Electronic Structure of the Atom

Chapter 3: Electronic Structure of the Atom

Quantum-Mechanical Model

• Rutherford's gold foil experiment led to the nuclear model of the atom.

  • However, classical mechanics couldn't explain why electrons wouldn't spiral inward towards the nucleus.

• Quantum mechanics is an extension of classical mechanics developed to describe matter at the atomic level.

• To understand the electronic structure of atoms, it is essential to understand the nature of electromagnetic radiation.

Wavelength vs. Amplitude

• Wavelength ($\lambda$):

  • The distance covered by one complete wave cycle.

  • Crest: The top of the wave.

  • Trough: The bottom of the wave.

  • Determines the color of visible light.

• Amplitude:

  • The height of the wave from the midline to the crest ( or trough).

  • Determines the intensity or brightness of the light.

Color of the Rainbow (ROYGBIV)

• White light is a mixture of all the colors of visible light.

• Color is what we perceive when certain wavelengths of light are not absorbed but reflected.

  • We see the reflected wavelengths, not the absorbed ones.

  • White objects reflect all wavelengths.

  • Black objects absorb all wavelengths.

Frequency and The Speed of Light

• Frequency ($\nu$):

  • The number of waves passing a given point per unit of time.

  • Unit: hertz (Hz), which is equivalent to $s^{-1}$ or $\frac{1}{s}$.

  • Frequency ($\nu$) and wavelength ($\lambda$) are inversely proportional.

• The total energy of a wave is proportional to both its amplitude and its frequency.

• All electromagnetic waves travel through space at the same constant speed, known as the speed of light ($c$).

  • $c = 3.00 \times 10^8 \text{ m/s}$.

  • Since $c$ is constant, frequency can be calculated from wavelength and vice versa using the formula: $c = \lambda \nu$.

Frequency vs. Wavelength Calculations

• Example: To find the frequency ($\nu$) of an electromagnetic wave with a wavelength ($\lambda$) of $457 \text{ nm}$:

  • First, convert wavelength to meters: $457 \text{ nm} = 457 \times 10^{-9} \text{ m}$.

  • Using $\nu = \frac{c}{\lambda} = \frac{3.00 \times 10^8 \text{ m/s}}{457 \times 10^{-9} \text{ m}} \approx 6.56 \times 10^{14} \text{ s}^{-1}$.

• Extra Practice Example: Calculate the wavelength in nanometers of red light with a frequency of $4.62 \times 10^{14} \text{ s}^{-1}$.

  • Using $\lambda = \frac{c}{\nu} = \frac{3.00 \times 10^8 \text{ m/s}}{4.62 \times 10^{14} \text{ s}^{-1}} \approx 6.49 \times 10^{-7} \text{ m}$.

  • Convert to nanometers: $6.49 \times 10^{-7} \text{ m} = 649 \text{ nm}$.

Wave Interference and Diffraction

• Diffraction: When traveling waves encounter an opening in a barrier, they bend around it.

• When two openings are present, the diffracted waves interact, producing patterns of alternating intensity (constructive and destructive interference).

Electromagnetic Spectrum

• The electromagnetic spectrum encompasses all wavelengths of electromagnetic radiation.

  • Visible light constitutes only a tiny fraction of this spectrum.

  • Radio waves: Have the lowest frequency ($\nu$), longest wavelength ($\lambda$), and lowest energy.

  • Gamma rays: Have the highest frequency ($\nu$), shortest wavelength ($\lambda$), and highest energy.

Challenges to Classical Mechanics

Several phenomena observed in the late 19th and early 20th centuries could not be explained by classical mechanics, leading to the development of quantum mechanics:

1. Atomic Emission (Johannes Rydberg)

2. Blackbody Radiation (Lord Rayleigh, Sir James Jeans, & Max Planck)

3. Photoelectric Effect (Albert Einstein & Heinrich Hertz)

4. Bohr Model of the Atom (Niels Bohr)

5. Wave-particle Duality (Louis de Broglie)

6. Uncertainty Principle (Werner Heisenberg)

1: Atomic Emission

• When an element absorbs energy, it emits a unique color of visible light.

  • Examples: Lasers, street lamps, and fireworks.

• Photoemission Spectrum:

  • When emitted light is passed through a prism, a line spectrum of component wavelengths is observed.

  • Physicists could not explain this phenomenon using classical physics but derived an equation to describe it.

1: Rydberg Equation

• Swedish physicist Johannes Rydberg analyzed the spectrum of hydrogen and found it could be described by specific equations:

  • In terms of wavelength: $\frac{1}{\lambda} = R<em>H\left(\frac{1}{n</em>f^2} - \frac{1}{n_i^2}\right)$

    • Where $R_H$ is the Rydberg constant for wavelength, $1.097 \times 10^7 \text{ m}^{-1}$.

  • In terms of frequency: $\nu = R<em>H\left(\frac{1}{n</em>f^2} - \frac{1}{n_i^2}\right)$

    • Where $R_H$ is the Rydberg constant for frequency, $3.29 \times 10^{15} \text{ Hz}$.

  • $n<em>i$ and $n</em>f$ are positive integers (whole numbers) representing the initial and final energy levels.

  • As the difference between $n<em>f$ and $n</em>i$ increases, the energy of the emitted radiation increases.

2: Blackbody Radiation

• A blackbody is a system that can absorb and emit radiation of all frequencies.

  • Example: A metal coil on an electric stove changes color as it is heated to hotter temperatures (incandescence).

• Incandescence: The emission of electromagnetic radiation due to heating.

• Classical theory predicted that the composition of the blackbody should not matter.

2: Ultraviolet Catastrophe

• The Rayleigh-Jeans law was an attempt to quantify the relationship between energy and frequency for blackbody radiation using classical physics.

• The ultraviolet catastrophe was a significant error in the Rayleigh-Jeans law, predicting infinite energy at high frequencies (ultraviolet range).

• This problem was solved by Max Planck, who proposed that energy is quantized.

2: Quantized Energy Analogy

• In 1900, German physicist Max Planck proposed that energy is not emitted continuously but in small, discrete packets called quanta.

  • This was a fundamental shift from classical mechanics, where energy was considered continuous.

2: Quantized Energy Equation

• The energy of each quantum is directly proportional to its frequency ($\nu$).

  • $E = h\nu = \frac{hc}{\lambda}$

  • $h$ is Planck's constant, $6.626 \times 10^{-34} \text{ J} \cdot \text{s}$.

  • $\nu$ is the frequency of the electromagnetic radiation absorbed or emitted.

  • $\lambda$ is the wavelength of the electromagnetic radiation absorbed or emitted.

• Planck also proposed that an atom would emit energy only if its electrons were moving at a minimum speed.

  • This explains why UV radiation is only emitted at high temperatures.

  • Planck correctly predicted that emitted radiation is dependent upon the electronic structure of the blackbody's atoms.

Wavelength vs. Energy Calculation Example

• Active Learning Example: A laser used in eye surgery produces radiation with a wavelength of $640.0 \text{ nm}$. Calculate the energy of this radiation.

  • Convert wavelength to meters: $640.0 \text{ nm} = 640.0 \times 10^{-9} \text{ m}$.

  • Using $E = \frac{hc}{\lambda} = \frac{(6.626 \times 10^{-34} \text{ J} \cdot \text{s})(3.00 \times 10^8 \text{ m/s})}{640.0 \times 10^{-9} \text{ m}} \approx 3.10 \times 10^{-19} \text{ J}$.

• Extra Practice: Calculate the energy (in J) of red light with a frequency of $4.62 \times 10^{14} \text{ s}^{-1}$.

  • Using $E = h\nu = (6.626 \times 10^{-34} \text{ J} \cdot \text{s})(4.62 \times 10^{14} \text{ s}^{-1}) \approx 3.06 \times 10^{-19} \text{ J}$.

3: Photoelectric Effect

• In 1887, German physicist Heinrich Hertz observed that when a metal absorbs sufficient energy, its electrons are ejected from the surface.

  • This phenomenon is known as the photoelectric effect.

  • Classical mechanics erroneously predicted that electron ejection depended only on the intensity of light, not its frequency.

• Hertz observed that a minimum frequency, called the threshold frequency ($\nu_0$), was required before electrons would be emitted, regardless of light intensity.

  • This observation validated Max Planck's theory of quantized energy.

3: Photons

• In 1905, German-born physicist Albert Einstein expanded on Planck's work, interpreting Hertz's photoelectric effect experiment.

  • While Planck focused on energy exchange between matter and radiation, Einstein focused on the energy itself.

• Einstein proposed that electromagnetic radiation existed as small, discrete particles called photons.

  • The energy of a single photon is given by: $E_{photon} = h\nu = \frac{hc}{\lambda}$.

  • When the intensity of light is above the threshold frequency ($\nu_0$):

    • The number of emitted electrons increases with the intensity of the light.

    • The kinetic energy (KE) of the emitted electrons increases with the frequency ($\nu$) of the light.

3: Photon Energy Worked Examples

• Example 1: A laser emits light with a frequency of $4.69 \times 10^{14} \text{ s}^{-1}$. What is the energy of one photon of this radiation?

  • $E = h\nu = (6.626 \times 10^{-34} \text{ J} \cdot \text{s})(4.69 \times 10^{14} \text{ s}^{-1}) \approx 3.11 \times 10^{-19} \text{ J/photon}$.

• Example 2: If the laser emits a pulse of energy containing $5.0 \times 10^{17}$ photons of this radiation, what is the total energy of that pulse?

  • Total Energy = (Energy of one photon) $\times$ (Number of photons)

  • Total Energy = $(3.11 \times 10^{-19} \text{ J/photon})(5.0 \times 10^{17} \text{ photons}) \approx 0.16 \text{ J}$.

3: Photoelectric Effect – Active Learning Example

• A nitrogen gas laser pulse with a wavelength of $337 \text{ nm}$ contains $3.83 \text{ mJ}$ of energy. How many photons does it contain?

  • Energy of one photon ($E_{photon}$):

    • Convert wavelength to meters: $337 \text{ nm} = 337 \times 10^{-9} \text{ m}$.

    • $E_{photon} = \frac{hc}{\lambda} = \frac{(6.626 \times 10^{-34} \text{ J} \cdot \text{s})(3.00 \times 10^8 \text{ m/s})}{337 \times 10^{-9} \text{ m}} \approx 5.89 \times 10^{-19} \text{ J/photon}$.

  • Total energy of the pulse = $3.83 \text{ mJ} = 3.83 \times 10^{-3} \text{ J}$.

  • Number of photons = $\frac{\text{Total Energy}}{\text{Energy of one photon}} = \frac{3.83 \times 10^{-3} \text{ J}}{5.89 \times 10^{-19} \text{ J/photon}} \approx 6.50 \times 10^{15} \text{ photons}$.

4: Bohr Model

• In 1914, Danish physicist Niels Bohr merged Rutherford's nuclear model with Planck's and Einstein's quantum theory to create the Bohr model of the atom.

  • This model suggested that electrons are only allowed to travel in fixed orbits around the nucleus.

  • Bohr orbits are assigned positive integer values ($n = 1, 2, 3, …$).

  • When $n=1$, this represents the Bohr radius, the closest an electron can be to the nucleus.

  • The Bohr model addressed the classical error that electrons should spiral inward towards the nucleus.

4: Bohr Energies and Coulomb's Law

• Bohr proposed that each electron orbit had a specific energy associated with it.

  • The energy of an electron in a specific orbit is given by: $E_n = -2.18 \times 10^{-18} \text{ J} \left(\frac{1}{n^2}\right)$.

  • The negative sign reflects the attractive force between the electrons and protons, which is explained by Coulomb's Law.

    • Coulomb's constant ($k$) is $8.99 \times 10^9 \text{ N} \cdot \text{m}^2 \text{/C}^2$.

    • When charges are opposite (attraction), potential energy ($E$) is negative.

    • When charges are the same (repulsion), potential energy ($E$) is positive.

  • The largest (least negative) energy corresponds to larger particles, smaller charges, and closest distance between particles.

4: Emission vs. Absorption

• Ground State: The lowest energy state for an electron (typically $n=1$).

• Photoemission (Emission): Occurs as an electron relaxes from a higher energy state ($n<em>i$) to a lower energy state ($n</em>f$), releasing energy (light).

• Absorption: Occurs when an electron gains energy and moves from a lower energy state ($n<em>i$) to a higher energy state ($n</em>f$).

• The Bohr frequency condition describes the energy change ($\Delta E$) for electron transitions:

  • $\Delta E<em>{electron} = E</em>{final} - E_{initial}$

  • $\Delta E = -2.18 \times 10^{-18} \text{ J} \left(\frac{1}{n<em>f^2} - \frac{1}{n</em>i^2}\right)$.

4: Emission Series of Hydrogen

• Different series of emitted light correspond to different final energy levels ($n_f$) for electrons relaxing to that orbit:

  • Balmer series ($n_f = 2$): This is the only series of hydrogen emission lines that falls within the visible spectrum.

  • When $n = \infty$ (the classical limit), the electron is no longer attracted to the nucleus, meaning the atom has been ionized.

4: Bohr Model – Active Learning & Extra Practice

• Indicate whether each of the following emits or absorbs energy:

  • $n = 3$ to $n = 1$: Emits energy (electron moves to a lower energy level).

  • $n = 2$ to $n = 4$: Absorbs energy (electron moves to a higher energy level).

• Calculate the energy released when an electron moves from $n = 6$ to $n = 2$:

  • $\Delta E = -2.18 \times 10^{-18} \text{ J} \left(\frac{1}{2^2} - \frac{1}{6^2}\right) = -2.18 \times 10^{-18} \text{ J} \left(\frac{1}{4} - \frac{1}{36}\right) = -2.18 \times 10^{-18} \text{ J} \left(\frac{9-1}{36}\right)$

  • $\Delta E = -2.18 \times 10^{-18} \text{ J} \left(\frac{8}{36}\right) = -2.18 \times 10^{-18} \text{ J} \left(\frac{2}{9}\right) \approx -4.84 \times 10^{-19} \text{ J}$.

  • The negative sign indicates energy is released.

• Extra Practice: Calculate the wavelength (in nm) of the radiation released in the previous question.

  • The energy released is $|\Delta E| = 4.84 \times 10^{-19} \text{ J}$.

  • Using $E = \frac{hc}{\lambda}$, so $\lambda = \frac{hc}{E} = \frac{(6.626 \times 10^{-34} \text{ J} \cdot \text{s})(3.00 \times 10^8 \text{ m/s})}{4.84 \times 10^{-19} \text{ J}} \approx 4.11 \times 10^{-7} \text{ m}$.

  • Convert to nanometers: $4.11 \times 10^{-7} \text{ m} = 411 \text{ nm}$.

5: Wave-Particle Duality

• In 1924, French physicist Louis de Broglie theorized that if light exhibits material (particle) properties, then matter (e.g., electrons) should also exhibit wave properties.

  • He predicted that the wavelength of matter is inversely proportional to its momentum.

  • De Broglie Wavelength Equation: $\lambda = \frac{h}{mv}$

    • $h$ is Planck's constant ( $6.626 \times 10^{-34} \text{ J} \cdot \text{s}$).

    • $m$ is the mass of the particle.

    • $v$ is the velocity of the particle.

• Because electrons are so small, their wave character is very significant and observable.

5: Double Slit Experiment – Particle vs. Wave

• If electrons behaved only like particles, a beam of electrons shined through two slits would produce two distinct bands on a screen.

• However, the double-slit experiment with electrons produced an interference pattern, which is characteristic of wave behavior, confirming de Broglie's hypothesis.

5: De Broglie Wavelength – Active Learning & Extra Practice

• Active Learning Example: What is the wavelength (in nm) of an electron moving at a speed of $100. \text{ km/s}$? The mass of an electron is $9.11 \times 10^{-28} \text{ g}$.

  • Convert mass to kg: $9.11 \times 10^{-28} \text{ g} = 9.11 \times 10^{-31} \text{ kg}$.

  • Convert velocity to m/s: $100. \text{ km/s} = 100. \times 10^3 \text{ m/s} = 1.00 \times 10^5 \text{ m/s}$.

  • Using $\lambda = \frac{h}{mv} = \frac{6.626 \times 10^{-34} \text{ J} \cdot \text{s}}{(9.11 \times 10^{-31} \text{ kg})(1.00 \times 10^5 \text{ m/s})} \approx 7.27 \times 10^{-9} \text{ m}$.

  • Convert to nanometers: $7.27 \times 10^{-9} \text{ m} = 7.27 \text{ nm}$.

• Extra Practice: Calculate the velocity of a neutron (mass is $1.67 \times 10^{-24} \text{ g}$) whose de Broglie wavelength is $500. \text{ pm}$.

  • Convert mass to kg: $1.67 \times 10^{-24} \text{ g} = 1.67 \times 10^{-27} \text{ kg}$.

  • Convert wavelength to meters: $500. \text{ pm} = 500. \times 10^{-12} \text{ m}$.

  • Using $v = \frac{h}{m\lambda} = \frac{6.626 \times 10^{-34} \text{ J} \cdot \text{s}}{(1.67 \times 10^{-27} \text{ kg})(500. \times 10^{-12} \text{ m})} \approx 794 \text{ m/s}$.

6: Uncertainty Principle

• In 1927, German theoretical physicist Werner Heisenberg stated the Uncertainty Principle:

  • It is impossible to know both the momentum ($mv$) and position ($x$) of a particle with absolute certainty simultaneously.

  • The principle is mathematically expressed as: $\Delta x \times \Delta mv \ge \frac{h}{4\pi}$.

    • $\Delta x$ represents the uncertainty in position.

    • $\Delta mv$ represents the uncertainty in momentum.

• Any experiment designed to observe the electron inevitably changes its behavior, making precise simultaneous measurements impossible.

Classical vs. Quantum Conclusion

• Classical Physics: Assumes velocity and position can be known simultaneously, leading to a deterministic (definite, predictable) future.

• Quantum Mechanics: States that velocity and position cannot be known simultaneously because they are complementary properties.

  • Indeterminacy: The future is indefinite; precise prediction is impossible.

  • Complementary properties: The more accurately one property (e.g., position) is known, the less accurately the other property (e.g., momentum) can be known.

• Consequently, statistics are used to calculate the probability of finding an electron in a specific region, rather than its exact location or trajectory.

Schrödinger and Wavefunctions

• In 1926, Austrian physicist Erwin Schrödinger proposed an equation that incorporated both the wave-like and particle-like behavior of electrons.

  • Schrödinger Equation: $H\psi = E\psi$

    • $H$ is the Hamiltonian operator, representing the total energy of the system.

    • $E$ is the total energy of the electron.

    • $\psi$ (psi) is the wavefunction. It has no direct physical meaning itself.

• Schrödinger's equation successfully calculated the energies associated with the hydrogen atom.

  • However, it does not work precisely for atoms with more than one electron.

Probability Density and Orbitals

• The square of the wave function, $\psi^2$, gives a probability density map.

  • This map indicates where an electron has a statistical likelihood of being found at any given instant in time.

• Solving the wave equation yields a set of wave functions, which are called orbitals, and their corresponding energies.

  • Orbitals describe a 3D distribution of electron density around the nucleus.

Quantum Numbers

• An orbital is fully described by a set of four quantum numbers, which act like an