Comprehensive Study Notes: Atomic Structure and the Bohr Model

12.1 Introduction to Atomic Hypothesis and Early Models

  • By the nineteenth century, significant evidence supported the atomic hypothesis of matter.

  • J. J. Thomson's Discovery (1897): Through experiments on electric discharge through gases, English physicist J. J. Thomson discovered that atoms of all elements contain identical negatively charged constituents called electrons.

  • Atomic Neutrality: Since atoms are electrically neutral, it was deduced they must contain a positive charge to neutralize the electrons. The primary question became the arrangement of these charges.

  • Thomson’s Plum Pudding Model (1898):     - This was the first proposed atomic model.     - It suggested the positive charge is uniformly distributed throughout the volume of the atom.     - Negatively charged electrons are embedded within this positive sphere like seeds in a watermelon.

  • Radiation and Atomic Interaction:     - Condensed matter (solids/liquids) and dense gases emit electromagnetic radiation with a continuous distribution of wavelengths across all temperatures.     - This is attributed to the oscillations of atoms and molecules governed by interactions with neighbors.     - Rarefied gases (e.g., neon signs, mercury vapor) emit discrete wavelengths, appearing as bright lines. In these gases, the large spacing between atoms allows radiation to be considered as the result of individual atoms rather than interactions.

  • Atomic Fingerprints: By the early nineteenth century, it was known that every element has a characteristic spectrum. For instance, hydrogen consistently produces a specific set of lines with fixed relative positions, suggesting a link between internal structure and emission.

  • Balmer's Formula (1885): Johann Jakob Balmer obtained a simple empirical formula to calculate the wavelengths of a group of lines in the hydrogen spectrum.

12.2 Alpha-Particle Scattering and Rutherford’s Nuclear Model

  • The Geiger-Marsden Experiment (1911): Suggested by Ernst Rutherford and performed by Hans Geiger and Ernst Marsden (then a 20-year-old student).

  • Experimental Setup:     - Source: A radioactive source of 83214Bi^{214}_{83}\text{Bi} emitted beam of 5.5MeV5.5\,\text{MeV} α\alpha-particles.     - Collimation: The beam was narrowed by passage through lead bricks.     - Target: A thin gold foil of thickness 2.1×107m2.1 \times 10^{-7}\,\text{m}.     - Detection: A rotatable detector with a zinc sulphide screen and a microscope. α\alpha-particles hitting the screen produced scintillations (light flashes).     - Environment: The entire apparatus was housed in a vacuum chamber.

  • Observations:     - Most α\alpha-particles passed through the foil without any collisions.     - Approximately 0.14%0.14\% of incident α\alpha-particles scattered by more than 11^{\circ}.     - About 1 in 8000 particles deflected by more than 9090^{\circ}.

  • Rutherford’s Deduction: To explain large-angle deflections (back-scattering), Rutherford proposed that a large repulsive force exists. This required the atom's positive charge and most of its mass to be concentrated in a tiny central region called the nucleus.

  • Dimensions and Scale:     - Kinetic theory established the atom's size as approximately 1010m10^{-10}\,\text{m}.     - Rutherford's experiments suggested the nucleus size is between 1015m10^{-15}\,\text{m} and 1014m10^{-14}\,\text{m}.     - This implies the nucleus is 10,00010,000 to 100,000100,000 times smaller than the atom, meaning an atom is largely empty space.

  • Quantitative Analysis of Alpha-Scattering:     - The target (gold nucleus) is assumed to be stationary as it is roughly 50 times heavier than the α\alpha-particle.     - α\alpha-particles carry a charge of 2e2e.     - The gold nucleus has a charge of ZeZe, where Z=79Z = 79.     - The trajectory is calculated using Newton's second law and Coulomb's law for electrostatic repulsion:

F=14πϵ0(2e)(Ze)r2F = \frac{1}{4\pi\epsilon_0} \frac{(2e)(Ze)}{r^2}

12.2.1 Alpha-particle Trajectory and Impact Parameter

  • Impact Parameter (bb): The perpendicular distance of the initial velocity vector of the α\alpha-particle from the center of the nucleus.

  • Scattering Relationship:     - Small bb: The particle is closer to the nucleus and suffers large scattering.     - Head-on collision (b=0b = 0): The particle rebounds (θ180\theta \approx 180^{\circ}).     - Large bb: The particle experiences small deflection (θ0\theta \approx 0^{\circ}).

  • Example 12.1 (The Solar System Analogy):     - In an atom, the ratio of orbital radius 1010m10^{-10}\,\text{m} to nucleus radius 1015m10^{-15}\,\text{m} is 10510^5.     - If applied to the solar system, where the sun's radius is 7×108m7 \times 10^8\,\text{m}, the Earth's orbit would need to be 105×7×108=7×1013m10^5 \times 7 \times 10^8 = 7 \times 10^{13}\,\text{m}.     - This is over 100 times the actual orbital radius (1.5×1011m1.5 \times 10^{11}\,\text{m}), indicating atoms have a much higher fraction of empty space than the solar system.

  • Example 12.2 (Distance of Closest Approach):     - For a 7.7MeV7.7\,\text{MeV} (1.2×1012J1.2 \times 10^{-12}\,\text{J}) α\alpha-particle, energy conservation dictates K=14πϵ02Ze2dK = \frac{1}{4\pi\epsilon_0} \frac{2Ze^2}{d}.     - Using 1/4\pi\epsilon_0 = 9.0 \times 10^9\,\text{N\,m^2/C^2} and e=1.6×1019Ce = 1.6 \times 10^{-19}\,\text{C}, the formula yields d=3.84×1016Zmd = 3.84 \times 10^{-16} Z\,\text{m}.     - For Gold (Z=79Z=79), d3.0×1014md \approx 3.0 \times 10^{-14}\,\text{m} (or 30fm30\,\text{fm}).     - Note: The actual radius of a gold nucleus is 6fm6\,\text{fm}. The α\alpha-particle reverses before touching it.

12.2.2 Electron Orbits in Rutherford's Model

  • Model Structure: Electrically neutral sphere with a massive positive nucleus and electrons in dynamically stable orbits.

  • Force Balance: The electrostatic force (FeF_e) provides the centripetal force (FcF_c):

Fe=Fc    14πϵ0e2r2=mv2rF_e = F_c \implies \frac{1}{4\pi\epsilon_0} \frac{e^2}{r^2} = \frac{mv^2}{r}

  • Orbit Radius vs. Velocity:

r=e24πϵ0mv2r = \frac{e^2}{4\pi\epsilon_0 m v^2}

  • Energy Relations in Hydrogen:     - Kinetic Energy (KK): K=e28πϵ0rK = \frac{e^2}{8\pi\epsilon_0 r}     - Potential Energy (UU): U=e24πϵ0rU = -\frac{e^2}{4\pi\epsilon_0 r}     - Total Energy (EE): E=K+U=e28πϵ0rE = K + U = -\frac{e^2}{8\pi\epsilon_0 r}     - The negative sign indicates the electron is bound to the nucleus.

  • Example 12.3:     - It takes 13.6eV13.6\,\text{eV} (2.2×1018J2.2 \times 10^{-18}\,\text{J}) to separate a hydrogen atom.     - Using E=2.2×1018JE = -2.2 \times 10^{-18}\,\text{J}, the orbital radius is computed as r=5.3×1011mr = 5.3 \times 10^{-11}\,\text{m}.     - The electron velocity is found to be v=2.2×106m/sv = 2.2 \times 10^6\,\text{m/s}.

12.3 Atomic Spectra and Rutherford’s Limitations

  • Hydrogen Spectrum: Displays discrete spectral lines (Emission Line Spectrum).

  • Absorption Spectrum: Produced when white light passes through a gas; the transmitted light shows dark lines at the same wavelengths as the emission lines.

  • Failures of Rutherford's Model:     - Instability: Classical electromagnetic theory dictates that an accelerating charged particle (revolving electron) must emit electromagnetic waves. This loss of energy would cause the electron to spiral inward and fall into the nucleus.     - Spectrum Contradiction: As the electron spirals, its frequency would change continuously, resulting in a continuous spectrum rather than the observed discrete line spectrum.

  • Example 12.4: According to classical theory, an electron in hydrogen (r=5.3×1011mr = 5.3 \times 10^{-11}\,\text{m}, v=2.2×106m/sv = 2.2 \times 10^6\,\text{m/s}) would have a revolution frequency of ν=v/2πr6.6×1015Hz\nu = v/2\pi r \approx 6.6 \times 10^{15}\,\text{Hz}, which should be the initial frequency of light emitted.

12.4 Bohr Model of the Hydrogen Atom

  • Niels Bohr combined classical and quantum concepts in 1913 through three postulates:
  1. Postulate 1 (Stationary States): Electrons can revolve in certain stable orbits without radiating energy. Each state has a definite total energy.
  2. Postulate 2 (Angular Momentum Quantisation): An electron revolves only in orbits where the angular momentum (LL) is an integral multiple of h/2πh/2\pi.

L=nh2πL = \frac{nh}{2\text{π}}

where n=1,2,3,n = 1, 2, 3, \dots (Principal Quantum Number) and h=6.6×1034Jsh = 6.6 \times 10^{-34}\,\text{J\,s}.

  1. Postulate 3 (Transition and Photons): An electron can transition from a higher energy orbit (EiE_i) to a lower one (EfE_f), emitting a photon.

hν=EiEfh\nu = E_i - E_f

  • Mathematical Results of Bohr Model:     - Orbital Radius (rnr_n): rn=n2h2ϵ0πme2r_n = \frac{n^2 h^2 \epsilon_0}{\pi m e^2}     - Total Energy (EnE_n): En=me48n2h2ϵ02E_n = -\frac{m e^4}{8 n^2 h^2 \epsilon_0^2}     - Simplified Energy Formula: En=2.18×1018n2J=13.6n2eVE_n = -\frac{2.18 \times 10^{-18}}{n^2}\,\text{J} = -\frac{13.6}{n^2}\,\text{eV}

12.4.1 Energy Levels

  • Ground State (n=1n=1): The lowest energy state (E1=13.6eVE_1 = -13.6\,\text{eV}). The minimum energy to remove the electron is the ionisation energy (13.6eV13.6\,\text{eV}).
  • Excited States (n>1n > 1): States with higher energy (lesser absolute value).     - First excited state (n=2n=2): E2=3.40eVE_2 = -3.40\,\text{eV}; excitation energy from ground = 10.2eV10.2\,\text{eV}.     - Second excited state (n=3n=3): E3=1.51eVE_3 = -1.51\,\text{eV}; excitation energy from ground = 12.09eV12.09\,\text{eV}.
  • Ionisation State (nn \rightarrow \infty): Energy is 0eV0\,\text{eV}. An electron can have any continuous energy value above this level.

12.6 De Broglie’s Explanation of Quantisation

  • Louis de Broglie in 1923 explained Bohr's second postulate by proposing that electrons have a wave nature.

  • Resonant Standing Waves: For an electron in a circular orbit, the orbit should accommodate an integral number of de Broglie wavelengths (λ\lambda) to form a standing wave:

2πrn=nλ2\pi r_n = n\lambda

  • Linking with Momentum: Since λ=h/p=h/mvn\lambda = h/p = h/mv_n:

2πrn=nhmvn    mvnrn=nh2π2\pi r_n = n \frac{h}{m v_n} \implies m v_n r_n = \frac{nh}{2\pi}

  • This derivation precisely matches Bohr's quantisation condition.

Limitations of the Bohr Model

  • Multi-electron Atoms: It applies only to hydrogenic atoms (one electron). It fails even for Helium due to complex electron-electron interactions (which are comparable in magnitude to the nucleus-electron interaction).
  • Spectral Intensities: It cannot explain why some spectral lines are stronger (more frequent transitions) than others.
  • Wave Mechanics Conflict: Its discrete orbital path violates the Heisenberg uncertainty principle.
  • Refinement: Replaced by Quantum Mechanics where orbits are replaced by regions of high probability characterized by four quantum numbers (n,l,m,sn, l, m, s).

Points to Ponder

  • Both Thomson and Rutherford models are classically unstable.
  • The Bohr model is a semiclassical bridge; it occasionally ignores logic (like non-radiating orbits) to achieve alignment with experimental predictions.
  • Hydrogen atom energy levels depend only on nn in a pure Coulomb potential, but in reality, states are more complex.
  • At high quantum numbers (nn1n \rightarrow n-1), the revolution frequency of the electron coincides with the emitted photon frequency.