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Quantum Mechanics

Quantum Mechanics. Through the Looking Glass. Democritus Dalton. Thomson. Rutherford. This is how the model of the atom has developed so far:. c =  where c =3.00 x 10 8 m/s.

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Quantum Mechanics

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  1. Quantum Mechanics Through the Looking Glass

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

  3. c =  where c =3.00 x 108 m/s

  4. 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 108 m/s  1 m 3.00 x 108 m/s = 589 nm 1x109nm  = 5.08 x 1014 s 1-

  5. Planck’s 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(plank’s constant) = h = 6.63 x 10-34 J-s

  6. Sample Problem: Calculate the smallest increment of energy that an object can absorb from yellow light whose wavelength is 589 nm we know from the previous problem: c = , that  = 5.08 x 1014 s 1- since E = h and h (plank’s constant) = 6.63 x 10-34 J-s E = (6.63 x 10-34 J-s )(5.08 x 1014 s 1- ) E = 3.37 x 10-19 J

  7. A Continuous Spectrum

  8. Light is a form of ... Electromagnetic Radiation

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

  10. Balmer’s Description of the Emission Spectrum of Hydrogen 1 1 22 n2 -  = C where n = 3, 4, 5, 6… and C = 3.29 x 1015 s-1

  11. Bohr’s Model of the Atom (1914) Limited the path of electrons to circular orbits with discrete energy (quantum energy levels) Explained the emissionspectrumof hydrogen

  12. Radii and Energies of the Three Lowest Energy orbits in the Bohr Model 1 1 n2  2 = 0 En = -RH n = 3 -0.242 x 10 -18 J n = 2 -0.545 x 10 -18 J En = -RH where RH = 2.18 x 10 -18J n = 1 -2.18 x 10 -18 J o 0 A o o 0.53 A 2.12 A 4.77 A radius = n2 (5.3 x 10-11m)

  13. Hydrogen’s Spectrum is Produced When photons produced energy is absorbed electrons transfer from an excited state Electrons appear in excited state Electrons return to their ground state Electrons are excited from their ground state

  14. Paschen Series • Infrared Balmer Series • Visible and Ultraviolet Lyman Series • Ultraviolet

  15. Explaining the Emission Spectrum of Hydrogen since E = E f - E i 1 1 - -RH thenE = ni2 nf2

  16. - RH E = 1 1 1 1 nf2 ni2 22 42 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. - 2.18 x 10-18J E = -4.09 x 10-19 J E = -4.09 x 10-19 J  = E = 6.63 x 10-34 J-s h  = 6.17 x 1014 s -1  = c = 3.00 x 108 m/s = 4.86 x 10-7 m = 486 nm (green)  6.17 x 1014 s -1

  17. ON TO PARTICLE WAVES

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