Electronic Structure of Atoms

Electronic Structure of Atoms

Electromagnetic Radiation (EMR)

• To understand the electronic structure of atoms, it's crucial to understand electromagnetic radiation (EMR).
• Electromagnetic waves possess characteristic wavelengths (\lambda) and frequencies (\nu).
• Wavelength (\lambda) is the distance between corresponding points on adjacent waves.
• Frequency (\nu) is the number of cycles that pass a point in one second.

Waves: Wavelength, Amplitude, and Frequency

• Wavelength: Distance between two corresponding points on a wave.
• Amplitude: Height of the wave, related to intensity/energy.
• Frequency: Number of cycles passing a point per unit time.

Wave Nature of Light

• All electromagnetic radiation (EMR) moves through a vacuum at the speed of light, denoted as c.
• c = 3.00 \times 10^8 \text{ m/s}
• c = \lambda \nu
• The SI unit for wavelength (\lambda) is the meter (m).
• The SI unit for frequency (\nu) is s$^{-1}$, which is also known as Hertz (Hz), i.e., 1 \text{ s}^{-1} = 1 \text{ Hz}.

Wavelength and Frequency Relationship

• Long wavelength (\lambda) corresponds to low frequency (\nu).
• Short wavelength (\lambda) corresponds to high frequency (\nu).
• Wavelength and frequency have an inverse relationship: \nu \propto \frac{1}{\lambda}.

Problem: Red Light Frequency

• Red light has a wavelength (\lambda) of 700 nm. What is its frequency (\nu)?

  1. Convert wavelength to meters:
     700 \text{ nm} \times \frac{1 \text{ m}}{1 \times 10^9 \text{ nm}} = 7.00 \times 10^{-7} \text{ m}
  2. Use the relationship \nu = \frac{c}{\lambda}:
     \nu = \frac{3.00 \times 10^8 \text{ m/s}}{7.00 \times 10^{-7} \text{ m}} = 4.29 \times 10^{14} \text{ s}^{-1} = 4.29 \times 10^{14} \text{ Hz}

Electromagnetic Radiation (EMR) Properties

• All types of electromagnetic radiation travel at the speed of light (c).
• c = \lambda \nu
• E = h\nu = \frac{hc}{\lambda}, where h is Planck’s constant (6.626 \times 10^{-34} \text{ J.s}).
• Relationships:
  • \nu \propto \frac{1}{\lambda}
  • E \propto \nu
  • E \propto \frac{1}{\lambda}
• Longer wavelength (\lambda) means lower frequency (\nu) and lower energy (E).

Electromagnetic Spectrum

• High-energy radiations like X-rays have much shorter wavelengths than low-energy radiations like radio waves.
• X-rays have high energy, which can cause tissue damage and cancer.

Common Wavelength Units

Sample Exercise: Calculating Frequency from Wavelength

• Yellow light from a sodium vapor lamp has a wavelength of 589 nm. What is its frequency?
• Given: c = 3.00 \times 10^8 \text{ m/s}, 1 \text{ m} = 1 \times 10^9 \text{ nm}
• Solution:
  • \nu = \frac{c}{\lambda} = \frac{3.00 \times 10^8 \text{ m/s}}{589 \times 10^{-9} \text{ m}} = 5.09 \times 10^{14} \text{ s}^{-1}

Practice Exercise

• (a) Laser radiation with \lambda = 640.0 \text{ nm}. Calculate the frequency.
• (b) FM radio station broadcasts at \nu = 103.4 \text{ MHz}. Calculate the wavelength.
• Answers:
  • (a) 4.688 \times 10^{14} \text{ s}^{-1}
  • (b) 2.901 m

Quantization of Energy

• A quantum is the smallest amount of energy (photon) that can be emitted or absorbed as electromagnetic radiation.
• The relationship between energy and frequency is: E = h\nu
• where h is Planck’s constant (6.626 \times 10^{-34} \text{ J.s}).

Photoelectric Effect

• The photoelectric effect is the emission of electrons from metal surfaces when light shines on them.
• It provides evidence for the particle nature of light and quantization.
• Einstein proposed that light travels in energy packets called photons.
• The energy of one photon is E = h\nu.
• Electrons are ejected only if photons have sufficient energy.

Problem: Laser Light Energy and Frequency

• Laser light with \lambda = 785 \text{ nm}. Determine energy and frequency.

• Solution:

  • Convert nm to m: 785 \text{ nm} = 7.85 \times 10^{-7} \text{ m}
  • Calculate frequency: \nu = \frac{c}{\lambda} = \frac{3.0 \times 10^8 \text{ m/s}}{7.85 \times 10^{-7} \text{ m}} = 3.82 \times 10^{14} \text{ Hz}
  • Calculate energy: E = h\nu = (6.63 \times 10^{-34} \text{ J s}) (3.82 \times 10^{14} \text{ s}^{-1}) = 2.53 \times 10^{-19} \text{ J}

Problem: Laser Pulse Photons

• Laser pulse with \lambda = 532 \text{ nm} contains 4.88 mJ of energy. How many photons?
• Solution:
  • Convert mJ to J: 4.88 \text{ mJ} = 4.88 \times 10^{-3} \text{ J}
  • Calculate energy per photon: E_{\text{photon}} = \frac{hc}{\lambda} = \frac{(6.626 \times 10^{-34} \text{ J.s})(3 \times 10^8 \text{ m/s})}{532 \times 10^{-9} \text{ m}} = 3.74 \times 10^{-19} \text{ J}
  • Calculate number of photons: \text{#Photons} = \frac{\text{Total energy}}{E_{\text{photon}}} = \frac{4.88 \times 10^{-3} \text{ J}}{3.74 \times 10^{-19} \text{ J/photon}} = 1.31 \times 10^{16} \text{ photons}

Sample Exercise: Energy of a Photon

• Calculate the energy of one photon of yellow light with \lambda = 589 \text{ nm}.

• Solution:

  • \nu = \frac{c}{\lambda} = \frac{3.0 \times 10^8 \text{ m/s}}{589 \times 10^{-9} \text{ m}} = 5.09 \times 10^{14} \text{ s}^{-1}
  • E = h\nu = (6.626 \times 10^{-34} \text{ J.s}) (5.09 \times 10^{14} \text{ s}^{-1}) = 3.37 \times 10^{-19} \text{ J}

Practice Exercise - Photon Energy

• (a) Laser emits light with \nu = 4.69 \times 10^{14} \text{ s}^{-1}. What is the energy of one photon?
• (b) If the laser emits a pulse of 5.0 \times 10^{17} photons, what is the total energy of the pulse?
• (c) If the laser emits 1.3 \times 10^{-2} \text{J} of energy, how many photons are emitted?
• Answers:
  • (a) 3.11 \times 10^{-19} \text{ J}
  • (b) 0.16 J
  • (c) 4.2 \times 10^{16} photons

Quantum Theory - Quantization of Energy

• Quantized: Electrons must gain a specific amount of energy (equal to the energy difference between levels) to move between energy levels.
• Energy can be gained from heating (thermal energy) or light (EMR).
• Quantum: Smallest amount of energy that can be emitted or absorbed as electromagnetic radiation.
• Photon: A quantum of electromagnetic radiation.
• Energy of photon: E = h\nu or E = \frac{hc}{\lambda}. h = 6.626 \times 10^{-34} \text{ J.s}.

Quantized vs. Unquantized States

• Quantized states: Discrete energy levels (like steps).
• Unquantized states: Smooth transition between levels (like a ramp).

Bohr’s Model

• Rutherford's model: electrons orbit the nucleus like planets orbit the sun. However, this model had issues as charged particles moving in a circle should lose energy, making the atom unstable.
• Bohr proposed that electrons are confined to specific energy states called orbits.
• Bohr noted the line spectra of certain elements.

Bohr’s Postulates

1. Electrons occupy only certain orbits corresponding to specific energies.
2. An electron in a permitted orbit has a specific energy and is in an