{ J}$ and $h = 6.626 \times 10^{-34} \text{ J}\cdot\text{s}$.
2. Rearrange Planck’s equation: $\nu = \frac{E}{h}$.
3. Solve: $$ \nu = \frac{3.37 \times 10^{-19} \text{ J}}{6.626 \times 10^{-34} \text{ J}\cdot\text{s}} = 5.09 \times 10^{14} \text{ s}^{-1} $$

The frequency is $5.09 \times 10^{14} \text{ s}^{-1}$.

Calculating Energy, Frequency, or Wavelength from One Variable

In most exams, you won't be given the frequency directly. You will be given a wavelength (in nanometers) and asked for energy, or vice versa. By combining the wave equation and Planck's equation, we get the most useful formula in this topic:

Case Study 1: Calculating Wavelength from Energy

Question: If the energy per photon of a certain EM radiation is $2.0 \times 10^{-19} \text{ J}$, what is its wavelength in nanometers?

Analysis:

• Rearrange the combined formula: $\lambda = \frac{hc}{E}$.
• Calculate $hc$ (a useful constant to memorize): $(6.626 \times 10^{-34})(3.00 \times 10^8) \approx 1.988 \times 10^{-25} \text{ J}\cdot\text{m}$.
• Divide by Energy: $$ \lambda = \frac{1.988 \times 10^{-25} \text{ J}\cdot\text{m}}{2.0 \times 10^{-19} \text{ J}} \approx 9.9 \times 10^{-7} \text{ m} $$
• Convert to nanometers ($1 \text{ m} = 10^9 \text{ nm}$): $$ 9.9 \times 10^{-7} \text{ m} \times 10^9 = 990 \text{ nm} $$

The answer is $9.9 \times 10^2 \text{ nm}$.

Case Study 2: Molar Energy Calculation (High Frequency Exam Question)

Question: Calculate the energy, in kJ/mol, of a mole of photons with a wavelength of 634 nm.

Strategy: This is a two-step process. First, find the energy of one photon. Second, multiply by Avogadro's number ($N_A = 6.022 \times 10^{23}$) to find the energy of a mole.

1. Convert $\lambda$: $634 \text{ nm} = 6.34 \times 10^{-7} \text{ m}$.
2. Energy of one photon: $$ E_{photon} = \frac{hc}{\lambda} = \frac{(6.626 \times 10^{-34})(2.998 \times 10^8)}{6.34 \times 10^{-7}} \approx 3.13 \times 10^{-19} \text{ J} $$
3. Energy of a mole ($E_{mol}$): $$ E_{mol} = (3.13 \times 10^{-19} \text{ J}) \times (6.022 \times 10^{23} \text{ mol}^{-1}) $$ $$ E_{mol} = 1.887 \times 10^5 \text{ J/mol} $$
4. Convert to kJ: Divide by 1000. Result: 189 kJ/mol.

Atomic Spectra: Bohr Model and the Rydberg Formula

The Bohr model explains that electrons exist in fixed energy levels ($n$). When an electron drops from a higher level ($n_i$) to a lower level ($n_f$), it emits a photon with energy equal to the difference between those levels.

The energy change is calculated using the Rydberg formula logic:

Where $R_H$ (Rydberg constant for Hydrogen energy) is $2.18 \times 10^{-18} \text{ J}$.

Practice Problem: Electron Transition

Question: What is the wavelength of the photon emitted when an electron in a hydrogen atom drops from $n=7$ to $n=4$?

Step-by-Step:

1. Calculate the fractions:
   • $n_f = 4 \rightarrow n_f^2 = 16$
   • $n_i = 7 \rightarrow n_i^2 = 49$
2. Calculate $\Delta E$: $$ \Delta E = 2.18 \times 10^{-18} \text{ J} \left( \frac{1}{16} - \frac{1}{49} \right) $$ $$ \Delta E = 2.18 \times 10^{-18} \text{ J} (0.0625 - 0.0204) $$ $$ \Delta E \approx 9.18 \times 10^{-20} \text{ J} $$
3. Convert Energy to Wavelength: Use $\lambda = \frac{hc}{E}$. $$ \lambda = \frac{1.986 \times 10^{-25} \text{ J}\cdot\text{m}}{9.18 \times 10^{-20} \text{ J}} \approx 2.16 \times 10^{-6} \text{ m} $$

The wavelength is $2.16 \times 10^{-6} \text{ m}$ (or 2160 nm, which is in the infrared spectrum).

Wave-Particle Duality: Calculating the de Broglie Wavelength

Louis de Broglie proposed that if light waves can behave like particles (photons), then matter particles (like electrons) can behave like waves. This is known as Wave-Particle Duality.

The wavelength of a moving particle is calculated as:

Where $m$ is mass in kg and $v$ is velocity in m/s.

Example: Wavelength of an Electron

Problem: Calculate the de Broglie wavelength of an electron with mass $9.11 \times 10^{-28} \text{ g}$ moving at a speed of $5.27 \times 10^6 \text{ m/s}$.

Solution:

1. Convert Mass:
   $9.11 \times 10^{-28} \text{ g} = 9.11 \times 10^{-31} \text{ kg}$.
2. Calculate Momentum ($p = mv$):
   $p = (9.11 \times 10^{-31} \text{ kg})(5.27 \times 10^6 \text{ m/s}) \approx 4.80 \times 10^{-24} \text{ kg}\cdot\text{m/s}$.
3. Apply Equation:
   $$ \lambda = \frac{6.626 \times 10^{-34} \text{ J}\cdot\text{s}}{4.80 \times 10^{-24} \text{ kg}\cdot\text{m/s}} $$ $$ \lambda \approx 1.38 \times 10^{-10} \text{ m} $$

The wavelength is $1.38 \times 10^{-10} \text{ m}$, which is roughly the size of an atom.

Key Takeaways: Atomic Structure Equations

• Light Speed: $c = \lambda \nu$. As wavelength increases, frequency decreases.
• Photon Energy: $E = h\nu$ or $E = \frac{hc}{\lambda}$. High frequency = High Energy.
• Units Matter: Always convert wavelength to meters and mass to kilograms before calculating.
• Molar Energy: Multiply single photon energy by Avogadro's number ($N_A$) to get J/mol.
• Hydrogen Spectrum: Use the Rydberg formula to find the energy gap ($\Delta E$), then convert that gap to a photon wavelength.