Chapter 7. Wave Behavior of Electrons

Transcription

1 Chapter 7 Wave Behavior of Electrons

2 2-Slit Interference

3 If electrons behave only like particles, there should only be two bright spots on the target

4 However, electrons actually present an interference pattern, demonstrating they behave like waves

5 Wave Behavior of Electrons Louis de Broglie ( ) Particles could have wave-like character. The wavelength of a particle was inversely proportional to its momentum (m x v). Because it is so small, the wave character of electrons is significant.

6 Wave Behavior of Electrons allowed forbidden

7

8 1. Calculate the wavelength of an electron traveling at 2.65 x 10 6 m/s (Masselectron = 9.11 x kg)

9 2. Determine the wavelength of a neutron traveling at 1.00 x 10 2 m/s (Massneutron = x g)

10 3. Determine the wavelength of a 80 kg human traveling at 1.00 m/s 8.0 x 10 2 = 8.3 x m

11 Complementary Properties When you try to observe the wave nature of the electron, you cannot observe its particle nature and vice-versa. wave nature = interference pattern particle nature = position The wave and particle nature of the electron are complementary properties. (As you know more about one you know less about the other.)

12 Uncertainty Principle Werner Karl Heisenberg ( ) The product of the uncertainties in both the position and speed of a particle was inversely proportional to its mass. x = position, Δx = uncertainty in position v = velocity, Δv = uncertainty in velocity m = mass (m)(δx)(δv) h/4π This means that the more accurately you know the position of a small particle, such as an electron, the less you know about its speed and vice-versa.

13 For an electron, m = 9 x kg, (Δx)(Δv) 6 x 10-5 Re = 3 x m; for an uncertainty in position of 1 radius, the uncertainty in velocity is approximately 20,000,000,000 m/sec. For an neutron, m = 2 x kg, (Δx)(Δv) 3 x 10-8 Rn = 1 x m; for an uncertainty in position of 1 radius, the uncertainty in velocity is approximately 300,000 m/sec. For an 80 kg man, m = 80 kg, (Δx)(Δv) 2 x Rhuman = 1 m; for an uncertainty in position of 1 radius, the uncertainty in velocity is approximately m/sec.

14 Determinacy vs. Indeterminacy According to classical physics, particles move in a path determined by the particle s velocity, position, and forces acting on it. Because we cannot know both the position and velocity of an electron, we cannot predict the path it will follow. The best we can do is to describe the probability an electron will be found in a particular region using statistical functions. Electron energy and position are complementary because KE = ½mv 2. Statistical Mechanics For an electron with a given energy, the best we can do is describe a region in the atom of high probability of finding it.

15 Trajectory vs. Probability

16 Schrödinger s Equation Erwin Schrödinger ( ) Schödinger s Equation allows us to calculate the probability of finding an electron with a particular amount of energy at a particular location in the atom. Solutions to Schödinger s Equation produce many wave functions, Ψ. A plot of distance vs. Ψ 2 represents an orbital, a probability distribution map of a region where the electron is likely to be found.

17 Solutions to the Wave Function, Ψ Calculations show that the size, shape, and orientation in space of an orbital are determined by three integer terms in the wave function. These integers are called quantum numbers. principal quantum number, n angular momentum quantum number, l magnetic quantum number, ml

18 The Quantum Numbers n, l, ml

19 Principal Quantum Number, n Corresponds to Bohr s energy level. The principal quantum number n can be any integer 1. The larger the value of n, the more energy the orbital has. Energies are defined as being negative. (An electron would have E = 0 when it just escapes the atom.) The larger the value of n, the larger the orbital. As n gets larger, the amount of energy between orbitals gets smaller.

20 Principal Energy Levels in Hydrogen

21 Angular Momentum Quantum Number, l The angular momentum quantum number determines the shape of the orbital. l can have integer values from 0 to (n 1). Each value of l is called by a particular letter that designates the shape of the orbital. s orbitals (l =0) are spherical p orbitals (l =1) are like two balloons tied at the knots d orbitals (l =2) are mainly like four balloons tied at the knot f orbitals (l = 3) are mainly like eight balloons tied at the knot

22 Magnetic Quantum Number, ml The magnetic quantum number is an integer that specifies the orientation of the orbital. Values are integers from l to +l including zero. Gives the number of orbitals of a particular shape. Ex: when l = 2, the values of ml are 2, 1, 0, +1, +2; which means there are five orbitals with l = 2

23 The Hierarchy of Quantum Numbers for Atomic Orbitals Name, Symbol (Property) Allowed Values Quantum Numbers Principal, n (size, energy) Positive integer (1, 2, 3,...) Angular momentum, l (shape) 0 to n Magnetic, m l (orientation) l,,0,,+l

24 Describing Orbitals

25 Describing an Orbital Each set of n, l, and ml describes one orbital. (the position of one electron) Orbitals with the same value of n are in the same principal energy level (or principal shell). Orbitals with the same values of n and l are said to be in the same sublevel (or subshell).

26 4. What are the quantum numbers and names of the orbitals in the n = 1 principal level? How many orbitals exist? n = 1 l : 0 n =1, l = 0 (s) ml : 0 1 orbital 1s total orbitals: 1 = 1 2 : n 2

27 5. What are the quantum numbers and names of the orbitals in the n = 2 principal level? How many orbitals exist? n = 2 l : 0,1 n = 2, l = 0 (s) ml : 0 1 orbital 2s n = 2, l = 1 (p) ml : 1,0,+1 3 orbitals 2p total of 4 orbitals: = 2 2 : n 2

28 6. What are the quantum numbers and names of the orbitals in the n = 3 principal level? How many orbitals exist? n = 3 l : 0,1,2 n = 3, l = 0 (s) ml : 0 1 orbital 3s n = 3, l = 1 (p) ml : 1,0,+1 3 orbitals 3p n = 3, l = 2 (d) ml : 2, 1,0,+1,+2 5 orbitals 3d total of 9 orbitals: = 3 2 : n 2

29 7. What are the quantum numbers and names of the orbitals in the n = 4 principal level? How many orbitals exist? n = 4 l : 0, 1, 2, 3 n = 4, l = 0 (s) n = 4, l = 1 (p) n = 4, l = 2 (d) n = 4, l = 3 (f) ml : 0 1 orbital 4s ml : 1,0,+1 3 orbitals 4p ml : 2, 1,0,+1,+2 5 orbitals 4d ml : 3, 2, 1,0,+1,+2,+3 7 orbitals 4f total of 16 orbitals: = 4 2 : n 2

30 Electron Arrangement of Elements

31 Quantum Mechanical Explanation of Atomic Spectra Each wavelength in the spectrum of an atom corresponds to an electron transition between orbitals. When an electron is excited, it transitions from an orbital in a lower energy level to an orbital in a higher energy level. When an electron relaxes, it transitions from an orbital in a higher energy level to an orbital in a lower energy level. When an electron relaxes, a photon of light is released whose energy equals the energy difference between the orbitals. Each line in the emission spectrum corresponds to the difference in energy between two energy states.

32 Energy Transitions in Hydrogen

33 Bohr Model of H Atoms

34 Quantum Leaps

35 Predicting the Spectrum of Hydrogen The wavelengths of lines in the emission spectrum of hydrogen can be predicted by calculating the difference in energy between any two states. For an electron in energy state n, there are (n 1) energy states it can transition to, therefore (n 1) lines it can generate. Both the Bohr and Quantum Mechanical Models can predict these lines very accurately for a 1- electron system.

36 Energy Transitions in Hydrogen The energy of a photon released is equal to the difference in energy between two energy levels. ΔEelectron = Efinal state Einitial state Eemitted photon = ΔEelectron It can be calculated by subtracting the energy of the initial state from the energy of the final state. RH

37 8. Calculate the wavelength of light emitted when the hydrogen electron transitions from n = 3 to n = 2 n i, n f ΔE atom E photon λ ΔE atom = E photon ( - ) x (1/4-1/9 = 0.139) Ephoton = ( x J) = x x JJ 3.03 x J 6.56

38 9. Calculate the wavelength of light emitted when the hydrogen electron transitions from n = 2 to n = 1 n i, n f ΔE atom E photon λ ΔE atom = E photon Ephoton = ( 1.64 x J) = 1.64 x J (1-¼ = 0.750) The unit is correct, the wavelength is in the UV, which is appropriate because it has more energy than 3 2 (in the visible).

39 9. Calculate the wavelength of light emitted when the hydrogen electron transitions from n = 6 to n = 5. n i, n f ΔE atom E photon λ ΔE atom = E photon Ephoton = ( x J) = x J (1/25-1/36 = 0.122) The unit is correct, the wavelength is in the infrared, which is appropriate because it has less energy than 3 2 (in the visible).

40 Wavelengths Associated with Hydrogen Electron Transitions 7.46 x10-6 m x10-7 m x10-7 m 2 1

41 The Shapes of Atomic Orbitals

42 The Shapes of Atomic Orbitals The l quantum number primarily determines the shape of the orbital. l can have integer values from 0 to (n 1) Each value of l is called by a particular letter that designates the shape of the orbital. s orbitals are spherical p orbitals are like two balloons tied at the knots d orbitals are mainly like four balloons tied at the knot f orbitals are mainly like eight balloons tied at the knot

43 l = 0, the s orbital Each principal energy level has one s orbital Lowest energy orbital in a principal energy state Spherical Number of nodes = (n 1)

44 1s, 2s and 3s Orbitals 1s 2s 3s

45 l = 1, p orbitals Each principal energy state above n = 1 has three p orbitals ml = 1, 0, +1 Each of the three orbitals points along a different axis px, py, pz 2 nd lowest energy orbitals in a principal energy state Two-lobed One node at the nucleus, total of n nodes

46 p orbitals

47 l = 2, d orbitals Each principal energy state above n = 2 has five d orbitals ml = 2, 1, 0, +1, +2 Four of the five orbitals are aligned in a different plane the fifth is aligned with the z axis, d z 2, d xy, d yz, d xz, d z2 y2 3 rd lowest energy orbitals in a principal energy level Mainly four-lobed Planar nodes higher principal levels also have spherical nodes

48 d orbitals

49 l = 3, f orbitals Each principal energy state above n = 3 has seven d orbitals ml = 3, 2, 1, 0, +1, +2, +3 4 th lowest energy orbitals in a principal energy state Mainly eight-lobed Planar nodes higher principal levels also have spherical nodes

50 f orbitals

51 Why are Atoms Spherical?