Four Chapter 8

EMISSION AND ABSORPTION OF SPECTRA

• Because of the photoelectric effect, we have realized that light has a dual nature.
  • Wave nature – which explains reflection, refraction, dispersion, polarization diffraction etc.
  • Particle nature – which explains thermionic emission and the photoelectric effect.
    • There are other effects that the particle nature explains which will be discussed later.
  • This particle nature has consequences in other areas, such as our understanding of the structure of matter.
  • One must realize, that light is generally produced by some process that occurs inside the atom. The particle nature of light, affects our understanding of these processes, and therefore our understanding of what an atom is.
  • Let us now consider the emission of light from substances.
  • Solids liquids and dense gases, when heated to high temperatures, give off a continuous spectrum.
  • Light given off by electrical discharges through a rarefied gas is quite different. (Produces bright lines and not a continuous spectrum)
  • A continuous spectrum is given off because of the interactions between each atom or molecule and its neighbors.
  • Rarefied gases have atoms very far apart, and therefore light cannot be a result of interactions between the atoms. Therefore a study of light given off by rarefied gases may give clues to atomic structure.
  • The light given off by electrically "excited" gases is:
    • Discrete (not continuous). i.e.: a bright line emission spectrum is produced.
    • Different for each different gas.
  • Emission Spectrum – The spectrum produced by a gas, when "excited" by some process such as an electrical discharge.
  • We produce an emission spectrum by placing electrodes at the ends of a tube containing a rarefied gas.
    • The discharge causes the gas to glow.
    • The light is passed through a spectroscope which causes a bright line spectrum.
    • By using l = (xd)/(nl) and the other equivalent equations, we can measure the wavelength for the light from each line.
    • Film sensitive to ultraviolet light and infrared light can be used to show that lines exit in the ultraviolet and infrared regions of the electromagnetic spectrum.
  • If white light passes through a cool gas and then through the spectroscope, then dark lines will appear in the continuous spectrum at the same places as the bright lines would be for the emission spectrum.
  • Absorption spectrum – The spectrum produced when white light is analyzed after passing through a cool gas.
  • From emission and absorption spectrums, it can be concluded that atoms absorb the same frequencies of light as they emit.
  • Any model of the atom must explain these observations.

Spectroscopy

• Previously we mentioned the use of a spectroscope to study the spectrum from an element.
• A spectroscope consists of
  • A slit that limits the light entering to a narrow beam
  • A prism or diffraction grating for dispersing the light into its parts
  • A viewing screen or telescope
    • The screen is usually circular
• The field of spectroscopy uses the spectroscope for analysis of light.
• Spectroscopy - The analytical study of the spectrum emitted by heated substances
• Each different element produces its own unique spectrum when "excited" to higher energy levels.
• The study of these eletromagnetic "finger prints" provide for indentifying substances.
• Both emission and absorption of spectra can be used to identify substances.
• Both emission and absorption spectrums can be used to identify substances
• Spectroscopy is used in analyzing
  • Impurities in steel making
  • The presence of pollutants
  • Light from distant objects in the universe in the field of Astronomy
• Many discoveries have been made by using spectroscopy
  • Astronomers have discovered the make up of stars and gas clouds
  • Astronomers have discovered that the universe is expanding
    • Hydrogen in stars in distant galaxies have a spectrum that is red shifted which indicates they are moving away from us (Doppler shift).
    • The speed of the motion can be calculated by the amount of shift
    • This led to the "Big Bang Theory" for the creation of the universe.
  • Used to identify materials by matching the spectral lines of the unknown material to the lines of known materials (now done by computers).
    • X-ray fluorescence - expose a sample to x-rays which causes it to fluoresce. Analysis of this light allows the sample to be identified.
  • Spectroscopy can be used on particles as well. We will see later in the course as to why this is possible.
  • Neutron spectroscopy resulted in the discovery of the structure of large molecules such as DNA and RNA.

MODELS OF THE ATOM

• 400 B.C. Democritus theorized that all matter was composed of particles called atoms.
• John Dalton, found the first evidence for the existence of atoms (1700's).
• Dalton thought of atoms as small balls that could join together in small whole number combinations to produce compounds.
• Thomson later suggested that atoms were spheres of positive charge, with negative electrons embedded throughout.
• By the 20th century, scientists believed atoms had internal structure and they were researching this structure.

RUTHERFORD'S MODEL

• By the end of the 19th century, it was known that the negative charge of the atom was carried by electrons.
• Electrons were known to have very small mass compared to the mass of the atom.
• Since atoms are neutral, there must be a positive part that takes up most of the mass of the atom. How the positive and negative charges in an atom were distributed, was not known.
• Rutherford had his assistants (Geiger and Marsden), perform a scattering experiment to determine internal structure.
• If the atom was like what Thomson suggested (water melon model), then a bullet fired at the atom (alpha particle), should travel mainly straight through, with some small deflections and small decreases in speed.
• Rutherford used high speed alpha particles from a radioactive polonium source. The alpha particles were directed in a narrow beam at a sheet of gold foil as shown below.

• When the alpha particles encountered the foil, they were scattered by the atoms in the foil. The scattered particles were observed with a scintillation counter.
• The scintillation counter was moved through angles of 0^o to plus or minus 180^o. The number of alpha particles at each angle was counted.
• Geiger and Marsden took 2 years to do the experiment and reported:
  • The great majority of alpha particles passed through the gold foil with virtually no deviation from their original path.
  • A very few alpha particles were scattered by sizable angles.
  • In extremely rare cases, an alpha particle was deflected by 180^o (back scattering).
  • After a period of exposure to the beam, the foil had acquired a positive charge.
• Rutherford reasoned that deflection of the positive alpha particles could be explained only if:
  • The atom has a positively charged very small but extremely dense, central region called a nucleus, that contains all of the atom's positive charge and most of its mass.
  • The greatest proportion of an atom's volume is empty space.
  • Within the empty space were a number of very light, negatively charged electrons moving in orbits around the nucleus.
• The scattering data can only be explained If there was a dense massive nucleus of positive charge, which occupied very little volume.
• The alpha particles that passed through the atoms, must periodically capture an electron. This causes the foil to become positively charged.
• The electrons had to be separate from the nucleus, otherwise the atom could not repel the alpha particles. To keep electrons separate from the nucleus, required some mechanism. It was suggested that the electrons could be kept from the nucleus if they were in orbits around the nucleus.
• Electrical forces between the nucleus and the electrons provided the centripetal force for the electrons.
• Rutherford's model became known as the planetary model.

TRAJECTORY OF ALPHA PARTICLES

• Rutherford was originally interested in the forces causing the scattering. Since the alpha particle is positively charged, and so is the nucleus in Rutherford's model, then possibly the scattering force could be electrostatic. According to Coulomb's law:

F = kQq/(r²)

• Rutherford was able to show mathematically, that electrostatic repulsion should result in a hyperbolic path for the particles. The initial and final directions of the particle would nearly be at the hyperbola's asymptotes. The nucleus would be at one of the hyperbola's foci.
• The eccentricity (amount of bending) would depend upon how direct a hit is involved.

• How direct the hit is, is determined by what is called the aiming error (b). Therefore, for a given value of b, an alpha particle of a given energy is deflected along a hyperbolic path, scattered by an angle U as in the diagram below.

• If the aiming error is b when the scattering angle is U, then if the aiming error is less than b, then the scatter angle will be greater than U. If b = 0, then the angle U = 180^o.
• Assuming the scattering force is a Coulomb force, then the magnitude of the angle U for any aiming error b, is dependent upon the initial energy of the alpha particle, and on the charge on the scattering nucleus.
• This will produce a graph with a shape like the one below

• b cannot be directly measured and therefore, an indirect method must be used to confirm:
  • The graph and its relationship.
  • That the scattering force is a Coulomb force.
• For a particle to scatter through an angle greater than U, the aiming error must be less than the corresponding b from the graph. In other words, the particle must approach the nucleus inside a cylinder of radius b, and cross sectional area of:

A = pb²

• For any beam, the number of particles scattered through angles greater than U, must be proportional to the target cylinder area. Smaller areas mean smaller numbers of particles will approach in the cylinder, and therefore smaller numbers will scatter through angles greater than U.
• Therefore we can write:

N_u µ b²

but u µ 1/b

or u² µ 1/b²

therefore b²µ 1/u²

or N_uµ 1/u²

This allows us to construct a graph of N versus U.

• This graph can be tested by counting the particles scattered at each angle. Geiger and Marsden performed the experiment to verify this graph.
• Their experiment also verified:
  • The assumptions upon which the computed curve is based.
  • That the scattering force obeyed Coulomb's law.

CHARGE ON THE NUCLEUS

• The methods used to account for scattering alpha particles can be used for any nucleus. Different nuclei differ only in mass and charge.
• Since the nucleus is assumed to be fixed in space, then the only factor affecting the Coulomb force is the charge on the nucleus. This means that the charge on the nucleus is the only factor affecting the scatter angle.

• Z₁ is less than Z₂ in the diagram above.
• A different charge causes different forces (F = kQq/r²), given b (r) is kept constant. Therefore a different charge causes a different scatter angle.
• It can be shown that N_u a Z² by using:
  • Newton's laws
  • Coulomb's laws
  • the geometry of the scattering interaction.
• Once these calculations are made, they can be verified experimentally by comparison with the scattering data collected by firing alpha particles at different foils made of different atoms (thickness of the foil kept constant).
• This means that if we didn't know the charge on the nucleus, we can calculate it from the scattering data.

THE SIZE OF THE NUCLEUS

• Back scattered particles allow us to get an upper limit to the size of the nucleus.
• Alpha particles of mass m and charge q, accelerate through a potential difference of V. For this circumstance we can write

qV = E_ko = (1/2)m(v_o)² = E_tot

• As the particle gets closer to the nucleus, the kinetic energy changes to potential energy or:

E_tot = E_k + E_p = E_ko = a constant

• A direct approach will stop the alpha particle at a distance r_o (the distance of closest approach). Therefore at closest approach:

E_tot = E_p = E_ko = (1/2)m(v_o)²

• Potential energy is calculated by calculating work. Two unlike charges attract with a force given by Coulomb's law.

F = (kQq)/(r²)

• To separate them, a Coulomb's force is applied over the distance between them. That is:

W = Fd (d = r)

W = (kQq/r²)r

E_p= (kQq)/r

• Given the equation above for electric potential energy, and the fact that the kinetic energy gets transformed to potential energy, we can write:

kQq/r_o = (1/2)m(v_o)²

or r_o = (2kQq)/(m(v_o)²)

• Given that an alpha particle is accelerated through a potential difference of 2.64 X 10⁶ V (the mass = 6.6 X10^-27 kg and q = 3.2 X 10^-196 C), then:

qV = (1/2)m(v_o)²

(3.2 X 10^-19 C)(2.64 X 10⁶ V) = (1/2)(6.6 X 10^-27kg)(v_o)²

v_o = 1.6 X 10⁷ m/s

• A gold nucleus has a charge of 1.3 X 10^-17 C. The distance of closest approach is therefore:

r_o = (2(8.99 X 10⁹Nm²/C²)(3.2 X 10^-19C)(1.3 X 10^-17C))/((6.6 X 10^-27kg)(1.6 X 10⁷m/s)²)

or r_o = 4.2 X 10^-14m

• This gives an upper limit for the size of a gold atom. If an alpha particle comes closer than
• 4.2 X 10^-14 m, it may contact the nucleus, and scattering would then not conform to the predictions based on Coulomb's law.
• Measurements (X-ray diffraction) gives us the distance between 2 gold atoms to be
• 2.5 X 10^-10 m. This means that the radius of a gold atom is 1.2 X 10^-10 m.
• The ratio of the volume of the atom to the volume of the nucleus, shows that the atom is mainly empty.

January 20, 2014