“continuous” spectrum.

1. H2 absorption spectrum H2 emission spectrum Line Spectra Ordinary white light is dispersed by a prism into a... “continuous” spectrum. From the gas in a nearly-evacuated gas-discharge tube, we instead get a... line spectrum. (if the H2 is absorbing light from an external source) (if the H2 has been energized and is emitting light)

2. e– found here e– never found here Niels Bohr took Planck’s quantum idea and applied it to the e– in atoms. -- e– could have only certain amounts of energy -- e– could be only at certain distances from nucleus Niels Bohr (1885–1962) planetary (Bohr) model N

3. -- Bohr assumed that the e– in his circular orbits had particular energies, given by: RH = Rydberg constant = 2.18 x 10–18 J n = 1, 2, 3, ... (the principal quantum number) The more/less (–) an electron’s energy is... i.e., the energy level the more/less stable the atom is. “barely” (if at all) stable When n is very large, En goes to zero, which is the most energy an e– can have because… “barely” (–) E - - - - - - - -0- - - - - - - - e– energy “well” e– “a bit” stable zero is larger than anything (–). “a bit” (–) E (e– has escaped completely from atom.) e– very stable very (–) E

4. ENERGY (HEAT, LIGHT, ELEC., ETC.) EMITTED LIGHT -- Bohr stated that e– could move from one level to another, absorbing light of a particular freq. to “jump up” and releasing light of a particular freq. to “fall down.” He: 1s2 He: 1s1 2s1 -- Bohr’s model (i.e., its specific equation) worked only for atoms with… a single e–.

5. Find the wavelength, frequency, and energy of a photon of the emission line produced when an e– in a hydrogen atom makes the transition from n = 5 to n = 2. For n = 5… = –8.72 x 10–20 J For n = 2… E2 = –5.45 x 10–19 J DE = 4.58 x 10–19 J = energy of photon E = h f = 6.91 x 1014 Hz c = f l = 4.34 x 10–7 m

6. The Wave Behavior of Matter Since light (traditionally thought of as a wave) was found to behave as both a wave and a particle, Louis de Broglie suggested that matter (traditionally thought of as particles) might also behave like both a wave and a particle. He called these… matter waves. Louis de Broglie (1892–1987) -- The wavelength of a moving sample of matter is given by: m = mass (kg) v = speed (m/s)

7. A proton moving at 1200 m/s would be associated with a matter wave of how many nanometers? In the absence of further info… mp ~ 1 amu = 1.66 x 10–27 kg (actual p+ mass is ~1.673 x 10–27 kg) = 3.3 x 10–10 m = 0.33 nm

8. A wave is smeared out through space, i.e., its location is not precisely defined. Since matter exhibits wave characteristics, there are limits to how precisely we can define a particle’s (e.g., an e–’s) location. -- the limitation also applies to a particle’s... momentum -- Heisenberg’s uncertainty principle: It is impossible to know simultaneously BOTH the exact momentum of a particle AND its exact location in space. D = “uncertainty in” h = Planck’s const. Werner Heisenberg (It is inappropriate to imagine an e– as a solid particle moving in a well-defined orbit.) (1901–1976)

9. Schrodinger’s wave equation (1926) accounts for both wave and particle behaviors of e–. where h-bar is the reduced Planck’s constant (=h/2p), V is the potential energy, and Y is the wave function) (for a non-relativistic particle of mass m w/no electric charge and zero spin, -- Solutions to the wave equation yield wave functions, symbolized by Y, which have no physical meaning, but Y2 at any point in space gives the probability that you’ll find an e– at that point. Y2 is called the probability density, which gives the electron density. Erwin Schrodinger (1887–1961)

10. -- Orbitals describe a specific distribution of electron density in space. Each orbital has a characteristic shape and energy.

11. (“Not.”) (Well, theory says that there ARE g orbitals, but they don’t look like this.) We also think g orbitals exist, and they look like this… The lanthanides and actinides contain f orbitals.