Properties of Waves. Wavelength ( ) is the distance between identical points on successive waves.
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1 0/9/2009 Properties of Waves Properties of Waves Quantum Theory and the Electronic Structure of Atoms Chapter 7 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 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. Frequency ( ) is the number of waves that pass through a particular point in second (Hz = cycle/s). The speed (u) of the wave = x 7. Maxwell (873), proposed that visible light consists of electromagnetic waves. Electromagnetic radiation is the emission and transmission of energy in the form of electromagnetic waves. A photon has a frequency of 6.0 x 0 4 Hz. Convert this frequency into wavelength (nm). Does this frequency fall in the visible region? x = c = c/ = 3.00 x 0 8 m/s / 6.0 x 0 4 Hz = 5.0 x 0 3 m = 5.0 x 0 2 nm Speed of light (c) in vacuum = 3.00 x 0 8 m/s All electromagnetic radiation x c Radio wave 7. Mystery #, Black Body Problem Solved by Planck in 900 Energy (light) is emitted or absorbed in discrete units (quantum). E = h x Planck s constant (h) h = 6.63 x 0-34 J s Mystery #2, Photoelectric Effect Solved by Einstein in 905 Light has both:. wave nature 2. particle nature Photon is a particle of light h = KE + BE KE = h - BE h KE e - 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 0.54 nm. E = h x E = h x c / E = 6.63 x 0-34 (J s) x 3.00 x 0 8 (m/s) / 0.54 x 0-9 (m) E =.29 x 0-5 J
2 0/9/2009 Bohr s Model of the Atom (93). e - can only have specific (quantized) energy values Line Emission Spectrum of Hydrogen Atoms 2. light is emitted as e - moves from one energy level to a lower energy level E n = -R H( ) n 2 n (principal quantum number) =,2,3, R H (Rydberg constant) = 2.8 x 0-8 J E = h E = h n i = 3 n i = 2 n i = 3 n f = 2 E photon = E = E f - E i E f = -R H ( ) n 2 f E i = -R H ( ) n 2 i E = R H( ) n 2 i n 2 f 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 = E = R H( ) n 2 E photon = 2.8 x 0-8 J x (/25 - /9) E photon = E = -.55 x 0-9 J E photon = h x c / i n 2 f = h x c / E photon n f f = = 6.63 x 0-34 (J s) x 3.00 x 0 8 (m/s)/.55 x 0-9 J = 280 nm Why is e - energy quantized? De Broglie (924) reasoned that e - is both particle and wave. 2 r = n = h/mu u = velocity of e - m = mass of e - What is the de Broglie wavelength (in nm) associated with a 2.5 g Ping-Pong ball traveling at 5.6 m/s? = h/mu h in J s m in kg = 6.63 x 0-34 / (2.5 x 0-3 x 5.6) =.7 x 0-32 m =.7 x 0-23 nm u in (m/s) In 926 Schrodinger wrote an equation that described both the particle and wave nature of the e - Wave function ( ) describes:. energy of e - with a given 2. probability of finding e - in a volume of space Schrodinger s equation can only be solved exactly for the hydrogen atom. Must approximate its solution for multi-electron systems
3 0/9/2009 fn(n, l, m l, m s ) principal quantum number n n =, 2, 3, 4,. distance of e - from the nucleus n= n=2 n=3 Where 90% of the e - density is found for the s orbital e - density (s orbital) falls off rapidly as distance from nucleus increases angular momentum quantum number l for a given value of n, l = 0,, 2, 3, n- n =, l = 0 n = 2, l = 0 or n = 3, l = 0,, or 2 l = 0 l = l = 2 l = 3 s orbital p orbital d orbital f orbital Shape of the volume of space that the e - occupies l = 0 (s orbitals) l = 2 (d orbitals) magnetic quantum number m l l = (p orbitals) for a given value of l m l = -l,., 0,. +l if l = (p orbital), m l = -, 0, or if l = 2 (d orbital), m l = -2, -, 0,, or 2 orientation of the orbital in space m l = - m l = 0 m l = spin quantum number m s m s = +½ or -½ 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. m l = -2 m l = - m l = 0 m l = m l = 2 m s = +½ m s = -½ Each seat is uniquely identified (E, R2, S8) Each seat can hold only one individual at a time 3
4 0/9/2009 Shell electrons with the same value of n Subshell electrons with the same values of n and l How many orbitals are there in an atom? n=2 l = If l =, then m l = -, 0, or + 3 orbitals Energy of orbitals in a single electron atom Energy only depends on principal quantum number n n=3 n=2 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) How many electrons can be placed in the 3d subshell? n=3 3d If l = 2, then m l = -2, -, 0, +, or +2 5 orbitals which can hold a total of 0 e - n= E n = -R H( ) n 2 An orbital can hold 2 electrons l = 2 Energy of orbitals in a multi-electron atom Energy depends on n and l Fill up electrons in lowest energy orbitals (Aufbau principle) The most stable arrangement of electrons in subshells is the one with the greatest number of parallel spins (Hund s rule). n=3 l = 2 n=3 l = 0 n=3 l =?? n=2 l = 0 n=2 l = Li Be B C electrons B Be Li s s 2 2s 2 2s 2 2 C O F Ne electrons Ne C N O F s 2 2 2s H He 2 electrons n= l = 0 He H s 2 Order of orbitals (filling) in multi-electron atom What is the electron configuration of Mg? Mg 2 electrons s < 2s < < 3s < 3p < 4s s 2 2s 2 6 3s = 2 electrons Abbreviated as [Ne]3s 2 [Ne] s 2 2s 2 6 Outermost subshell being filled with electrons s < 2s < < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s What are the possible quantum numbers for the last (outermost) electron in Cl? Cl 7 electrons s 2 2s 2 6 3s 2 3p 5 Last electron added to 3p orbital s < 2s < < 3s < 3p < 4s = 7 electrons n = 3 l = m l = -, 0, or + m s = ½ or -½ 7.8 4
5 0/9/2009 Paramagnetic unpaired electrons Diamagnetic all electrons paired 7.8 5
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