Atomic Spectra, Photon Energy, and the Bohr Model

Wave-Particle Nature of Light

Wavelength, Frequency, and Energy Relationships

Light exhibits both wave-like and particle-like properties. The fundamental characteristics of light include its wavelength (λ), frequency (ν), and energy (E). These properties are interrelated and are essential for understanding atomic emission and absorption phenomena.

• Wavelength (λ): The distance between successive peaks of a wave, typically measured in meters (m) or nanometers (nm).

• Frequency (ν): The number of wave cycles that pass a given point per second, measured in hertz (Hz) or s-1.

• Energy (E): The energy of a photon is directly proportional to its frequency and inversely proportional to its wavelength.

The relationships are given by the following equations:

• $c = \lambda \nu$ — where c is the speed of light ($3.00 \times 10^8$ m/s).

• $E = h \nu$ — where h is Planck's constant ($6.626 \times 10^{-34}$ J·s).

• $E = \dfrac{hc}{\lambda}$ — combines the above relationships to relate energy directly to wavelength.

Key Point: As wavelength decreases, frequency and photon energy increase, and vice versa.

Calculating Photon Energy: Example Problems

Red and Green Laser Pointers

To determine the frequency and energy of photons emitted by lasers of different wavelengths, use the above equations. The process involves converting wavelength to meters, calculating frequency, and then finding energy.

• Example 1: Red Laser (650 nm)

  • Convert wavelength: $650\ \text{nm} \times 10^{-9}\ \text{m/nm} = 6.50 \times 10^{-7}\ \text{m}$

  • Calculate frequency: $\nu = \dfrac{c}{\lambda} = \dfrac{3.00 \times 10^8\ \text{m/s}}{6.50 \times 10^{-7}\ \text{m}} = 4.6 \times 10^{14}\ \text{s}^{-1}$

  • Calculate energy: $E = h\nu = (6.626 \times 10^{-34}\ \text{J·s})(4.6 \times 10^{14}\ \text{s}^{-1}) = 3.1 \times 10^{-19}\ \text{J}$

• Example 2: Green Laser (532 nm)

  • Convert wavelength: $532\ \text{nm} \times 10^{-9}\ \text{m/nm} = 5.32 \times 10^{-7}\ \text{m}$

  • Calculate frequency: $\nu = \dfrac{3.00 \times 10^8\ \text{m/s}}{5.32 \times 10^{-7}\ \text{m}} = 5.6 \times 10^{14}\ \text{s}^{-1}$

  • Calculate energy: $E = h\nu = (6.626 \times 10^{-34}\ \text{J·s})(5.6 \times 10^{14}\ \text{s}^{-1}) = 3.7 \times 10^{-19}\ \text{J}$

Conclusion: The green laser (shorter wavelength) emits more energetic photons than the red laser (longer wavelength).

Atomic Emission and Spectra

Continuous vs. Line Spectra

When atoms and molecules emit light, they do not produce a continuous spectrum (a full rainbow of colors) as white light does. Instead, they emit light at specific, discrete wavelengths, resulting in a line spectrum. Each element has a unique line spectrum, which can be used for identification.

• Continuous Spectrum: Produced by white light; contains all visible wavelengths.

• Line Spectrum: Produced by excited atoms; contains only specific wavelengths unique to each element.

Example: Neon and hydrogen gas tubes emit different colors and line spectra when excited electrically.

The Hydrogen Atom and the Bohr Model

Rydberg Equation and Energy Levels

The emission spectrum of hydrogen can be described mathematically using the Rydberg equation, which relates the wavelengths of emitted light to integer values corresponding to electron energy levels.

• Rydberg Equation: $\dfrac{1}{\lambda} = R_H \left( \dfrac{1}{n_1^2} - \dfrac{1}{n_2^2} \right)$

  • $R_H$ is the Rydberg constant ($1.097 \times 10^7\ \text{m}^{-1}$)

  • $n_1$ and $n_2$ are integers, with $n_2 > n_1$

• Bohr Model: Niels Bohr explained the hydrogen spectrum by proposing that electrons occupy only certain allowed orbits (energy levels) around the nucleus. Transitions between these levels result in absorption or emission of photons with energy $E = h\nu$.

Key Postulates of the Bohr Model

• Electrons exist only in specific, quantized energy levels (stationary states).

• Energy is absorbed or emitted only when an electron transitions between energy levels.

• The energy of each level in hydrogen is given by: $E_n = -\dfrac{2.18 \times 10^{-18}\ \text{J}}{n^2}$ where $n$ is the principal quantum number (n = 1, 2, 3, ...).

• The change in energy for a transition is: $\Delta E = E_{n_f} - E_{n_i}$

Example: Calculating Photon Energy and Wavelength for Hydrogen Transitions

• For a transition from $n_i = 4$ to $n_f = 1$:

  • $\Delta E = -2.18 \times 10^{-18}\ \text{J} \left( \dfrac{1}{1^2} - \dfrac{1}{4^2} \right) = -2.04 \times 10^{-18}\ \text{J}$

  • Photon energy: $E_{photon} = |\Delta E| = 2.04 \times 10^{-18}\ \text{J}$

  • Frequency: $\nu = \dfrac{E}{h} = \dfrac{2.04 \times 10^{-18}\ \text{J}}{6.626 \times 10^{-34}\ \text{J·s}} = 3.08 \times 10^{15}\ \text{s}^{-1}$

  • Wavelength: $\lambda = \dfrac{c}{\nu} = \dfrac{2.998 \times 10^8\ \text{m/s}}{3.08 \times 10^{15}\ \text{s}^{-1}} = 9.72 \times 10^{-8}\ \text{m} = 97.2\ \text{nm}$

  • This wavelength lies in the ultraviolet (UV) region of the electromagnetic spectrum.

Summary Table: Key Equations and Constants

Quantity	Symbol	Equation	Value/Unit
Speed of Light	c		$3.00 \times 10^8$ m/s
Planck's Constant	h		$6.626 \times 10^{-34}$ J·s
Rydberg Constant	$R_H$		$1.097 \times 10^7$ m-1
Photon Energy	E	$E = h\nu = \dfrac{hc}{\lambda}$	Joules (J)
Wavelength-Frequency	λ, ν	$c = \lambda \nu$	m, s-1
Hydrogen Energy Levels	$E_n$	$E_n = -\dfrac{2.18 \times 10^{-18}}{n^2}$	Joules (J)

Key Takeaways

• Light can be described as both a wave and a particle (photon).

• Energy, frequency, and wavelength are mathematically related.

• Atoms emit light at discrete wavelengths, producing line spectra unique to each element.

• The Bohr model explains the hydrogen atom's line spectrum by quantizing electron energy levels.

• Transitions between energy levels result in absorption or emission of photons with specific energies and wavelengths.