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Chapter 4. Properties of Waves Wavelength, (lambda) - the distance between identical points on successive waves. Amplitude - the vertical distance from.

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Presentation on theme: "Chapter 4. Properties of Waves Wavelength, (lambda) - the distance between identical points on successive waves. Amplitude - the vertical distance from."— Presentation transcript:

1 Chapter 4

2 Properties of Waves Wavelength, (lambda) - the distance between identical points on successive waves. Amplitude - the vertical distance from the midline of a wave to the peak or trough. Frequency, (nu) - the number of waves that pass through a particular point in 1 second (Hz = 1 cycle/s). The speed (u) of the wave = x

3 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 ELECTROMAGNETIC RADIATION All electromagnetic radiation x c

4 TYPES OF ELECTROMAGNETIC RADIATION

5 x = c = c/ = 3.00 x 10 8 m/s / 6.0 x 10 4 Hz = 5.0 x 10 3 m 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 nm

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7 PLANCKS QUANTUM THEORY Quantum – the smallest quantity of energy that can be emitted (or absorbed) in the form of electromagnetic radiation. When solids are heated, they emit electromagnetic radiation over a wide range of wavelengths. Radiant energy emitted by an object at a certain temperature depends on its wavelength. Atoms and molecules could emit (or absorb) energy (light) in discrete units (quantum).

8 E = h x c E = h c h = Plancks constant = 6.63 x Js frequency of the radiation

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10 Light has both: 1.wave nature 2.particle nature h = KE + W Photoelectric Effect Photon is a particle of light KE = h - W h KE e - W = is the work function and depends how strongly electrons are held in the metal KE = kinetic energy of the ejected electron

11 E = h x E = 6.63 x (J s) x 3.00 x 10 8 (m/s) / x (m) E = 1.29 x J E = h x c / EXAMPLE: 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 nm.

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13 Line Emission Spectrum of Hydrogen Atoms EMISSION SPECTRA

14 EMISSION SPECTRA OF VARIOUS ELEMENTS

15 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 Bohrs 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 J

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

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18 E photon = 2.18 x J x (1/25 - 1/9) E photon = E = x J = 6.63 x (Js) x 3.00 x 10 8 (m/s)/1.55 x 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 =

19 QUANTUM NUMBERS Atomic orbital – the wave function of an electron in an atom 3 quantum numbers required to describe the distribution of electrons in hydrogen and other atoms. i) The Principle Quantum Number (n) ii)The Angular Momentum Quantum Number (l) iii)The Magnetic Quantum Number (m l ) Describes the behavior of a specific electron i) The Spin Quantum Number (m s ) Quantum Numbers, = (n, l, m l, m s )

20 Principal Quantum Number ( n) n = 1, 2, 3, 4, …. n=1 n=2 n=3 - distance of e - from the nucleus - the larger n value, the greater average distance of an e - in the orbital from the nucleus, and the larger the orbital

21 Angular Momentum Quantum Number (l) for a given value of n, l = 0, 1, 2, 3, … n-1 n = 1, l = 0 n = 2, l = 0 or 1 n = 3, l = 0, 1, or 2 - Shape of the volume of space that the e - occupies /orbitals l = 0 s orbital l = 1 p orbital l = 2 d orbital l = 3 f orbital

22 Where 90% of the e - density is found for the 1s orbital

23 l = 0 (s orbitals) l = 1 (p orbitals)

24 l = 2 (d orbitals)

25 Magnetic Quantum Number (m l ) for a given value of l m l = -l, …., 0, …. +l - orientation of the orbital in space if l = 1 (p orbital), m l = -1, 0, or 1 if l = 2 (d orbital), m l = -2, -1, 0, 1, or 2 * Each subshell of quantum number (l) contains (2l + 1) orbitals.

26 m l = -1, 0, or 1 3 orientations is space

27 m l = -2, -1, 0, 1, or 2 5 orientations is space

28 Spin Quantum Number (m s ) m s = +½ or -½ m s = -½m s = +½

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30 Quantum Numbers: (n, l, m l, m s ) Shell – electrons with the same value of n Subshell – electrons with the same values of n and l Orbital – electrons with the same values of n, l, and m l How many electrons can an orbital hold? If n, l, and m l are fixed, then m s = ½ or - ½ = (n, l, m l, ½) or = (n, l, m l, -½) An orbital can hold 2 electrons

31 How many 2p orbitals are there in an atom? 2p2p n=2 l = 1 If l = 1, then m l = -1, 0, or +1 3 orbitals How many electrons can be placed in the 3d subshell? 3d3d n=3 l = 2 If l = 2, then m l = -2, -1, 0, +1, or +2 5 orbitals which can hold a total of 10 e -

32 Existence (and energy) of electron in atom is described by its unique wave function. Pauli exclusion principle - no two electrons in an atom can have the same four quantum numbers. Schrodinger Wave Equation quantum numbers: (n, l, m l, m s )

33 Energy of orbitals in a single electron atom Energy only depends on principal quantum number n E n = -R H ( ) 1 n2n2 n=1 n=2 n=3

34 Energy of orbitals in a multi -electron atom Energy depends on n and l n=1 l = 0 n=2 l = 0 n=2 l = 1 n=3 l = 0 n=3 l = 1 n=3 l = 2

35 Fill up electrons in lowest energy orbitals (Aufbau principle) H 1 electron H 1s 1 He 2 electrons He 1s 2 Li 3 electrons Li 1s 2 2s 1 Be 4 electrons Be 1s 2 2s 2 B 5 electrons B 1s 2 2s 2 2p 1 ??

36 Order of orbitals (filling) in multi-electron atom 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s

37 C 6 electrons The most stable arrangement of electrons in subshells is the one with the greatest number of parallel spins (Hunds rule). C 1s 2 2s 2 2p 2 N 7 electrons N 1s 2 2s 2 2p 3 O 8 electrons O 1s 2 2s 2 2p 4 F 9 electrons F 1s 2 2s 2 2p 5 Ne 10 electrons Ne 1s 2 2s 2 2p 6

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41 Electron configuration is how the electrons are distributed among the various atomic orbitals in an atom. 1s 1 principal quantum number n angular momentum quantum number l number of electrons in the orbital or subshell Orbital diagram H 1s 1

42 Electronic Configurations of Some Second Row Elements

43 What is the electron configuration of Mg? Mg 12 electrons 1s < 2s < 2p < 3s < 3p < 4s 1s 2 2s 2 2p 6 3s = 12 electrons Abbreviated as [Ne]3s 2 [Ne] 1s 2 2s 2 2p 6 What are the possible quantum numbers for the last (outermost) electron in Cl? Cl 17 electrons1s < 2s < 2p < 3s < 3p < 4s 1s 2 2s 2 2p 6 3s 2 3p = 17 electrons Last electron added to 3p orbital n = 3l = 1m l = -1, 0, or +1m s = ½ or -½

44 Outermost subshell being filled with electrons

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46 Paramagnetic unpaired electrons 2p Diamagnetic all electrons paired 2p

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