Comprehensive Study Notes: Rutherford and Bohr Atomic Models

Rutherford’s Atomic Model and the Gold Foil Experiment

• The Experiment: Gold Foil Experiment

  • Ernest Rutherford conducted a historic experiment involving the bombardment of alpha rays on a thin gold foil.

  • Alpha Rays: These are positively charged particles used as projectiles in the bombardment.

• Experimental Observations:

  • Majority of Particles: The vast majority of alpha particles passed through the gold foil without any deflection. Deflection is defined as the deviation or shifting from the body's original straight-line path.

  • Minor Deflections: A small number of alpha particles suffered minor deflections upon passing through the foil.

  • Major Deflections (180°): An extremely small fraction of particles—approximately $1$ out of $8000$, or roughly $0.0125\%$ —experienced a major deflection of $180^\circ$, essentially bouncing back in the direction they came from.

• Conclusions and the Rutherford Model:

  • Empty Space: Because most alpha particles passed through without interaction, Rutherford concluded that the major portion (the "major chunk") of the atom is empty space. If there were significant matter throughout, the particles would have interacted and deflected.

  • Existence of Orbits: Electrons are revolving in certain orbits, energy levels, or shells around a central core.

  • The Nucleus: From the particles that bounced back ($180^\circ$), Rutherford deduced that there is a center called the nucleus.

  • Positive Charge: The nucleus must be positively charged because it repelled the positively charged alpha particles.

  • Extreme Density: The nucleus is extremely dense and small. Because only $1$ in $8000$ particles hit it, the nucleus must occupy a tiny volume compared to the whole atom.

• Mass Concentration Metaphor (Sodium Case Study):

  • Atoms consist of protons, neutrons, and electrons.

  • Mass Values:

  • Mass of Proton/Neutron: $\approx 1.6 \times 10^{-27}\,kg$

  • Mass of Electron: $\approx 9 \times 10^{-31}\,kg$

  • Comparison: A proton or neutron is approximately $1800$ times heavier than an electron.

  • Sodium ($Na, 11/23$) Example:

  • Nucleus contains $11$ protons and $12$ neutrons (Total $23$ particles).

  • Shells contain $11$ electrons.

  • If an electron mass is assumed to be $1\,kg$ for simplicity, its proton/neutron counterpart would be $1800\,kg$.

  • Nucleus mass: $23 \times 1800 = 41,400\,kg$.

  • Total electron mass: $11 \times 1 = 11\,kg$.

  • This illustrates that $99.999\%$ of an atom's mass is concentrated in the tiny nucleus.

  • Nucleus Size: The nucleus is measured in Fermameters (order of $10^{-15}\,m$).

Drawbacks of the Rutherford Model

• The Stability Problem (Maxwell’s Laws):

  • According to classical electromagnetism (Maxwell’s laws), any accelerated charge must emit electromagnetic radiations continuously.

  • An electron moving in a circular orbit around the nucleus undergoes centripetal acceleration.

  • Therefore, the electron should continuously lose energy as radiation (photons/energy packets).

  • This energy loss should cause the electron to spiral inward, eventually falling into the nucleus, making the atom unstable. However, atoms are observed to be stable.

• The Spectrum Problem:

  • Since the electron is supposed to emit energy continuously, the resulting electromagnetic spectrum should be a Continuous Spectrum.

  • Continuous Spectrum: A spectrum where colors or wavelengths blend without distinct boundaries or demarcations.

  • Reality (Line Spectrum): Experimental observations of atoms (like Hydrogen) show a Line Spectrum.

  • Line Spectrum: A spectrum containing discrete lines separated by dark regions (where no energy is emitted). Rutherford could not explain why atoms produced line spectra instead of continuous ones.

Atomic Spectra and Identification Tools

• Spectrometer: A device used to obtain and study the spectrum of an atom by plotting emitted wavelengths.

• Energy Calculation: Energy is related to wavelength by the formula $E = \frac{hc}{\lambda}$.

• Electromagnetic Wave Mnemonic: "Rahul's Mother Is Visiting Uncle Xavier's Garden" represents the order: Radio, Micro, Infrared, Visible, Ultraviolet, X-rays, Gamma rays. Only the visible portion is seen by the naked eye.

• Identification: In the 1800s, atomic spectra served as the "salt analysis" or identification tool for elements (e.g., distinguishing Hydrogen from Lithium or Platinum) based on their unique line patterns.

Bohr’s Atomic Model and Postulates

• Niels Bohr’s Proposal: Bohr introduced his model to solve the defects of Rutherford's model, often called the Bohr-Rutherford Model.

• Postulate 1: Stationary Energy Orbits

  • Electrons revolve only in specific orbits called Stationary Energy Orbits or fixed energy levels.

  • While in these orbits, electrons do not radiate or lose energy. Their energy is quantized/fixed.

• Postulate 2: Transitions and Energy Change

  • Energy is only emitted or absorbed when an electron undergoes a Transition (shifting between levels).

  • De-excitation: When an electron moves from a higher energy level to a lower one, it releases energy (represented by a negative sign).

  • Excitation: When an electron moves from a lower level to a higher one, it absorbs/gains energy (represented by a positive sign).

  • The energy of the emitted or absorbed photon is exactly equal to the difference between the levels: $\Delta E = E_{final} - E_{initial}$.

• Postulate 3: Quantization of Angular Momentum

  • An electron can only exist in an orbit where its angular momentum ($L = mvr$) is an integral multiple of $\frac{h}{2\pi}$.

  • Formula: $mvr = \frac{nh}{2\pi}$, where $n$ is the principal quantum number (integer $1, 2, 3, \dots$).

  • Value of $\frac{h}{2\pi}$: Approximately $1.05 \times 10^{-34}\,J\,s$.

Mathematical Derivations: Radius and Speed

• Calculating the Radius of the $n$-th Orbit ($r_n$):

  • Bohr linked the Coulombic force of attraction to the centripetal force required for circular motion.

  • Coulomb Force: $F = \frac{k q_1 q_2}{r^2} = \frac{Ze^2}{4\pi\epsilon_0 r^2}$, where $Z$ is the atomic number (number of protons).

  • Centripetal Force: $F_c = \frac{mv^2}{r}$.

  • Equating both and substituting Bohr's angular momentum condition leads to the radius formula:     $r_n = n^2 \times r_0$

  • Bohr Radius ($r_0$): The radius of the first orbit of Hydrogen ($n=1, Z=1$).     $r_0 = 0.53\,\text{Å} = 0.53 \times 10^{-10}\,m$

  • Implication: The radius of orbits increases by the square of the shell number ($n^2$). The 2nd orbit is $4$ times wider than the 1st; the 3rd is $9$ times wider; the 4th is $16$ times wider.

• Calculating the Speed of the Electron ($v_n$):

  • Using $mvr = \frac{nh}{2̀̀̀̀̀̀̀̀̀̀̀̀̀̀̀̀̀̀̀̀2\pi}$, and substituting the derived radius, we find the speed:     $v_n = \frac{v_1}{n}$

  • Speed in the 1st Shell ($v_1$): $2.2 \times 10^6\,m/s$.

  • Inverse Relationship: Speed is inversely proportional to the shell number ($v \propto \frac{1}{n}$). As the shell number increases, the speed of the electron decreases.

Questions & Discussion

• Q: How can we tell if a specific location is an allowed orbit?

  • A: Calculate the angular momentum at that spot. If the result divided by $\frac{h}{2\pi}$ ($1.05 \times 10^{-34}\,J\,s$) yields an integer, it is a valid shell. For example, an angular momentum of $5.25 \times 10^{-34}\,J\,s$ divided by $1.05 \times 10^{-34}$ equals $5$, meaning it is the $5$-th allowed shell.

• Q: To which atoms does the Bohr model apply?

  • A: It applies to "Hydrogen-like atoms" which possess only one electron in their shell. This includes:

  • Hydrogen ($H$)

  • Singly charged Helium Ion ($He^+$)

  • Doubly charged Lithium Ion ($Li^{2+}$)

• Q: What is the ratio of the radius of the 7th orbit to the 4th orbit?

  • A: Since $r \propto n^2$, the ratio is $\frac{R_7}{R_4} = \frac{7^2}{4^2} = \frac{49}{16}$.

• Q: Calculate the speed of an electron in the 2nd shell ($n=2$).

  • A: $v_2 = \frac{2.2 \times 10^6}{2} = 1.1 \times 10^6\,m/s$.

• Q: What is the constant value for Bohr's Radius?

  • A: $0.53\,\text{Å}$. In meters, it is $0.53 \times 10^{-10}\,m$.