1. Quantum Theory and the Electronic Structure of Atoms Chapter 7 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. PowerPoint Lecture Presentation by J. David Robertson University of Missouri

2. Properties of Waves Wavelength ( ) is the distance between identical points on successive waves. Amplitude is the vertical distance from the midline of a wave to the peak or trough. 7.1

3. Properties of Waves Frequency ( ) is the number of waves that pass through a particular point in 1 second (Hz = 1 cycle/s). The speed (u) of the wave = x 7.1

4. Maxwell (1873), proposed that visible light consists of electromagnetic waves. Electromagnetic radiation is the emission and transmission of energy in the form of electromagnetic waves. Speed of light (c) in vacuum = 3.00 x 10 8 m/s All electromagnetic radiation x  c 7.1

6. x = c = c/ = 3.00 x 10 8 m/s / 6.0 x 10 4 Hz = 5.0 x 10 3 m Radio wave A photon has a frequency of 6.0 x 10 4 Hz. Convert this frequency into wavelength (nm). Does this frequency fall in the visible region? = 5.0 x 10 12 nm 7.1

7. Mystery #1, “Black Body Problem” Solved by Planck in 1900 Energy (light) is emitted or absorbed in discrete units (quantum). The energy (E) of a single quantum of energy is given by: E = h x Planck’s constant (h) h = 6.63 x 10 -34 J s v = frequency of radiation 7.1 Energy is always emitted in multiples of hv

8. Light has both: 1.wave nature 2.particle nature h = KE + BE Mystery #2, “Photoelectric Effect” Solved by Einstein in 1905 Photon is a “particle” of light KE = h - BE h KE e - 7.2 Photoelectric effect: electrons are ejected from the surface of certain metals exposed to light of a minimum frequency, called the threshold frequency Einstein suggested that a beam of light is really a stream of particles Each photon possesses energy E E= hv, w here v is the frequency of light

9. http://phet.colorado.edu/en/simulation/photoelectric Photoelectric Effect Animation

10. E = h x E = 6.63 x 10 -34 (J s) x 3.00 x 10 8 (m/s) / 0.154 x 10 -9 (m) E = 1.29 x 10 -15 J E = h x c /  7.2 When copper is bombarded with high-energy electrons, X rays are emitted. Calculate the energy (in joules) associated with the photons if the wavelength of the X rays is 0.154 nm.

11. Bohr’s Theory of the Hydrogen Atom Emission spectra: either continuous or line spectra of radiation emitted by substances. –Ex. Sun and “glow” of heated metal (continuous spectra) –All wavelengths of visible light are represented Atoms in the gas phase produce line spectra Line spectra: the light emission only at specific wavelengths Every element has a unique emission spectrum, which can be used to identify unknown atoms

12. 7.3 Line Emission Spectrum of Hydrogen Atoms

13. 7.3

14. 1.e - can only have specific (quantized) energy values 2.light is emitted as e - moves from one energy level to a lower energy level Bohr’s Model of the Atom (1913) E n = -R H ( ) 1 n2n2 n (principal quantum number) = 1,2,3,… R H (Rydberg constant) = 2.18 x 10 -18 J 7.3 n=1 is the ground state

15. E = h 7.3

16. E photon =  E = E f - E i E f = -R H ( ) 1 n2n2 f E i = -R H ( ) 1 n2n2 i i f  E = R H ( ) 1 n2n2 1 n2n2 n f = 1 n i = 2 n f = 1 n i = 3 n f = 2 n i = 3 7.3

17. http://www.mhhe.com/physsci/chemistry/essentialchemistry /flash/linesp16.swf

18. E photon = 2.18 x 10 -18 J x (1/25 - 1/9) E photon =  E = -1.55 x 10 -19 J = 6.63 x 10 -34 (Js) x 3.00 x 10 8 (m/s)/1.55 x 10 -19 J = 1280 nm Calculate the wavelength (in nm) of a photon emitted by a hydrogen atom when its electron drops from the n = 5 state to the n = 3 state. E photon = h x c /  = h x c / E photon i f  E = R H ( ) 1 n2n2 1 n2n2 E photon = 7.3 E= hv and v= c/λ, so

19. De Broglie (1924) reasoned that e - is both particle and wave. The relationship between the circumference of an allowed orbit and the wavelength of the electron is 2  r = n = h/mu u = velocity of e - m = mass of e - Why is e - energy quantized? 7.4

20. = h/mu = 6.63 x 10 -34 / (2.5 x 10 -3 x 15.6) = 1.7 x 10 -32 m = 1.7 x 10 -23 nm What is the de Broglie wavelength (in nm) associated with a 2.5 g Ping-Pong ball traveling at 15.6 m/s? m in kgh in J su in (m/s) 7.4 1J = 1 kg m 2 / s 2

21. Chemistry in Action: Element from the Sun In 1868, Pierre Janssen detected a new dark line in the solar emission spectrum that did not match known emission lines In 1895, William Ramsey discovered helium in a mineral of uranium (from alpha decay). Mystery element was named Helium

22. Chemistry in Action: Laser – The Splendid Light Laser light is (1) intense, (2) monoenergetic, and (3) coherent

23. Chemistry in Action: Electron Microscopy STM image of iron atoms on copper surface e = 0.004 nm