1. Atomic and Molecular Spectra

2.  Atoms are very small,  Atoms are stable  Atoms contain negatively charge electrons, but are neutral(because matters are neutral)  Atoms emit and absorb electromagnetic radiation(Black body radiation, photoelectric effect)

3. Blackbody Radiation and Photoelectric Effect gave the impulse to understand the nature of light However, there was no understanding on how electrons “exist” inside matter –It was known that a toms that form matter contain electrons ( Thompson, 1897) and that atoms are neutral Atomic structure?

4. Thomson Model Model of J.J. Thomson: –Electrons (Corpuscular) are embedded within positive charge sphere (so-called “plum pudding” model) –Both the positive charge and the mass of the atom would be more or less uniformly distributed over its size Approximately 10 -10 meters across = 1 Å = 0.1 nm

5. Challenges to Thomson’s Model How does the atom emit radiation? This model soon came into conflict with experiments by Rutherford

6. The Rutherford Experiments Rutherford discovered α (alpha)-particles –α-particles are the nuclei of helium atoms, which were products of nuclear disintegration –He used α-particles in various studies In 1909, he with co-workers (Hans Geiger and Ernest Marsden) experimented with streams of alpha particles passing through a thin gold foil

7. Rutherford Experiment

8. The Rutherford Experiments Conclusions: 1.Most α-particles must have passed through the atoms without being deflected. An atom must be mostly empty space 2.The α-particles occasionally strick a small strongly scattering region. This region could be a concentrated positive charge 3.Observed trajectory suggested the latter is the case! 1.Can calculate atomic volume and area using Avagadro’s number 2.Can calculate cross sectional area of nucleus relative to atomic cross section based on fraction that scatterer and the distribution of scattering angles which is consistent with trajectories of the charged a- particle scattering from a positive core Observations: 1.Almost all α-particles went through the gold foil with very slight deflections 2.A very, few α-particles were reflected – scattered by more than of 90°

9. Rutherford’s Experiments Since atoms have no overall electrical charge (atoms are neutral), there must be just enough negatively-charged particles outside the nucleus to just balance the positively- charged nucleus Therefore, The Rutherford's experiment suggested that the atom consists of mostly empty space with a very small positively- charged nucleus, outside which are just enough negative charges to equal the positive charge in the nucleus Since the electrons do not crash into the nucleus but exist in stable atoms, this suggests a planetary model with negatively charged electrons circulating around a positive core

10. Estimate of Nuclear Cross Section I Approximately one in 10 4  particles with energies of ~ 10 7 eV are scattered backward from a one micron thick sheet of metal foil If atomic spacing is ~ 1 A, then there are 10 4 atomic layers so the probability of scattering by a single layers is 10 -8 This implies the nuclear cross section is one part in 10 8 of the atomic cross section Since the area goes as the square of the radius the nuclear radius is smaller than atomic radius by a factor of 10 -4 Since the atomic diameter is ~ 1 A, the nuclear diameter is ~ 10 -4 A

11. Estimate of Nuclear Cross Section II The model of a small positively charge nuclear core was confirmed by a comparison with the angular distribution to scattered  -particles The incident  -particles are assumed to have a uniform distribution in space. Initial trajectories that are headed towards the nuclei of atoms in the foil with random separations from the nearest nuclear core Full trajectory calculated for each distance of the trajectory line from the center of the nearest nucleus to give the scattering angle From the distribution of distances of the  -particles from the positive core calculate distribution of scattering angles Agreement found with measured scattering angle distribution

12. The Rutherford Model “Planetary Model” –Positive charge in the center of the atom with almost all mass concentrated within this positive charge - nuclei –Electrons - negative charge- are attracted to the nucleus about which they orbit (just as planets orbit the sun due to attractive 1/r 2 force) –Sizes nuclei ~ 10 -14 m (calculated from fraction of  -particles that scatter more than 90 0 in a foil of given thickness) atom ~ 10 -10 m (from the mass density and number of atoms in a mole – Avogadro’s number)

13. Difficulties with the Rutherford Model Since electron travels in a circular orbit, it is constantly accelerated (even though its speed is constant.) Thus, the electron emits EM radiation, which carriers away energy. The energy of the atoms is reduced. Thus the atom has a lower potential energy and moves closer to the nucleus

14. Thus, classically, the Rutherford Atom is Unstable

15. Difficulties of the Rutherford Model Another problem is that the spectrum of the emitted EM radiation would be continuous. Classical approach gives the following expression As r decrease, the emission wavelength changes continuously, so this model predicts that the emission spectrum of atoms is broad But sharp spectral lines are observed, not a continuum

16. The Spectrum of Hydrogen At room temperature, hydrogen gas does not emit light When heated to high temperatures, hydrogen emits visible radiation –distinct spectral lines are observed rather than the continuous radiation spectrum expected classicallydistinct Example of visible part of the spectrum

17. The Spectrum of Hydrogen Several families of such lines were observed They can be fit empirically by the Rydberg-Ritz Formula R H is known as the Rydberg constant k and n are integers, and n > k The visible hydrogen emission is known as the Balmer series with k = 2

18. The Bohr’s Atom Assumption: Electron has, for some yet unknown reason, only certain energies in the hydrogen atom Bohr called these allowed energy levels The atom can occasionally jump between energy levels, emitting a photon when it makes a transition to a lower energy state and absorbing a photon when jumping to a higher energy level The difference in the energy between the two levels is  E = h So have discrete lines observed in hydrogen spectrum

19. Bohr’s Postulates Bohr started from the assumption that the electron moves in circular orbits around the proton under the influence of the attractive electric field. Postulate 1: Only certain orbits are stable. These are stationary or more precisely quasi-stationary states. An electron does not emit EM radiation when in one of these states (orbits)

20. Bohr’s Postulates Postulate 2. If the electron is initially in an allowed orbit (stationary state), i, having the energy, E i, goes into another allowed orbit, f, having energy, E f (< E i ), EM radiation is emitted, with energy and frequency,

21. Bohr’s Postulates Postulate 3. The electron can only have an orbit for which the angular momentum of the electron, L, takes on discrete values (the orbits are quantized): Orbits characterized by angular momentum since this depends on both the distance of the electron from nucleus and its velocity

22. Electron Orbits in the Bohr Atom

23. Electron Orbits in The Bohr’s Atom The radii (orbits) are said to be discrete (‘quantized’) r min = a 0 = 0.53×10 -10 m = 0.53 Å, is called the Bohr radius In this model, the electron and the nucleus in the hydrogen atom cannot get closer than the distance a 0

24. The Wavelength of the Emitted Electromagnetic radiation Following Postulate 2:

25. The Spectrum of Hydrogen

26. Ionization Energy For hydrogen atom, with only one electron, the ionization energy has a clear meaning: –This is the energy required to remove the electron from a hydrogen atom: H 0 + 13.6 eV  H + + e - An atom is now ionized, since now have H + instead of H 0 Similarly for any other atom, we can introduce the ionization energy –The minimum energy required to remove the most energetic electron from the atom in its lowest energy state The energy required to remove the second electron is the “second ionization energy” –Question: Is it larger or smaller than “first” ionization energy?

27. Discrete Spectrum Bohr’s Postulates Predicted a Discrete Spectrum Consistent with the spectrum of hydrogen Direct proof found in measurements by James Frank and Gustav Hertz in 1914 Results consistent with spectra

28. The Franck-Hertz Experiment The heated filament ejects electrons into the tube, which can be either evacuated (vacuum) or filled with Hg vapor A variable (positive) accelerating voltage, V A, is applied to the grid (a wire mesh) –The electrons acquire kinetic energy K = eV A There is (negative) retarding voltage (V r ) between grid and collector Only electrons having enough [kinetic] energy will overcome this potential and reach the Collector and contribute to current measured with an Electrometer

29. The Franck-Hertz Experiment If V r > V A no electrons can reach the Collector, so no current would be measured. If V r < V A, then, if the tube is highly evacuated, most of the electrons would reach the Collector, and have energy |e|(V A - V r )

30. The Franck-Hertz Experiment If the tube contains some gas, the electrons can loose energy via collisions with the gas atoms –Such collisions are inelastic, i.e. electrons lose energy, which is transferred to internal energy of atoms in the gas Thus, even in the case when V r < V A, it is possible that the electrons would not be able to reach Collector, and contribute to the current

31. The Franck-Hertz Experiment Franck and Hertz observed the Collector current as a function of V A (>V r ) when tube was filled with various gases (result for mercury gas is shown here) At first, the current increased as was expected for a typical vacuum tubes, but at ~4.9 V current suddenly dropped Then, the increase resumed until 9.8 V, an so on VAVA

32. The Franck-Hertz Experiment This occurs only if the electrons undergo inelastic collisions Thus, when V A = 4.9 × n Volts (n = 1, 2, 3, …) the electrons undergo inelastic collisions with Hg atoms! In inelastic collision kinetic energy of electrons becomes internal energy of Hg atoms – Hg atoms absorb the energy of electrons! The current drops because fewer electrons reach the Collector VAVA

33. The Franck-Hertz Experiment Why do we see the drop only at specific voltages? If distribution of energy levels of Hg atom is continuous, then kinetic energy should be transferred to Hg atoms regardless of the energy of electrons However, If we assume that Hg spectrum is discrete, then only when electrons reach certain energies do they undergo inelastic collision with Hg atoms –Thus energy spectrum of Hg atoms is such that an electron energy level lies ~4.9 eV above the ground state

34. The Franck-Hertz Experiment Explained Why did Frank and Hertz not observe dips in the current at other voltage? –Their experiment was not sufficiently sensitive –Since as soon as electrons gain energy of 4.9 eV they transfer it to Hg atoms, only small fraction of the electrons could have higher energies, (e.g., 6.65 eV), making it difficult to observe current dips associated with other (higher) voltages