Topic 6 - Quantum Theory of the Atom - Mr Crosby's Chemistry 主題6的原子的量子理論的克羅斯比先生的化學(xué)

1、.16Topic 6 Quantum Theory of the AtomWAVE NATURE OF LIGHTA. Properties of waves as they relate to light1. WaveA constantly repeating change or oscillation in matter or in a physical field2. Wavelengtha. DefinitionThe distance between identical points on successive wavesb. Symbol lc. Units(1) Usually

2、 nanometers (nm) 1 nm = 1 x 10-9 m(2) Sometimes meters (m)3. Frequencya. DefinitionThe number of waves that pass through a given point in one unit of time - usually one secondb. Symbol nc. Units(1) Reciprocal secondss-1(2) HertzHz = 1 = 1 = 1 s-14. Speed a. DefinitionThe distance a wave travels in o

3、ne unit of time - usually one secondb. Symbolcc. Speed of light(1) Depends on the medium(2) values2.9979 x 108 m/s2.9979 x 1017 nm/s(3) Mathematical relationshipc = lnB. Light as waves1. Light is one form of electromagnetic radiation which also includesradio wavesmicrowavesinfrared radiationvisible

4、lightultraviolet radiationX raysgamma rays2. Electromagnetic radiationa. Has an electric field component and a magnetic field component hence the term “electromagnetic”b. The electric field component and the magnetic field component(1) Travel in mutually perpendicular planes(2) Have the same wavelen

5、gth, frequency, and speedc. Electromagnetic spectrumThe whole range of wavelengths or frequencies of electromagnetic radiationC. Examples1. Finding wavelengthA laser used to weld detached retinas has a frequency of 4.69 x 1014 s-1. What is the wavelength of its light?c = lnl= c/nl = = 639 nm2. Findi

6、ng frequencyThe light given off by a sodium lamp has a wavelength of 589 nm. What is the frequency of this light?c = lnn = c/ln = = 5.09 x 1014 s-1FROM CLASSICAL PHYSICS TO QUANTUM THEORYA. Plancks theory1. The reason for its developmenta. Classical physics assumed that all energy changes were conti

7、nuous. (1) This means that there are no restrictions on the amount of one form of energy which can be converted to another form of energy.Example:A ball rolling downhill can change any amount of potential energy into kinetic energy.(2) This meant that atoms and molecules should be able to emit or to

8、 absorb any arbitrary amount of energy.b. The spectrum of light emitted by a hot object blackbody radiation could not be explained by classical physics.2. Planck proposalsa. That energy could only be released or absorbed in chunks of some minimum size.b. That atoms and molecules could only emit or a

9、bsorb energy in discrete quantitiesc. Two comparisons between continuous and discrete(1) Violin and piano A violin can play every pitch between the two notes B and C - this is continuous.A piano can only play the pitch for the note B or the pitch for the note C - this is discrete.(2) Inclined plane

10、and stairsA ball can roll down in inclined plane and have any possible height on the plane between the top and the bottom - this is continuous.A ball rolling down a flight of stairs can only have certain heights corresponding to the height of the stair it is on at the time - this is discrete.3. Plan

11、cks quantum theorya. The smallest possible increment of energy that can be gained or lost was called a “quantum” (the plural is “quanta”).b. Radiant energy is always emitted or absorbed in whole number multiples of a constant “h” times the frequency.DE = hn, 2 hn, 3 hn, c. “h” is known as “Plancks c

12、onstant” and has the value of 6.6262 x 10-34 J·sB. The photoelectric effect1. The photoelectric effect could not be explained by classical physics.a. The photoelectric effect described(1) The photoelectric effect is the ejection of electrons from the surface of certain metals when light of at l

13、east a certain minimum frequency (called the threshold frequency) was shined upon them.(2) Whether or not current flowed depended on the frequency of the light NOT its intensity.(3) Increasing the intensity of the light increased the amount of the current.(a) Dim low frequency lightß ß

14、0; No flow of electrons(b) Intense low frequency light ßß¾ No flow of electrons(c) Dim high frequency lightßß¾ Small flow of electrons(d) Intense high frequency lightßß¾ Greater flow of electronsb. Classical physics predicted that intensity should determi

15、ne the amount of current and that frequency should be irrelevant.2. Einsteins use of Plancks theory to explain the photoelectric effecta. Einsteins three assumptions(1) He assumed that a beam of light is really a stream of particles (now called “photons”).(2) He assumed that each photon ejects one e

16、lectron when it strikes the metal.(3) He also assumed that this photon must have at least enough energy to free the electron from the forces that hold it in the atom.b. Using Plancks constant he calculated the energy of a photon from its frequency.E = hnc. Einsteins explanation using these assumptio

17、ns:It does not matter how many photons strike the metal surface if none of them have enough energy to kick out an electron.If a photon with at least the minimum energy strikes the metal, then that photon is absorbed by the electron.A certain minimum amount of energy is needed to free the electron.Th

18、e excess energy, if any, goes into the kinetic energy of the electron.3. Examplesa. Calculating energy from frequencyWhat is the energy of a X-ray photon with a frequency of 6.00 x 1018 s-1?E = hn = (6.6262 x 10-34 J·s)(6.00 x 1018 s-1) = 3.98 x 10-15 Jb. Calculating the energy from wavelengthW

19、hat is the energy of an infrared photon with a wavelength of 5.00 x 104 nm?E = hnSubstituting for n gives usE = Since c = 2.9979 x 1017 nm/sE = E = 3.97 x 10-21 JC. The emission spectrum of hydrogen1. The prediction of classical physics and Rutherfords model of the atom a. Has the electrons orbiting

20、 around the nucleus where the attractive force of the nucleus is exactly balanced by the acceleration due to the circular motion of the electron.b. The two observed problems with the Rutherford model(1) The stability of the atomA charged particle, such as the electron, moving around the nucleus shou

21、ld lose energy and spiral down into the nucleusin about 10-10 s.(2) The line spectrum of atomsThe electron should be able to lose energy inany amounts Which should produce a continuous spectrum (all colors like a rainbow) Rather than a line spectrum (a set of lines of specific colors)2. The new mode

22、l of the atom would use quantum theory.BOHRS THEORY OF THE HYDROGEN ATOMA. Bohrs Postulates1. The electron moves in a circular orbit around the proton.a. These orbits can only have certain radii corresponding to certain definite energies.b. These energies are given by the equationEn = -RHRydberg dev

23、eloped this from his study of the line spectra of many elements.RH = 2.179 x 10-18 J (the Rydberg constant) n = 1, 2, 3, (indicates the energy level of the electron)The “-” sign indicates that the energy of the electron in the atom is lower than the energy of a free electron.c. As the electron gets

24、closer to the nucleus, En increases in absolute value, but becomes more negative.Think about a ball rolling down a staircase.When it reaches the lowest step (n = 1) it has its lowest potential energy and it is the most stable.d. Ground state and excited state.n = 1 is the ground state for the hydrog

25、en electronn = 2, 3, are the excited states for the hydrogen electron2. Transitions of the electron occurs between specific energy states and involves the emission or absorption of a photon of a specific energy and frequency.B. Bohrs explanation of the emission spectrum of hydrogen. A photon is emit

26、ted when an electron transitions from one energylevel to a lower one. DE = Efinal - Einitial Efinal = -RH and Einitial = -RH DE = - DE = - +DE = hn = RH n = c/l hc/l = RH1/l = C. Bohrs model completely explained the observed emission spectra of hydrogen, including the Paschen (IR), the Balmer (visib

27、le), and the Lyman (UV) series.D. ExampleWhat is the wavelength of the photon emitted when a hydrogen electron transitions from the n = 4 state to the n = 2 state?RH = 2.179 x 10-18 Jh = 6.6262 x 10-34 J·sc = 2.9979 x 1017 nm/sninitial = 4nfinal = 21/l = 1/l = 1/l = -2.0567 x 10-3 l = - 486.2 n

28、mThe negative sign comes from the calculation of the energy, indicating that light is emitted. l = 486.2 nmTHE DUAL NATURE OF THE ELECTRONA. The history of deBroglies proposal1. Physicists accepted Bohrs model but were puzzled as to why the energy level of the hydrogen atom should be quantized.2. Ei

29、nstein has shown that light has both wave properties and particle properties. 3. deBroglie proposed that particles such as electrons can also posses wave properties under the proper circumstances.B. deBroglies two proposals1. The first proposal was the standing wave model of the electrona. An electr

30、on bound to the nucleus behaves like a standing wave.(1) A standing wave is similar to plucking the string of a guitar.(2) These waves get their name from the fact that they are stationary they do not travel along the string.b. If the electron behaves like a standing wave then the length of the wave

31、 must fit the circumference of the circle (the orbit) exactly, otherwise it would partially cancel itself out. c. Since 2p r = the circumference of the orbit, then the wavelengths that will fit are: 2p r = l 2p r = 2l 2p r = 3l 2p r = nl2. The second proposal was that very small particles moving ver

32、y fast would exhibit wavelike properties particularly a wavelength.a. This would NOT be observable in the macroscopic world due to the insignificant wavelength.b. In the submicroscopic world the wavelength could be calculated using:l = h = Plancks constant = 6.6262 x 10-34 J·sbut since 1 J = 1

33、J·s = ·sh = 6.6262 x 10-34 m = mass in kgv = velocity in m/s3. ExampleWhat is the wavelength associated with an electron with amass of 9.11 x 10-31 kg and a velocity of 4.19 x 106 m/s? l = = = 1.74 x 10-10 mC. Heisenbergs Uncertainty Principle1. Because an electron can behave like a wave i

34、t is difficult to determine exactly where an electron is.2. The precise location of a wave cannot be specified because a wave extends out in space.3. To describe this problem Heisenberg formulated his uncertainty principlea. It is impossible to know simultaneously both the exact position and the exa

35、ct momentum of a particle.b. There is always a limit to how precisely we can know both values at the same time.c. DxDp ³QUANTUM MECHANICSA. Formulated by Ernst Schroedinger for the hydrogen atom.B. The Schroedinger equation incorporates both particle behavior and wave behavior.C. Solving the Sc

36、hroedinger equation for a hydrogen atom 1. Requires advanced calculus even for so simple a system2. Produces a wave function y3. Each wave functiona. Specifies the possible energy states that an electron can occupy in a hydrogen atomb. Is characterized by a set of quantum numbersD. Electron density1

37、. The square of the wave function y2 gives the probability of finding the electron in a certain region of space at a given instant.2. Regions of high electron density are areas where there is a high probability of finding the electron.3. Regions of low electron density are areas where there is a low

38、 probability of finding the electron.E. Orbitals1. The complete set of solutions to the Schroedinger equation yields a set of wave functions and a corresponding set of energies.2. Each of these allowed wave functions is called an orbital.3. Each orbital describes a specific distribution of electron

39、density in space.4. An atomic orbital describes the region of space where there is a high probability of finding the electron.5. Each orbital has a. A characteristic sizeb. A characteristic shapec. A characteristic orientation in spaceF. The many-electron atom1. The Schroedinger equation for them ca

40、nnot be solved.2. The energies and wave functions of the hydrogen atom are a good approximation of the behavior of the electrons in more complex atoms.QUANTUM NUMBERSA. The principal quantum number1. Describes the size of the orbital2. Is symbolized by “n”3. “n” can have the value of any non-zero in

41、tegern = 1, 2, 3, 4. These are sometimes referred to as “shells”.5. Is the quantum number on which the energy of an electron principally but NOT exclusively depends6. The larger the principal quantum numbera. The greater the average distance of the electron from the nucleusb. The greater the energy

42、of the electron (generally)c. The less tightly the electron is bound to the nucleusd. The less stable the condition of having the electron in that orbitalB. The angular momentum quantum number1. Gives the shape of the orbital2. Is symbolized by “l(fā) ”3. “l(fā) ”can have integer values from 0 to n - 14. Th

43、ere are “n” different values for “l(fā) ”.5. Different values of “l(fā) ” are usually denoted by letters.These are from the old spectroscopic terminologya. 0 = s(sharp, NOT spread out)b. 1 = p(principal, very strong)c. 2 = d(diffuse, rather spread out)d. 3 = f(fundamental)e. 4 = g6. These are sometimes refe

44、rred to as “sublevels” or as “subshells”. 7. Subshells are designated as the “n” followed by the letter of the subshellExamples:1s4f8. To whatever extent the energy of an orbital does not depend on the principal quantum number it will depend on the angular momentum quantum number.C. The magnetic qua

45、ntum number1. Gives the orientation of the orbital of the same energy and shape2. Is symbolized by “ml ”.3. ml can have integer values from - l to + l .4. There are 2l + 1 different values for ml .5. These are referred to as “orbitals”.D. The spin quantum number1. Describes which of the two possible spin orientations of t