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Quantum Mechanics Through the Looking Glass

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This is how the model of the atom has developed so far: Rutherford Thomson Democritus Dalton

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c = where c =3.00 x 10 8 m/s

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Sample Problem: The yellow light given off by a sodium lamp has a wavelength of 589 nm. What is the frequency of this radiation? c =, where c =3.00 x 10 8 m/s 3.00 x 10 8 m/s = 589 nm 1 m 1x10 9 nm = 5.08 x 10 14 s 1-

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Plancks Theory: Energy is released incrementally as individual packets of energy called quanta, where the change in energy of a system is E = h, 2h,…n h and h(planks constant) = h = 6.63 x 10 -34 J-s

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we know from the previous problem: c =, that = 5.08 x 10 14 s 1- since E = h and h (planks constant) = 6.63 x 10 -34 J-s E = (6.63 x 10 -34 J-s )(5.08 x 10 14 s 1- ) E = 3.37 x 10 -19 J Sample Problem: Calculate the smallest increment of energy that an object can absorb from yellow light whose wavelength is 589 nm

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A Continuous Spectrum

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Light is a form of... Electromagnetic Radiation

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An Emission Spectrum... … is produced when a gas is placed under reduced pressure......as a high voltage is applied

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Balmers Description of the Emission Spectrum of Hydrogen = C 12 - 1 n2n2 where n = 3, 4, 5, 6… and C = 3.29 x 10 15 s -1

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Bohrs Model of the Atom (1914) Limited the path of electrons to circular orbits with discrete energy (quantum energy levels) Explained the emission spectrum of hydrogen

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0 A o 2.12 A o 4.77 A o n = 1 n = 2 n = 3 -2.18 x 10 -18 J 0 -0.545 x 10 -18 J -0.242 x 10 -18 J Radii and Energies of the Three Lowest Energy orbits in the Bohr Model radius = n 2 (5.3 x 10 -11 m) 0.53 A E n = -R H 1 n2n2 where R H = 2.18 x 10 -18 J E n = -R H 1 2 =

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Hydrogens Spectrum is Produced When Electrons are excited from their ground state Electrons appear in excited state electrons transfer from an excited state photons produced Electrons return to their ground state energy is absorbed

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Lyman Series Balmer Series Paschen Series Ultraviolet Visible and Ultraviolet Infrared

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Explaining the Emission Spectrum of Hydrogen since E = E f - E i then E = -R H 1 nf2nf2 - 1 ni2ni2

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Sample Problem: Calculate the wavelength of light that corresponds to the transition of the electron from the n = 4 to the n=2 state of the hydrogen atom. RHRH 1 ni2ni2 - 1 nf2nf2 E = 2.18 x 10 -18 J 1 4242 - 12 E = -4.09 x 10 -19 J E = = E h = -4.09 x 10 -19 J 6.63 x 10 -34 J-s = 6.17 x 10 14 s -1 = c = 3.00 x 10 8 m/s 6.17 x 10 14 s -1 = 4.86 x 10 -7 m = 486 nm (green)

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