Rutherford's Nuclear Model of Atom

 

Experimental design of α-particle scattering on gold foil

Lord Rutherford realized that the information regarding the internal structure of the matter could be revealed by bombarding matter with high speed α-particles. The α-particles are doubly ionized helium atoms, i.e. He²⁺. Geiger and Marsden carried out the experiment under the suggestion of Rutherford by bombarding a very thin foil (thickness of the order of 10^-7 m) of gold (Z = 79) with a narrow beam of α-particles of high energy incoming from radium, a radioactive material. Behind the target foil, fluorescent screen was placed to detect the fate of α-particles after bombardment. It was found that close to 99% of the incident particles passed the target gold foil being undeflected. Only a very small fraction got deflected by quite large angles and a very few of them were deflected by 90^o or more.

Deflection by these large angles could not be explained from Thomson's atomic model in which positive charges were assumed to remain uniformly distributed throughout the volume of atomic dimension. The massive positive charge matter is responsible to deflect He²⁺. Hence, if the model were correct we could have the observations as:

(i) The fate of all the incident α-particles would be the same as the interacting positive charge matter is uniformly distributed throughout the space.

(ii) The α-particles should experience only a very weak electrostatic repulsive barrier from the positive charge matter, as it is diluted due to its uniform distribution throughout the volume. Such with interactions can never produce deflection by large angles such as 90° or more.

The Thomson's atomic model could not survive in the experimental verification and Rutherford proposed his nuclear model in 1911 to explain the experimental observations. According to this model:

(i) All the positive charges of an atom along with most of the atomic mass are concentrated within a very small central region of atom. The central region is called the nucleus.

(ii) The electrons which balance the positive charge a revolving around the nucleus in different circular orbits just like the planets around the sun. Due to the revolving nature, they acquire a centrifugal force (= mu²/r, where m and u are the mass and velocity of electron respectively, and r is the radius of the orbit) which balances is the electrostatic attractive force towards the nucleus. Thus it saves them from falling in the nucleus and ensures the stability of the atom.

The proposed model explains the observations of α-particles scattering. When the α-particles collide with the nucleus or pass through the vicinity of the nucleus which occupies only a very small region in the whole atomic dimension are deflected by large angles. In this model, most of the space is vacant and transparent to the α-particles. The negatively charged very light particles, i.e. the electrons, have nothing to do with the high energy massive α-particles. They fail to attract the high energy massive α-particles.

Here it is worth mentioning that the foil must be sufficiently thin so that,

(i) cross sections of adjacent nuclei do not overlap,

(ii) an α-particle suffers its entire deflection from an encounter with a single nucleus,

(iii) it does not reduce the velocity of the α-particle which traverses it.

In the treatment, it is assumed that the scattering nucleus remains at rest and this approximation is only reasonably valid for the heavier nucleus like that of gold.

Also, there is a reason why gold foil is chosen for this particular experiment. The reasons are:

Condition of malleability: In the α-particle scattering experiment, the foil was required to be very thin (10^-7 m). The reasons behind the requirement of such a thin foil have been explained already. To make such a thin foil, the material must be highly malleable. The metals which possess the fcc structure are highly malleable because of the existence of large number of slip systems. Gold having the fcc structure satisfies the condition of malleability.

Condition of massive scattering nucleus: During the interaction between the approaching α-particle and scattering nucleus, the scattering nucleus must be sufficiently massive to satisfy the condition.

Condition of high positive charge of the scattering nucleus: To detect the α-particles, by a large angle the scattering nucleus should have high positive charge and this condition is satisfied for gold having Z = 79.

 Rutherford's theory of α-particle scattering:

Trajectory of α-particle Scattering

An α-particle of mass m_a and charge q approaching with a velocity u along AO, when a nucleus of charge Ze is placed at D. Let the position of the particle α-particle at a particular instant at A be denoted by (r, Ф) where r gives the distance between the particle and the nucleus. The repulsive force is given by

F = Zeq/4πε₀r² (SI unit)

F = Zeq/ r², (CGS unit)

It can be mathematically proved that under the repulsive force, the α-particle will move on a hyperbolic path along ABC with nucleus at the focus D. If the perpendicular distance (i.e. shortest distance) between the nucleus (D) and the original line of approach (AO) of the α-particle is given by b, called impact parameter, then we have:

Cot(θ/2) = 4πε₀m_au²b/Zeq (SI unit)

              = m_au²b/Zeq (CGS unit)

The above equations indicate that the value of impact parameter (b) increases as θ decreases. In fact, when b→0, then θ→180^o, i.e., the particle moving in the direction of the nucleus will be deflected back in the same direction.

So it appears that, all the α-particles approaching a target nucleus with an impact parameter in the range zero to b, will be deflected by an angle ≥ θ. The circular area (πb²) of radius b is called the cross- section for interaction. Any particle approaching towards the nucleus within this area will suffer deflection by θ or more. If, the thickness of the foil target is t, and the number of nuclei is n per unit volume then the number of nuclei present in the cylinder of length t and cross-section πb² is πb²nt. As we are dealing with a very large number of α-particles, it is reasonably expected that the fraction (f) of the α-particles deflected by θ or more is given by:

f = πb²nt

Nuclear Dimension from the α-particle Scattering Experiment

Impact parameter and α-particle Scattering

It is known that the α-particle follows the trajectory ABC of a hyperbola due to the electrostatic repulsive force between the target nucleus and the positively charged α-particle. The nucleus D is placed at the outer focus. Finally, the scattering α-particle follows the asymptote OC, after suffering a deflection by θ, the angle between the initial path AO which is the path of asymptote of the hyperbolic trajectory and the final path OC.

The shortest distance between the nucleus D and the original path AO of the particle is called impact parameter (b) which has been already defined. The physical significance of b is that all the α-particles lying in the cross section of area πb² will suffer a deflection by θ or more. When b becomes zero, θ becomes 180^o. Under this condition the kinetic energy of the α-particle is equal to the electrostatic repulsive energy at the closest distance r₀ between the α-particle and the target nucleus.

Hence,

1/2 m_au² = Zeq/4πε₀r₀

 r₀ = Zeq/[4πε₀(1/2)m_au²]

If the maximum kinetic energy (=1/2 m_au²) is 7.7 MeV (7.7 MeV = 7.7 × 10⁶ eV × 1.6 × 10^-19 joules/eV = 1.23 ×10^-12 joules; ε₀ = 8.854 × 10^-12; q = 2e for α-particles), then by substituting these values we get, r₀ = 3.8 × 10^-16 Zm. For the gold foil, Z = 79, and r₀ = 3.8 × 10^-16 ×79 m, as r₀ gives the minimum possible distance between the α-particle (of 7.7 MeV) and then the target nucleus. It indicates that the nucleus is about 1/10⁴ th of the dimension of the atom.

Drawbacks of Rutherford's Model:

(i) Instability of the Rutherford atom: Though it was attempted to explain the stability of the atoms by considering to strong foundation stones of the classical physics, i.e. Newton's law of motion and Coulomb's law of electrostatic force, it could not be supported by the electromagnetic theory, a revolving electron under an acceleration will radiate its energy in the form of electromagnetic radiation. Thus by losing its energy, it will gradually spiral down into the nucleus and the atom will get destroyed. As a matter of fact, if it occurs so, it can be shown mathematically that the revolving electron will require about one hundred-millionth of a second to get plunged into the nucleus.

Fate of the revolving electron according to electromagnetic theory in Rutherford model

(ii) No explanation of atomic line spectra: From Rutherford's model, according to the electromagnetic theory, the energy radiated in the form of electromagnetic radiation from the revolving electron should lead to a continuous spectrum covering all frequencies in a certain range. But experimentally, some well-defined spectral lines corresponding to certain frequencies are noted instead of continuous spectrum.

Reference

1) Concise inorganic chemistry by J. D. Lee.

2) Inorganic Chemistry by James E. Huheey, Ellen A Keither, Richard L. Keither, Okhil K. Medhi.

3) Shriver and Atkins Inorganic Chemistry.