Class 26: Rutherford-Bohr atom

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1 Class 6: Rutherford-Bohr atom Atomic structure After the discovery of the electron in 1897 by J. J. Thomson, Thomson and others developed models for the structure of atoms. It was known that atoms contained electrons and were electrically neutral. Also atoms were known to have sizes of about 0.1 nm, and atoms of individual elements were associated with characteristic sets of spectral lines. Thomson s model was a sphere of uniform positive charge density with electrons embedded in it to give overall charge neutrality. The electrons had to be placed such that they were in mechanical equilibrium and had normal modes of vibration with frequencies equal to those of the observed spectral lines. It was found impossible to find equilibrium positions for the electrons. Hence the electrons had to move, and since they had to stay in the atom, accelerate, which would give rise to continuous radiation which is not observed. The Thomson model was replaced by one developed by Rutherford. Rutherford has shown by experiment that uranium emitted at least two types of radiation which he called and β. He then determined by scintillation experiments that the radiation was particles of charge. By capturing particles in an evacuated glass vial, he showed spectroscopically that they were ionized helium atoms. Rutherford realized that these relatively massive particles would be useful for probing the structure of atoms of other elements. Along with Geiger and Marsden, Rutherford carried out a number of experiments in which particles were scattered off atoms of gold in a thin foil. Most of the particles went straight through or were scattered though a small angle of 1 or so. This was not inconsistent with what was expected form the Thomson model of the atom. However, to Rutherford s surprise, some of the particles were scattered through angles greater than 90, which is inconsistent with the Thomson model.

2 Rutherford scattering The problem with the Thomson model is that it gives too small a force to produce a large deflection. However if the positive charge is concentrated in a much smaller region, for near impacts a much larger force results and gives rise to a large deflection. We can estimate the conditions for significant deflection of the particle by comparing its kinetic energy to its potential energy at closest approach to the region of positive charge (which we assume to be much more massive than the particle). Denote the initial velocity of the particle by v, and let the impact parameter, i.e. the distance of closest approach in the absence of the Coulomb force, by b. Let the velocity at closest approach be vc and the distance of closest approach be rc. Conservation of angular momentum about the target gives bv = r v (5.1) c c. Also, we have from conservation of mechanical energy 1 1 Z Zte m v = m vc + k, (5.) r where m is the mass of the particle, Z e is its charge and Zte is the charge of the target. Here k is a constant which depends on the system of units. Solving for rc, we get c where The ratio of potential energy to kinetic energy at closest approach is c c c c r c w w 1, b = + + (5.3) Z Z e w = k. (5.4) m v t b ( ) U kzazte v b rc = = w = w = w w + w + 1. (5.5) K m v r v r b We find the energies are equal when w = 1 8, and r = b. Hence if w 0.3, most of the energy will be kinetic at closest approach, and the deflection will be small. In Rutherford s experiment, the kinetic energy of the particles was about 5 MeV. Setting the distance of closest approach to the size of a gold atom (about 0.14 nm), we find w Hence if the positive charge were spread out over a sphere of atomic radius, the deflection of the particles would be small. A large deflection would require the positive charge to be inside a sphere of radius less than 0.1 atomic radii. c

3 In an exact treatment of scattering of charged particle Rutherford showed that the scattering angle, θ, is related to the properties of the incident particle by In terms of the parameter w this is Z Z e θ b = k cot. (5.6) m v a t θ w = tan. (5.7) We see that scattering through an angle 90 occurs for w = 1, and w = 0.03 results in scattering through an angle 3. Scattering cross sections Consider a beam of particles incident on a thin fixed target. The target particles will scatter some of the incident particles. The ratio of scattered to unscattered particles will depend on the surface density of target particles and the effective size of the target particles. The effective size of the target particles is the scattering cross section, usually denoted by σ. Cross section σ Suppose the surface density of target particles is ntar (this is the volume density multiplied by the thickness of the target). The fraction of the incident beam that hits a target and is scattered is N N sct inc = n tar σ. (5.8) The differential scattering cross section Scattered particles will have a range of scattering angles. The fraction of incident particles that are scattered into a cone of solid angle dω ( sinθdθ dφ ) polar angles θ, φ is = about a direction given by spherical 1 dnsct dω = ntar dω, N dω dω inc (5.9) where dω is the differential scattering cross section.

4 By convention, the z-axis is taken in the direction of the incident beam. Scattering is often azimuthally symmetric about this direction, which much simplifies the calculation of the differential cross section. Calculating the differential cross section The differential cross section can be calculated by considering a collision between two particles. The impact parameter, b, is equal to the distance of closest approach of the two particles if there were no interaction. For the azimuthally symmetric case, the scattering angle is a function of b alone. Consider the figure on right. The small circle indicates the initial position of the target particle. b dθ All particles passing through the annulus of radius b and width db are scattered between angles θ and θ + dθ. The area of the ring is equal to the cross section for a cone of solid angle dω = π sin θdθ. Hence π bdb b db = =, dω π sinθ dθ sinθ dθ (5.10) where the absolute values signs are put in to make the differential cross section positive. Rutherford scattering For Rutherford scattering Z Z e θ b = k cot. (5.11) m v a t Hence the differential cross section for Rutherford scattering is a Zt e 1 4 b db Z = = k dω sinθ dθ m v sin θ ( ). (5.1)

5 Note that the differential cross section is divergent as θ 0. This leads to the total cross section being infinite. This is due to the long range nature of the Coulomb force. Even at large distances, small deflections occur that contribute to the total cross-section. In atoms, the nuclei are shielded by the atomic electrons, which effectively shut off the Coulomb force at distances greater than the size of the atom. Combining equation (5.1) with equation (5.8), the fraction of incident particles that are scattered at angle θ into a detector of solid angle dω is a Zt e 1 4 m v Z f ( θ ) = ntar dω = ntar k dω. dω sin ( θ ) (5.13) This equation was first derived by Rutherford in 1911 and experimentally verified in 1913 by Geiger and Marsden. The Rutherford model of the atom Based on the experimental results, Rutherford concluded that the atom consisted of a positively charged compact nucleus with the electrons outside the nucleus. By scattering particles off an aluminum target and noting that backscattering requires a head on collision, Rutherford determined from the distance of closest approach that the size of the nucleus has to be less than m, which is a factor 5000 smaller than the atomic size.

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