2. Quantum Mechanics Through the Looking Glass

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

4. Understanding Radiant Energy c = where c =2.99 x 10 8 m/s

5. 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 =2.99 x 10 8 m/s 2.99 x 10 8 m/s = 589 nm x1 m 1 x 10 9 nm x = 5.08 x 10 14 s 1-

6. Planck’s Theory: Energy is released incrementally as 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) = 6.63 x 10 -34 J-s 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 10 14 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 10 14 s 1- )  E = 3.37 x 10 19 J

7. A Continuous Spectrum

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

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

10. Balmer’s 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

11. Bohr’s 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

12. 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 2 =

13. Hydrogen’s 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

14. Lyman Series Balmer Series Pashen Series Ultraviolet Visible and Ultraviolet Infrared

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

16. 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)