Calculations

Calculation Type 1: The Speed of Light Equation

  • Formula: \(\mathbf{c=\lambda \nu }\)

  • When to use it: When a question gives you wavelength (\(\lambda \)) and asks for frequency (\(\nu \)), or vice versa.

đź’ˇ The Secret to this Calculation: Unit Conversion

The speed of light (\(c\)) is always given in meters per second (\(m/s\)). However, word problems will almost always give you the wavelength in nanometers (\(nm\)). You must convert nanometers to meters before doing any math.

  • Conversion factor: \(1\text{ nm} = 1 \times 10^{-9}\text{ m}\)

Example Problem:

Find the frequency of green light that has a wavelength of \(500\text{ nm}\).

  1. Convert to meters: \(500\text{ nm} = 500 \times 10^{-9}\text{ m}\) (or \(5.00 \times 10^{-7}\text{ m}\))

  2. Rearrange the formula to solve for frequency (\(\nu \)):
    \(\nu =\frac{c}{\lambda }\)

  3. Plug in the constants and solve:
    \(\nu =\frac{3.00\times 10^{8}\text{\ m/s}}{5.00\times 10^{-7}\text{\ m}}=\mathbf{6.00\times 10}^{\mathbf{14}}\text{\ s}^{\mathbf{-1}}\text{\ (or\ Hz)}\)

Calculation Type 2: Energy of Light Equations

  • Formulas: \(\mathbf{E=h\nu }\) or \(\mathbf{E=}\frac{\mathbf{hc}}{\mathbf{\lambda }}\)

  • When to use it: When a question asks for the energy (\(E\)) of a single photon.

đź’ˇ The Secret to this Calculation: Picking the Shortcut

  • If the problem gives you frequency (\(\nu \)), use the short version: \(E = h\nu\).

  • If the problem gives you wavelength (\(\lambda \)), use the combined version: \(E = \frac{hc}{\lambda}\) so you don't have to do two separate math steps.

Example Problem:

What is the energy of a single photon of light with a frequency of \(6.00 \times 10^{14}\text{ Hz}\)?

  1. Pick the formula: Since we have frequency, use \(E = h\nu\).

  2. Plug in Planck's constant (\(h\)) and solve:
    \(E=(6.626\times 10^{-34}\text{\ J}\cdot \text{s})\times (6.00\times 10^{14}\text{\ s}^{-1})\)
    \(E=\mathbf{3.98\times 10}^{\mathbf{-19}}\text{\ Joules}\)

Calculation Type 3: Matter Wave Equation (de Broglie)

  • Formula: \(\mathbf{\lambda =}\frac{\mathbf{h}}{\mathbf{mv}}\)

  • When to use it: When the problem involves a physical object with a mass (like an electron, a baseball, or a car) moving at a certain velocity, and asks for its wavelength.

đź’ˇ The Secret to this Calculation: Watch the "v" and the Units

  • Don't confuse the variables: In this formula, \(v\) stands for velocity (speed in \(m/s\)), not frequency (\(\nu \)).

  • Mass must be in kilograms (\(kg\)): Planck's constant (\(h\)) uses Joules, and a Joule is structurally made of \(kg \cdot m^2/s^2\). If the problem gives you the mass of an electron or atom in grams (\(g\)), you must convert it to \(kg\) first.

Example Problem:

An electron has a mass of \(9.11 \times 10^{-31}\text{ kg}\) and is traveling at a velocity of \(2.2 \times 10^6\text{ m/s}\). What is its de Broglie wavelength?

  1. Plug everything directly into the formula:
    \(\lambda =\frac{6.626\times 10^{-34}\text{\ kg}\cdot \text{m}^{2}/\text{s}}{(9.11\times 10^{-31}\text{\ kg})\times (2.2\times 10^{6}\text{\ m/s})}\)

  2. Solve the math:
    \(\lambda =\frac{6.626\times 10^{-34}}{2.004\times 10^{-24}}=\mathbf{3.3\times 10}^{\mathbf{-10}}\text{\ meters}\)

Summary Conceptual Guide (For Multiple Choice Questions)

  • High Energy light (like Gamma rays or X-rays) = High Frequency = Short, squished wavelengths.

  • Low Energy light (like Radio waves) = Low Frequency = Long, stretched wavelengths.

  • Quantized means light energy isn't a smooth ramp. If an electron needs exactly a \(3.0 \times 10^{-19}\text{ J}\) photon to jump to the next level (like climbing a stair step), hitting it with a \(2.9 \times 10^{-19}\text{ J}\) photon does absolutely nothing. It's all or nothing.