Atomic Physics and Bohr's Model Study Notes

Thomson Model of Atom

• Model Postulates:

  • According to the Thomson model, every atom consists of a positively charged sphere of radius of the order of $10^{-10}\,\text{m}$, in which the entire mass and positive charge of the atom are uniformly distributed.
  • Inside this sphere, the electrons are embedded like seeds in a watermelon or like plums in a pudding.
  • The number of electrons is such that their negative charge is equal to the positive charge of the atom. Thus, the atom is electrically neutral.

• Limitations of Thomson Atom Model:

  • It could not explain the origin of spectral series of hydrogen and other atoms observed experimentally.
  • It could not explain the large angle scattering of $\alpha$-particles from thin metal foils, as observed by Rutherford.

Rutherford's $\alpha$-Ray Scattering Experiment

• Experimental Setup:

  • The experimental setup used by Rutherford and his collaborators, Geiger and Marsden, consists of:
  • Radioactive Source 'S': A piece of radioactive source contained inside a thick lead cavity.
  • Lead Collimator: Lead slits used to collimate the emitted $\alpha$-particles into a narrow beam.
  • Gold Foil: A thin gold foil of thickness of the order of $2.1 \times 10^{-7}\,\text{m}$ (or approximately $10^{-8}\,\text{m}$).
  • Rotatable Detector: Consists of a zinc sulphide ($\text{ZnS}$) screen attached to a microscope.
  • The collimated $\alpha$-particle beam is allowed to fall on the gold foil. The scattered $\alpha$-particles produce bright flashes (scintillations) on the $\text{ZnS}$ screen, which are observed and counted through the microscope at various scattering angles $\theta$.
  • Scattering Angle ($\theta$): The angle of deviation of an $\alpha$-particle from its original direction of motion.

• Observations:

  • Most of the $\alpha$-particles pass straight through the gold foil without suffering any collision or deviation. Only about $0.14\%$ of incident $\alpha$-particles scatter by more than $1^\circ$.
  • About $1$ in every $8000$ $\alpha$-particles is deflected by an angle greater than $90^\circ$ (some even retrace their path at $180^\circ$).

Distance of Closest Approach ($r_0$)

• Concept:

  • When an $\alpha$-particle approaches a target nucleus directly along the center line, its Kinetic Energy ($\text{K.E.}$) continuously decreases due to electrostatic repulsion, while its Electrical Potential Energy ($\text{P.E.}$) increases.
  • At a specific distance $r_0$ from the nucleus, the kinetic energy of the $\alpha$-particle reduces to zero. The particle stops momentarily and turns back through $180^\circ$.
  • This minimum distance $r_0$ is defined as the distance of closest approach.

• Derivation:

  • Electrical potential at distance $r_0$ due to a nucleus with atomic number $Z$:     $V = \frac{1}{4\pi\varepsilon_0} \frac{Ze}{r_0}$
  • Potential energy of an $\alpha$-particle (charge $q = 2e$) at distance $r_0$:     $\text{P.E.} = V \times (2e) = \frac{1}{4\pi\varepsilon_0} \frac{2Ze^2}{r_0}$
  • Initial kinetic energy of an $\alpha$-particle of mass $m$ moving with velocity $v$:     $\text{K.E.} = \frac{1}{2} m v^2$
  • Conservation of energy at the point of distance of closest approach ($\text{K.E.} = \text{P.E.}$):     $\frac{1}{2} m v^2 = \frac{1}{4\pi\varepsilon_0} \frac{2Ze^2}{r_0}$$r_0 = \frac{1}{4\pi\varepsilon_0} \frac{4Ze^2}{m v^2} = \frac{1}{4\pi\varepsilon_0} \frac{2Ze^2}{\left(\frac{1}{2} m v^2\right)}$

Impact Parameter ($b$)

• Definition:

  • Impact parameter ($b$) is defined as the perpendicular distance of the initial velocity vector of the $\alpha$-particle from the central line of the nucleus when the particle is far away from the atom.

• Dependence of Scattering Angle on $b$:

  • When the impact parameter $b$ is large, the repulsive force is weak, and the $\alpha$-particle deviates through a small scattering angle $\theta$.
  • When the impact parameter $b$ is small, the repulsive force is strong, causing the $\alpha$-particle to scatter through a large angle $\theta$.
  • For a head-on collision ($b = 0$), the $\alpha$-particle rebounds back ($\theta = 180^\circ$).

• Analytical Relation:

  • Rutherford derived the relation between impact parameter $b$ and scattering angle $\theta$ as:     $b = \frac{1}{4\pi\varepsilon_0} \frac{Ze^2 \cot\left(\frac{\theta}{2}\right)}{\text{K.E.}} = \frac{1}{4\pi\varepsilon_0} \frac{Ze^2 \cot\left(\frac{\theta}{2}\right)}{\frac{1}{2} m v^2}$

Rutherford's Atom Model

• Key Features:
  1. Central Nucleus: Every atom consists of a tiny central core called the atomic nucleus, containing all the positive charge and almost the entire mass of the atom.
  2. Size Comparison: The nuclear radius is of the order of $10^{-15}\,\text{m}$, which is extremely small compared to the overall size of the atom ($10^{-10}\,\text{m}$).
  3. Charge Neutrality: The nucleus is surrounded by electrons such that total negative charge of electrons equals total positive charge of the nucleus, making the atom electrically neutral.
  4. Planetary Motion: Electrons revolve around the nucleus in circular orbits. The required centripetal force is supplied by the electrostatic force of attraction between the positive nucleus and negative electrons.

Energy of Electron in Rutherford's Orbit

• Derivation for Hydrogen Atom ($Z = 1$):
  • Equating centripetal force ($F_c$) to electrostatic attraction force ($F_e$):     $F_c = F_e$$\frac{m v^2}{r} = \frac{1}{4\pi\varepsilon_0} \frac{e \cdot e}{r^2} = \frac{e^2}{4\pi\varepsilon_0 r^2}$$m v^2 = \frac{e^2}{4\pi\varepsilon_0 r}$
  • Kinetic Energy ($\text{K.E.}$):     $\text{K.E.} = \frac{1}{2} m v^2 = \frac{e^2}{8\pi\varepsilon_0 r}$
  • Potential Energy ($U$):     $U = \frac{1}{4\pi\varepsilon_0} \frac{(e)(-e)}{r} = -\frac{e^2}{4\pi\varepsilon_0 r}$
  • Total Energy ($E$):     $E = \text{K.E.} + U = \frac{e^2}{8\pi\varepsilon_0 r} - \frac{e^2}{4\pi\varepsilon_0 r} = -\frac{e^2}{8\pi\varepsilon_0 r}$
  • Physical Meaning: The total energy is negative, proving that the electron is bound to the nucleus and cannot leave its orbit freely.

Limitations of Rutherford Atom Model

• Inability to Explain Atomic Stability:

  • According to classical electromagnetic theory, an accelerating charge (revolving electron) continuously radiates energy as electromagnetic waves. Losing energy causes the orbit radius to decrease continuously, forcing the electron to spiral inwards and collapse into the nucleus. However, atoms are stable.

• Inability to Explain Line Spectra:

  • As the spiraling electron's orbital radius decreases, its angular velocity and frequency of revolution change continuously. Hence, the atom should emit a continuous spectrum of light, whereas experimental observation shows discrete line spectra.

Bohr Model of Hydrogen Atom

• Postulate 1 (Circular Orbits & Centripetal Force):

  • An atom consists of a central nucleus containing all positive charge and nearly all mass. Electrons revolve in circular orbits where centripetal force is provided by Coulomb electrostatic attraction:     $\frac{m v^2}{r} = \frac{1}{4\pi\varepsilon_0} \frac{(Ze)(e)}{r^2} = \frac{K Z e^2}{r^2}$     where $K = \frac{1}{4\pi\varepsilon_0}$.

• Postulate 2 (Bohr Orbital Quantization Condition):

  • Electrons can revolve only in specific non-radiating circular orbits called stationary orbits. In these orbits, the orbital angular momentum ($L$) of the electron is an integral multiple of $\frac{h}{2\pi}$:     $L = m v r = \frac{n h}{2\pi}, \quad n = 1, 2, 3, \dots$     where $n$ is the principal quantum number and $h$ is Planck's constant. Electrons in non-radiating orbits do not radiate energy.

• Postulate 3 (Energy Transition & Frequency Condition):

  • Emission or absorption of energy occurs only when an electron jumps from one non-radiating stationary orbit to another. The frequency $\nu$ of the emitted or absorbed photon is:     $h \nu = E_2 - E_1$     where $E_2$ is the energy of the higher outer orbit and $E_1$ is the energy of the lower inner orbit.

Derivation of Radius of Bohr Orbit in Hydrogen Atom

• Step-by-Step Derivation:
  • From force balance condition:     $\frac{m v^2}{r} = \frac{K Z e^2}{r^2} \implies r = \frac{K Z e^2}{m v^2} \quad \text{--- (1)}$
  • From Bohr's angular momentum quantization condition:     $m v r = \frac{n h}{2\pi} \implies v = \frac{n h}{2\pi m r} \quad \text{--- (2)}$
  • Substituting equation (2) into equation (1):     $r = \frac{K Z e^2}{m \left(\frac{n h}{2\pi m r}\right)^2} = \frac{K Z e^2}{m \frac{n^2 h^2}{4\pi^2 m^2 r^2}} = \frac{4\pi^2 m r^2 K Z e^2}{n^2 h^2}$$1 = \frac{4\pi^2 m r K Z e^2}{n^2 h^2}$$r = \frac{n^2 h^2}{4\pi^2 m K Z e^2}$
  • For Hydrogen atom ($Z = 1$):     $r = \frac{n^2 h^2}{4\pi^2 m K e^2}$
  • Since $h, \pi, m, K, e$ are constant properties, $r \propto n^2$.
  • The radii of consecutive stationary orbits in hydrogen are in the ratio $1 : 4 : 9 : 16 : 25 : \dots$

Derivation of Velocity of Electron in Bohr Orbit

• Step-by-Step Derivation:
  • From electrostatic force balance:     $r = \frac{K Z e^2}{m v^2} \quad \text{--- (1)}$
  • From quantization condition:     $r = \frac{n h}{2\pi m v} \quad \text{--- (2)}$
  • Comparing equation (1) and equation (2):     $\frac{K Z e^2}{m v^2} = \frac{n h}{2\pi m v}$$v = \frac{2\pi K Z e^2}{n h}$
  • For Hydrogen atom ($Z = 1$):     $v = \frac{2\pi K e^2}{n h}$
  • Relationship with principal quantum number: $v \propto \frac{1}{n}$.
  • For ground state ($n = 1$), $v = \frac{c}{137}$ where $c$ is the speed of light.

Derivation of Total Energy of Electron in Bohr Orbit

• Step-by-Step Derivation:

  • Equating forces gives $m v^2 = \frac{K Z e^2}{r}$.
  • Kinetic Energy ($\text{K.E.}$):     $\text{K.E.} = \frac{1}{2} m v^2 = \frac{1}{2} \frac{K Z e^2}{r}$
  • Potential Energy ($U$):     $U = \frac{K (-e)(Z e)}{r} = -\frac{K Z e^2}{r}$
  • Total Energy ($E$):     $E = \text{K.E.} + U = \frac{1}{2} \frac{K Z e^2}{r} - \frac{K Z e^2}{r} = -\frac{1}{2} \frac{K Z e^2}{r} \quad \text{--- (3)}$
  • Substitute Bohr radius $r = \frac{n^2 h^2}{4\pi^2 m K Z e^2}$ into equation (3):     $E = -\frac{1}{2} K Z e^2 \left(\frac{4\pi^2 m K Z e^2}{n^2 h^2}\right) = -\frac{2\pi^2 m K^2 Z^2 e^4}{n^2 h^2}$
  • For Hydrogen atom ($Z = 1$):     $E = -\frac{2\pi^2 m K^2 e^4}{n^2 h^2}$
  • Substituting numerical values of physical constants ($m, e, K, h$):     $E = -\frac{13.6}{n^2}\,\text{eV}$

• Energies of Stationary States in Hydrogen:

  • Ground State ($n = 1$): $E_1 = -\frac{13.6}{1^2} = -13.6\,\text{eV}$
  • First Excited State ($n = 2$): $E_2 = -\frac{13.6}{2^2} = -3.4\,\text{eV}$
  • Second Excited State ($n = 3$): $E_3 = -\frac{13.6}{3^2} = -1.51\,\text{eV}$
  • Third Excited State ($n = 4$): $E_4 = -\frac{13.6}{4^2} = -0.85\,\text{eV}$
  • Fifth State ($n = 5$): $E_5 = -\frac{13.6}{5^2} = -0.54\,\text{eV}$
  • State Nomenclature:
  • $n = 1$: Ground state / 1st energy state
  • $n = 2$: 1st excited state / 2nd energy state
  • $n = 3$: 2nd excited state / 3rd energy state
  • $n = 4$: 3rd excited state / 4th energy state

Hydrogen Spectral Lines & Rydberg Formula

• Origin of Spectral Lines:
  • Room-temperature hydrogen atoms reside in ground state ($n = 1$). When energy is absorbed via heating or collisions, electrons jump to outer excited states ($n_2 > n_1$).
  • Excited states are unstable, so electrons fall back to lower orbits ($n_1$), emitting photons of energy h\nu = E_2 - E_1$.\n\n* **Mathematical Derivation of Wave Number:**\n  h u = E_2 - E_1 = -\frac{2\pi^2 m K^2 Z^2 e^4}{n_2^2 h^2} - \left(-\frac{2\pi^2 m K^2 Z^2 e^4}{n_1^2 h^2}\right)\n  h u = \frac{2\pi^2 m K^2 Z^2 e^4}{h^2} \left[\frac{1}{n_1^2} - \frac{1}{n_2^2}\right]\n  Since u = \frac{c}{\lambda}:\n  \frac{h c}{\lambda} = \frac{2\pi^2 m K^2 Z^2 e^4}{h^2} \left[\frac{1}{n_1^2} - \frac{1}{n_2^2}\right]\n  \bar{ u} = \frac{1}{\lambda} = \frac{2\pi^2 m K^2 Z^2 e^4}{c h^3} \left[\frac{1}{n_1^2} - \frac{1}{n_2^2}\right]\n  where \bar{ u} = \frac{1}{\lambda} is the wave number (number of waves per unit length).\n  Rydberg Constant (R):\n  R = \frac{2\pi^2 m K^2 e^4}{c h^3} = 1.097 \times 10^7\,\text{m}^{-1}\n  Rydberg Formula for Hydrogen (Z = 1):\n  \bar{ u} = \frac{1}{\lambda} = R \left[\frac{1}{n_1^2} - \frac{1}{n_2^2}\right]\n  Frequency equation:\n   u = R c \left[\frac{1}{n_1^2} - \frac{1}{n_2^2}\right]\n\n# Spectral Series of Hydrogen Atom\n\n* **1. Lyman Series:**\n * Transition: Lower level n_1 = 1$, upper levels$n_2 = 2, 3, 4, \dots\n * Formula: \bar{ u} = \frac{1}{\lambda} = R \left[\frac{1}{1^2} - \frac{1}{n_2^2}\right]\n * Spectral Region: **Ultraviolet (UV) region**.\n\n* **2. Balmer Series:**\n * Transition: Lower level n_1 = 2$, upper levels$n_2 = 3, 4, 5, \dots\n * Formula: \bar{ u} = \frac{1}{\lambda} = R \left[\frac{1}{2^2} - \frac{1}{n_2^2}\right]\n * Spectral Region: **Visible region**.\n\n* **3. Paschen Series:**\n * Transition: Lower level n_1 = 3$, upper levels$n_2 = 4, 5, 6, \dots\n * Formula: \bar{ u} = \frac{1}{\lambda} = R \left[\frac{1}{3^2} - \frac{1}{n_2^2}\right]\n * Spectral Region: **Infrared region**.\n\n* **4. Brackett Series:**\n * Transition: Lower level n_1 = 4$, upper levels$n_2 = 5, 6, 7, \dots\n * Formula: \bar{ u} = \frac{1}{\lambda} = R \left[\frac{1}{4^2} - \frac{1}{n_2^2}\right]\n * Spectral Region: **Infrared region**.\n\n* **5. Pfund Series:**\n * Transition: Lower level n_1 = 5$, upper levels$n_2 = 6, 7, 8, \dots\n * Formula: \bar{ u} = \frac{1}{\lambda} = R \left[\frac{1}{5^2} - \frac{1}{n_2^2}\right]\n * Spectral Region: **Infrared region**.\n\n# Energy Level Diagram\n\n* **Characteristics:**\n * Represents total energies of electrons in different stationary orbits using parallel horizontal lines.\n * Values: E_1 = -13.6\,\text{eV}$,$E_2 = -3.4\,\text{eV}$,$E_3 = -1.51\,\text{eV}$,$E_4 = -0.85\,\text{eV}$,$E_5 = -0.54\,\text{eV}$,$E_6 = -0.37\,\text{eV}$,$\dots$,$E_\infty = 0\,\text{eV}.\n * As n increases, energy values become less negative and energy levels spacing comes closer together.\n * At n = \infty$,$E_\infty = 0\,\text{eV}, representing an electron completely removed from the atom at rest.\n\n# Worked Numerical Examples\n\n* **Calculation of Wavelength of H_\alpha Line in Balmer Series:**\n * Given: R = 1.097 \times 10^7\,\text{m}^{-1}$,$n_1 = 2$,$n_2 = 3\n * Calculation:\n    \frac{1}{\lambda} = R \left[\frac{1}{2^2} - \frac{1}{3^2}\right] = 1.097 \times 10^7 \times \left[\frac{1}{4} - \frac{1}{9}\right] = 1.097 \times 10^7 \times \frac{5}{36}\n    \frac{1}{\lambda} = 0.1524 \times 10^7\,\text{m}^{-1}\n    \lambda = \frac{1}{0.1524 \times 10^7} = 6563 \times 10^{-10}\,\text{m} = 6563\,\text{\AA}\n\n* **Calculation of de-Broglie Wavelength in n = 2 Orbit:**\n * Given: Ground state radius r_1 = 0.53\,\text{\AA}$, level$n = 2\n * Formula for orbit radius: r_n = n^2 r_1\n * For n = 2$:$r_2 = (2)^2 \times 0.53\,\text{\AA} = 2.12\,\text{\AA}\n * From quantization condition 2\pi r_2 = n \lambda:\n    2\pi (2.12\,\text{\AA}) = 2 \lambda \implies \lambda = 2.12 \pi\,\text{\AA} = 6.66\,\text{\AA}\n\n* **Maximum Number of Spectral Lines in De-excitation:**\n * Formula for maximum number of emitted lines when de-exciting from n_2$to$n_1:\n    \text{Maximum lines} = \frac{(n_2 - n_1)(n_2 - n_1 + 1)}{2}\n * Example for de-excitation from n = 3$to$n = 1:\n    \text{Maximum lines} = \frac{(3 - 1)(3 - 1 + 1)}{2} = \frac{2 \times 3}{2} = 3\n * Emission transitions: 3 \rightarrow 2$,$2 \rightarrow 1$, and$3 \rightarrow 1.\n\n# Excitation and Ionization Definitions\n\n* **Excitation Energy:**\n * The energy required by an electron to jump from its ground state to any excited state.\n * First Excitation Energy of Hydrogen (n = 1 \rightarrow 2$):$E_2 - E_1 = -3.4 - (-13.6) = 10.2\,\text{eV}\n * Second Excitation Energy of Hydrogen (n = 1 \rightarrow 3$):$E_3 - E_1 = -1.51 - (-13.6) = 12.09\,\text{eV}\n\n* **Ionization Energy:**\n * The energy required to completely remove an electron from the atom from its ground state (n = 1 \rightarrow \infty).\n * Ionization Energy of Hydrogen: E_\infty - E_1 = 0 - (-13.6) = 13.6\,\text{eV}\n\n* **Excitation Potential:**\n * The accelerating potential that provides a bombarding electron with enough energy to excite a target atom.\n * First Excitation Potential of Hydrogen: 10.2\,\text{V}\n * Second Excitation Potential of Hydrogen: 12.09\,\text{V}\n\n* **Ionization Potential:**\n * The accelerating potential that provides a bombarding electron with enough energy to ionize a target atom.\n * Ionization Potential of Hydrogen: 13.6\,\text{V}\n\n# Proof of Bohr's Quantization Condition using De-Broglie Hypothesis\n\n* **Derivation:**\n * According to de-Broglie hypothesis, the wavelength \lambda$of a electron moving with velocity$v is:\n    \lambda = \frac{h}{m v} \quad \text{--- (1)}\n * For a electron moving in a circular orbit of radius r to form a stable standing wave, the circumference must contain an integral number of wavelengths:\n    2\pi r = n \lambda \quad \text{--- (2)}\n * Substituting equation (1) into equation (2):\n    2\pi r = n \left(\frac{h}{m v}\right)\n    m v r = \frac{n h}{2\pi}$$
  • This explicitly proves Bohr's orbital quantum condition for non-radiating stable orbits.