Rutherford's Model and Atomic Spectra

Alpha-Particle Scattering And Rutherford’S Nuclear Model Of Atom

Before the early 20th century, the prevailing model of the atom was J.J. Thomson's "plum pudding model," which suggested that the atom was a sphere of positive charge with negatively charged electrons embedded within it, like plums in a pudding. This model was challenged by the results of experiments conducted by Ernest Rutherford and his colleagues, Hans Geiger and Ernest Marsden, around 1909-1911. Their experiments, involving the scattering of alpha ($\alpha$) particles by thin metal foils, led to the development of the nuclear model of the atom.

Geiger-Marsden (Alpha-Particle Scattering) Experiment

Setup: A beam of high-speed alpha particles (which are positively charged helium nuclei, $_2^4He^{++}$) emitted from a radioactive source (like Radium or Polonium) was directed onto a very thin gold foil (thickness about $10^{-8}$ m). Alpha particles were chosen because they are relatively heavy and carry a double positive charge, making them suitable probes to interact with the charged particles within the atom. The scattered alpha particles were detected by a movable detector consisting of a zinc sulphide (ZnS) screen and a microscope. When an alpha particle struck the ZnS screen, it produced a tiny flash of light (scintillation), which could be observed through the microscope. By moving the detector, the number of alpha particles scattered at different angles could be counted.

Schematic diagram of the alpha-particle scattering experiment.

Observations of Alpha-Particle Scattering

Geiger and Marsden expected that the alpha particles, being energetic and massive compared to electrons, would pass straight through the thin gold foil with little or no deflection, consistent with the plum pudding model where the positive charge was thought to be spread out uniformly. However, the results were quite surprising:

Most alpha particles passed straight through the gold foil undeflected. This indicated that most of the space within an atom is empty.
A small fraction of the alpha particles were deflected through significant angles. Some were scattered at angles greater than $90^\circ$.
A very small fraction (about 1 in 8000 for gold) of the alpha particles were scattered backwards, i.e., deflected through angles greater than $150^\circ$ or even $180^\circ$. This was a completely unexpected result. Rutherford famously described this as being "as incredible as if you had fired a 15-inch shell at a piece of tissue paper and it came back and hit you."

Rutherford’s Nuclear Model of Atom

Based on these scattering results, Rutherford proposed a new model of the atom, called the nuclear model or Rutherford's model, in 1911. This model explained the observations by suggesting that:

Most of the mass and all of the positive charge of an atom is concentrated in a very small region at the center, called the nucleus. The nucleus is extremely dense.
The electrons, which are negatively charged, revolve around the nucleus in orbits, much like planets revolve around the Sun.
Most of the space within the atom is empty, allowing most alpha particles to pass straight through.

The large deflections and backscattering of alpha particles were explained as being due to the strong electrostatic repulsion between the positively charged alpha particles and the concentrated positive charge of the nucleus. A head-on collision with the nucleus would cause an alpha particle to be scattered backwards. Particles passing close to the nucleus would be deflected through large angles, while those passing far away would experience negligible deflection.

Alpha-Particle Trajectory (Impact Parameter)

The trajectory of an alpha particle scattered by a nucleus in Rutherford's model is determined by the Coulomb force between the alpha particle and the nucleus. The shape of the trajectory is a hyperbola. The scattering angle depends on the initial velocity of the alpha particle, the charge of the nucleus, and a parameter called the impact parameter.

The impact parameter ($b$) is defined as the perpendicular distance of the initial velocity vector of the alpha particle from the center of the nucleus, when the particle is far away from the nucleus and its velocity is directed towards the nucleus.

Diagram illustrating the trajectory of an alpha particle and the impact parameter

Trajectory of an alpha particle scattered by a nucleus, showing the impact parameter $b$ and scattering angle $\theta$.

Relationship between scattering angle ($\theta$) and impact parameter ($b$):

For a large impact parameter ($b$), the alpha particle passes far from the nucleus and experiences a weak repulsive force. The scattering angle is small ($\theta \approx 0$).
For a small impact parameter ($b$), the alpha particle passes close to the nucleus and experiences a strong repulsive force. The scattering angle is large.
For a head-on collision ($b = 0$), the alpha particle is scattered backwards ($\theta = 180^\circ$).

The number of alpha particles scattered at a particular angle is related to the range of impact parameters leading to that angle. Rutherford derived a formula for the number of scattered particles as a function of scattering angle, which agreed well with the experimental data, confirming the validity of the nuclear model.

Electron Orbits

In Rutherford's model, electrons are proposed to revolve around the nucleus in orbits. The electrostatic attraction between the positively charged nucleus and the negatively charged electrons provides the necessary centripetal force for circular motion.

Consider an electron of mass $m_e$ and charge $-e$ revolving around a nucleus of charge $+Ze$ (where $Z$ is the atomic number) in a circular orbit of radius $r$ with velocity $v$.

The electrostatic force of attraction between the electron and the nucleus is $F_e = \frac{1}{4\pi\epsilon_0} \frac{(Ze)(e)}{r^2}$. This force provides the centripetal force $F_c = \frac{m_e v^2}{r}$.

$ \frac{1}{4\pi\epsilon_0} \frac{Ze^2}{r^2} = \frac{m_e v^2}{r} $

From this, the kinetic energy of the electron is $K = \frac{1}{2} m_e v^2 = \frac{1}{8\pi\epsilon_0} \frac{Ze^2}{r}$.

The potential energy of the electron in the field of the nucleus is $U = \frac{1}{4\pi\epsilon_0} \frac{(Ze)(-e)}{r} = -\frac{1}{4\pi\epsilon_0} \frac{Ze^2}{r}$.

The total energy $E$ of the electron in the orbit is the sum of kinetic and potential energies:

$ E = K + U = \frac{1}{8\pi\epsilon_0} \frac{Ze^2}{r} - \frac{1}{4\pi\epsilon_0} \frac{Ze^2}{r} $

$ E = -\frac{1}{8\pi\epsilon_0} \frac{Ze^2}{r} $

The total energy is negative, indicating that the electron is bound to the nucleus.

Drawbacks of Rutherford's Model

Despite successfully explaining the scattering experiments and proposing the nuclear model, Rutherford's model had significant limitations based on classical physics:

Stability of the Atom: According to classical electromagnetic theory (Maxwell's equations), an electron revolving in a circular orbit is an accelerating charge. An accelerating charge should continuously radiate energy in the form of electromagnetic waves. As the electron loses energy, its orbit should spiral inwards, and it should eventually fall into the nucleus. This would make the atom unstable, which contradicts the observed stability of atoms.
Atomic Spectra: If the electron continuously loses energy, its angular velocity and orbital radius would change continuously. Consequently, the frequency of the radiated light would also change continuously, leading to a continuous spectrum of emitted radiation. However, atoms are observed to emit light at specific, discrete wavelengths, producing line spectra. Rutherford's model could not explain these observed line spectra.

These drawbacks highlighted the need for a new model of the atom based on quantum principles, which was later developed by Niels Bohr.

Atomic Spectra (Emission and Absorption)

When atoms are heated or excited by other means (like electric discharge), they emit light. When this emitted light is passed through a prism or spectrometer, it is observed to consist of a set of discrete, bright lines on a dark background, each corresponding to a specific wavelength (or frequency). This is called an emission spectrum. Each element has a unique emission spectrum, acting like its fingerprint.

Diagram showing the emission spectrum of hydrogen

Line emission spectrum of hydrogen.

Conversely, when white light (containing all wavelengths) is passed through a gas or vapour of an element, and the transmitted light is analysed, a spectrum is observed with dark lines appearing at specific wavelengths against a continuous background. These dark lines correspond to the wavelengths of light that have been absorbed by the atoms of the gas. This is called an absorption spectrum. The wavelengths of the dark lines in the absorption spectrum of an element are found to be exactly the same as the wavelengths of the bright lines in its emission spectrum.

Spectral Series (Lyman, Balmer, Paschen, etc.)

The atomic spectrum of the simplest atom, hydrogen, consists of several series of lines. The wavelengths of the lines in each series can be described by empirical formulas. The first series to be discovered in the visible region was the Balmer series.

Johannes Rydberg, in 1890, developed a general formula to describe the wavelengths of all the spectral lines of hydrogen:

$ \frac{1}{\lambda} = R_H \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) $

Where:

$\lambda$ is the wavelength of the emitted or absorbed line.
$R_H$ is the Rydberg constant for hydrogen ($R_H \approx 1.097 \times 10^7 \, m^{-1}$).
$n_1$ and $n_2$ are positive integers ($n_1 < n_2$).

Different spectral series correspond to different values of $n_1$:

Lyman series: $n_1 = 1$, $n_2 = 2, 3, 4, ...$. These lines are in the ultraviolet (UV) region. This series corresponds to electron transitions ending in the ground state ($n=1$).
Balmer series: $n_1 = 2$, $n_2 = 3, 4, 5, ...$. These lines are in the visible region (H-alpha, H-beta, etc.). This series corresponds to electron transitions ending in the first excited state ($n=2$).
Paschen series: $n_1 = 3$, $n_2 = 4, 5, 6, ...$. These lines are in the infrared (IR) region. This series corresponds to electron transitions ending in the second excited state ($n=3$).
Brackett series: $n_1 = 4$, $n_2 = 5, 6, 7, ...$. These lines are in the infrared (IR) region.
Pfund series: $n_1 = 5$, $n_2 = 6, 7, 8, ...$. These lines are in the infrared (IR) region.
And so on for $n_1 = 6, 7, ...$

Diagram showing the energy levels of hydrogen and the spectral series transitions

Energy levels of hydrogen atom and transitions corresponding to different spectral series.

The existence of these discrete spectral lines and series was a mystery in classical physics. It suggested that atoms could only emit or absorb energy in specific amounts, corresponding to transitions between distinct energy states. This observation was a key motivation for the development of quantum models of the atom, starting with Bohr's model.

Example 1. Calculate the wavelength of the first line (longest wavelength) in the Balmer series of the hydrogen spectrum. (Rydberg constant $R_H = 1.097 \times 10^7 \, m^{-1}$).

Answer:

Given:

Rydberg constant, $R_H = 1.097 \times 10^7 \, m^{-1}$

For the Balmer series, $n_1 = 2$. The first line corresponds to the transition from the lowest possible higher energy level, which is $n_2 = 3$.

We use the Rydberg formula: $ \frac{1}{\lambda} = R_H \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) $

Substitute $n_1 = 2$ and $n_2 = 3$:

$ \frac{1}{\lambda} = (1.097 \times 10^7 \, m^{-1}) \left(\frac{1}{2^2} - \frac{1}{3^2}\right) $

$ \frac{1}{\lambda} = (1.097 \times 10^7) \left(\frac{1}{4} - \frac{1}{9}\right) \, m^{-1} $

$ \frac{1}{\lambda} = (1.097 \times 10^7) \left(\frac{9 - 4}{36}\right) \, m^{-1} $

$ \frac{1}{\lambda} = (1.097 \times 10^7) \left(\frac{5}{36}\right) \, m^{-1} $

$ \frac{1}{\lambda} = \frac{5.485}{36} \times 10^7 \, m^{-1} \approx 0.15236 \times 10^7 \, m^{-1} $

$ \frac{1}{\lambda} \approx 1.5236 \times 10^6 \, m^{-1} $

Now, find the wavelength $\lambda = 1 / (1/\lambda)$:

$ \lambda = \frac{1}{1.5236 \times 10^6 \, m^{-1}} \approx 0.6563 \times 10^{-6} \, m $

$ \lambda = 6.563 \times 10^{-7} \, m = 656.3 \, nm $

The wavelength of the first line in the Balmer series (H-alpha line) is approximately 656.3 nm. This is in the red part of the visible spectrum.