1. Chapter 7 Quantum Theory and Atomic Structure

2. Electro-Magnetic radiation includes visible light, microwave, TV, radio, x-ray, etc. Radiation is a combination of vibrating electric and magnetic fields in repeatable waveforms Wavelength, λ (lambda): distance from crest to crest Frequency,  (nu) : # crests to pass a point in 1 second, units are #/s or Hz Wave velocity = λ All E-M radiation in a vacuum has constant velocity called the speed of light: c = 2.998 x 10 8 m/s. Therefore c = λ  (Memorize formula and c) Long λ => short & vice-versa

3. Figure 7.1 Frequency and Wavelength c =  The Wave Nature ofLight

4. Figure 7.2 Amplitude (intensity) of a wave.

5. Figure 7.3 Regions of the electromagnetic spectrum. Wavelength above in nanometers, Frequency below in Hertz or #/second Frequency unit is #/s or s -1. All waves travel at the same speed through vacuum but differ in frequency and wavelength

6. Sample Problem 7.1a SOLUTION:PLAN: Interconverting Wavelength and Frequency wavelength in units given wavelength in m frequency (s -1 or Hz) = c/ Use c =  = 1.000 x 10 -10 m == 2.998 x 10 8 m/s 1.000 x 10 -10 m = 2.998 x 10 18 s -1 PROBLEM: A dental hygienist uses x-rays ( = 1.000A) to take a series of dental radiographs. What is the frequency (in s -1 ) of the electromagnetic radiation? (Assume that the radiation travels at the speed of light, 2.998 x 10 8 m/s.) 1 A = 10 -10 m 1 cm = 10 -2 m 1 nm = 10 -9 m o 1.000A o 10 -10 m 1A o

7. E-M radiation was considered to be a wave/energy phenomenon and not matter Max Planck developed a new physics when classical physics could not be used to interpret data from blackbody radiation ( Blackbody is an object that absorbs all radiation incident on it) Blackbody radiation is emitted by solid bodies that are heated to high T and become incandescent Classical physics had to assume continuous radiation, and it could not resolve the data that there was discrete radiation Planck developed theory of Packets of Energy called quanta The energy associated with quanta was proportional to the frequency of the radiation: E = h h = Planck’s constant 6.626 x 10 -34 J. s

8. If c =, and E = h, then with rearranging and substituting: E = hc/ What is the energy of a photon with a wavelength of 399.0 nm?

9. Figure 7.6 Blackbody radiation (4 th ed.) E = n h  E =  n h  E = h when n = 1 smoldering coal electric heating elementlight bulb filament Solid heated To 1000 K It emits visible light) At 1500 K At 2000K

10. Figure 7.6 Demonstration of the photoelectric effect.

11. Wave model could not explain photoelectric effect. Flow of current when monochromatic light of suff frequency falls on metal plate. Photoelectric Effect: electrons are ejected from a metal's surface if it is exposed to uv radiation Each metal required a characteristic minimum uv frequency to start ejecting e-s Called Threshold freq, o - As increases more e-s ejected with higher vel (KE) These data also defied classical physical explanation Einstein reviewed data, recalled Planck's quanta The "incident" radiation consists of quanta of energy, E = h, called photons(small bundle of electromagnetic energy) - thus the PHOTOELECTRIC Effect In order to eject an e-, a min KE is required, E = h o If E>h o then excess KE is supplied to the e-, increasing its velocity For Na metal, o = 5.51 x 10 14 Hz

12. Sample Problem 7.2 SOLUTION: PLAN: Calculating the Energy of Radiation from Its Wavelength PROBLEM:A cook uses a microwave oven to heat a meal. The wavelength of the radiation is 1.20 cm. What is the energy of one photon of this microwave radiation? After converting cm to m, we can use the energy equation, E = h combined with = c/ to find the energy. E = hc/ E = (6.626 x 10 -34 J*s)2.998 x 10 8 m/s 1.20 cm x 10 -2 m cm = 1.66 x 10 -23 J

13. Experiments with "excited" atoms of H produced emission spectra - always a discrete set of lines at certain wavelengths White light dispersed by a prism or diffraction grating: - we see ROYGBIV – a continuous spectrum from 750 nm to 400 nm When a gas-filled tube is charged with current, only certain EM 's are detected - called a line spectrum or emission spectrum The gas particles split into individual atoms The e-s are excited by the current into a higher energy level. When they drop down, they emit energy of a certain λ, with energy gaps at distinct intervals

14. Figure 7.7 The line spectra of several elements.

15. Hydrogen atomic line spectra – also called emission spectra – - worked out mathematically (by several scientists) to define the energy of the light emitted & relationships between the lines Balmer: red, green, blue, and violet lines (656.3, 486.1, 434.0, 410.1nm) 1/λ = R y (1/2 2 - 1/n 2 ) If n > 2 R y = Rydberg constant = 1.096776 x 10 7 /m If n = 3 get red, n = 4 get green, n = 5 get blue, n = 6 get violet

16. = RRydberg equation* - 1 1 n22n22 1 n12n12 R is the Rydberg constant = 1.096776x10 7 m -1 Figure 7.8 Three series of spectral lines of atomic hydrogen. Balmer is in the visible region, and the other series, which have names also, are in uv or ir area. for the visible series, n 1 = 2 and n 2 = 3, 4, 5,... *Memorize!

17.  1. H atoms have only certain allowable energy levels called stationary states.  2. Atom does not radiate energy while in a stationary state.  3. Atoms changes to another stationary state by absorbing or emitting a photon. Energy=EstateA-EstateB=h

18. Bohr found E n = - Rhc/n 2 R = 1.097 x 10 7 /mh = 6.626 x 10 -34 J. s c = speed of light Rhc = 2.178 x 10 -18 J (since they are all constant) Then E n = -2.178 x 10 -18 J/n 2 All E is therefore < 0, and has discrete values only Nucleus (proton) & e- are so far apart there's no attraction anymore Negative E is more stable than zero energy n = 1 is the ground state, all above are excited states

19. Figure 7.9 Quantum staircase.

20. Figure 7.10 The Bohr explanation of the three series of spectral lines.

21.  Instrumental techniques used to obtain information about atomic or molecular energy levels  Emission: electrons in an atom are excited to a higher energy state and then emit photons as they return to lower energy states  Absorption: electrons in an atom absorb photons of certain wavelengths and jump to higher energy states; photons NOT absorbed are observed!  See figure 7.11 in text: why chlorophyll looks green

22. Figure B7.2 Figure B7.1 (4 th ed.) Emission and absorption spectra of sodium atoms. Flame tests. strontium 38 Srcopper 29 Cu

23. A hydrogen atom has an e- excited up to level 4, and it drops back to level 2. (a) determine the wavelength of the photon emitted and (b) the energy difference. (a) Use 1/ = R y (1/2 2 – 1/4 2 ) = 4.8617 x 10 -7 m (b) Use  E = hc/ E = 4.086 x 10 -19 J Follow-up: Answer the same questions for the e- excited up to level 6.

24. Stationary wave: - fixed at both ends - has "nodes" - never moves on those spots with distance = length/2 Only certain λ's are possible for a standing wave

25. Figure 7.12 Wave motion in restricted systems.

26. Einstein remembered for E = mc 2 m = E/c 2 = (hc/λ)/c 2 = h/λc This appears to say that a photon of a certain wavelength has mass! Proved by Arthur Compton in 1922 E-M radiation is both waves & little packets of energy and matter called photons De Broglie 1923: if light has wave-particle duality, then matter, which is particle-like, must also be wavelike under certain conditions Rearranged m = h/ c to get λ = h/mv This is called the deBroglie wavelength It means that all matter exhibits both particle and wave properties

27. Sample Problem 7.4 SOLUTION: PLAN: Calculating the de Broglie Wavelength of an Electron PROBLEM:Find the deBroglie wavelength of an electron with a speed of 1.00 x 10 6 m/s (electron mass = 9.11 x 10 -31 kg; h = 6.626 x 10 -34 J. s). Knowing the mass and the speed of the electron allows to use the equation = h/m u to find the wavelength. = 6.626 x 10 -34 ( kg*m 2 /s 2 )s 9.11 x 10 -31 kgx1.00 x 10 6 m/s = 7.27 x 10 -10 m

28. Bohr’s Theory: 1 e- in H atom occupying certain energy states - a certain quanta Spherical orbitals around the nucleus With de Broglie's hypothesis: e- must have a certain λ to make a complete revolution - like a standing wave An integral # of complete λ's to fit the sphere's circumference Circumference = 2  r, therefore nλ = 2  r, n = 1, 2, 3....

29. CLASSICAL THEORY Matter particulate, massive Energy continuous, wavelike Since matter is discontinuous and particulate perhaps energy is discontinuous and particulate. ObservationTheory Planck: Energy is quantized; only certain values allowed blackbody radiation Einstein: Light has particulate behavior (photons)photoelectric effect Bohr: Energy of atoms is quantized; photon emitted when electron changes orbit. atomic line spectra Figure 7.14 Summary of the major observations and theories leading from classical theory to quantum theory.

30. Since energy is wavelike perhaps matter is wavelike ObservationTheory deBroglie: All matter travels in waves; energy of atom is quantized due to wave motion of electrons Davisson/Germer: electron diffraction by metal crystal Since matter has mass perhaps energy has mass ObservationTheory Einstein/deBroglie: Mass and energy are equivalent; particles have wavelength and photons have momentum. Compton: photon wavelength increases (momentum decreases) after colliding with electron Figure 7.14 continued QUANTUM THEORY Energy same as Matter particulate, massive, wavelike

31. It is impossible to know the exact position andmomentum of a particle simultaneously. Uncertainty: (Δx)(mΔv) > h/4  Δx is the location of the electron mΔv is its momentum More accurately we know the position of the particle less accurately we know the speed. Need Δ because we can’t know both at the same time

32. Sample Problem 7.4 (4 th ed) SOLUTION: PLAN: Applying the Uncertainty Principle PROBLEM:An electron moving near an atomic nucleus has a speed 6 x 10 6 ± 1% m/s. What is the uncertainty in its position (  x )? The uncertainty (  x ) is given as ±1%(0.01) of 6 x 10 6 m/s. Once we calculate this, plug it into the uncertainty equation.  u = (0.01)(6 x 10 6 m/s) = 6 x 10 4 m/s  x * m  u ≥ h 44 x ≥x ≥ 4  (9.11 x 10 -31 kg)(6 x 10 4 m/s) 6.626 x 10 -34 ( kg*m 2 /s 2 ). s = 9.52 x 10 -9 m 1 J = 1 kg*m 2 /s 2

33. Schrodinger developed Wave Functions, Ψ(psi), where Ψ 2 is the probability of finding e- in a given space Led to 4 quantum numbers that describe the e-'s position in a complex equation: 1. Only certain wave functions are allowed 2. Each Ψ n corresponds to an allowed energy for e- in atom 3. Thus energy of e- is quantized 4. Ψ has no physical meaning, but Ψ 2 give the probability density 5. Allowed energy states are called orbitals 6. 3 integer #'s req'd to solve Ψ 2 for 3-D space: n, l, m l

34. An atomic orbital is specified by three quantum numbers. n = principal quantum number, a positive integer = 1, 2, 3,... - determines total E of e- in its electron shell - gives measure of prob distance from nucleus (orbital size) - 2 or more e-s can be in same electron shell l = angular momentum or shape = < n - 1, = 0,1,2,... - subshells w/in main shell, characterized by certain wave shapes 0 = s, 1 = p, 2= d, 3 = f, etc. m l = magnetic q.n. = + l, + l -1, + l - 2, … 0,... - l - specifies which orbital w/in a subshell e- is in (later we’ll do m s = spin q.n., +½ or -½ for each e-) Watch: YouTube - ‪The Quantum Number Rag‬‏YouTube - ‪The Quantum Number Rag‬‏

35. Table 7.2 The Hierarchy of Quantum Numbers for Atomic Orbitals Name, Symbol (Property) Allowed ValuesQuantum Numbers Principal, n (size, energy) Angular momentum, l (shape) Magnetic, m l (orientation) Positive integer (1, 2, 3,...) 0 to n-1 - l,…,0,…,+ l 1 0 0 2 01 0 3 0 12 0 0 +1 0+1 0 +2-2

36. Sample Problem 7.5 SOLUTION: PLAN: Determining Quantum Numbers for an Energy Level PROBLEM: What values of the angular momentum ( l ) and magnetic (m l ) quantum numbers are allowed for a principal quantum number (n) of 3? How many orbitals are allowed for n = 3? Follow the rules for allowable quantum numbers found in the text. l values can be integers from 0 to n-1; m l can be integers from - l through 0 to + l. For n = 3, l = 0, 1, 2 For l = 0 m l = 0 For l = 1 m l = -1, 0, or +1 For l = 2 m l = -2, -1, 0, +1, or +2 There are 9 m l values and therefore 9 orbitals with n = 3.

37. Sample Problem 7.6 SOLUTION: PLAN: Determining Sublevel Names and Orbital Quantum Numbers PROBLEM:Give the name, magnetic quantum numbers, and number of orbitals for each sublevel with the following quantum numbers: (a) n = 3, l = 2(b) n = 2, l = 0(c) n = 5, l = 1(d) n = 4, l = 3 Combine the n value and l designation to name the sublevel. Knowing l, we can find m l and the number of orbitals. n l sublevel namepossible m l values# of orbitals (a) (b) (c) (d) 3 2 5 4 2 0 1 3 3d 2s 5p 4f -2, -1, 0, 1, 2 0 -1, 0, 1 -3, -2, -1, 0, 1, 2, 3 5 1 3 7

38. Ψ 2 1s – the 1s orbital is spherical. Ψ 2 2s – the 2s orbital has some density close to nucleus and then another sphere farther away – a sphere within a sphere Ψ 2 2p – the 2p orbitals have no probabilty of e- at the nucleus - called nodal plane Can be oriented in 3 directions of 3-D graph - x, y, z. 2p x, 2p y, 2p z have the 3 m l “names” +1, 0 and -1 Ψ 2 3d – the 3d orbitals have 5 m l values, and each has 2 nodal surfaces, so they are in four sections. 3d xy, 3d xz, 3d yz, 3d x2- y2, 3d z2 Ψ 2 4f – the 4f orbitals have 7 m l values, 3 nodal surfaces

39. Figure 7.15 Electron probability in the ground-state H atom.

40. Figure 7.16 The 1s, 2s, and 3s orbitals 1s2s 3s

41. Figure 7.17 The 2p orbitals.

42. Figure 7.18 The 3d orbitals.

43. Figure 7.18 continued

44. Figure 7.19 One of the seven possible 4f orbitals.

45.  Be able to draw 1s, 2s, 2p, and 3d orbitals.  Practice now!