Clicker question Which of the following electron transitions in a hydrogen atom will result in the emission of light with the shortest shortest wavelength wavelength A n 1 to n 3 a B n 3 to n 1 e C n 4 to n 3 e D n 2 to n 3 a E n 3 to n 2 e Ephoton hc wavelength delta E of the transition Ephoton Smallest wavelength high Ephoton largest delta E 1 Upcoming deadlines Finish the Week 1 Lectures 1 3 homework due tomorrow at recitation See Homework Expectations document on ANGEL Take the ANGEL Quiz 1 due tomorrow by 11 55PM Read Lessons 03 1 to 03 4 including the reading guides signs before class on Fri 9 4 example problems and stop Finish up ALEKS due Sun 9 6 Take the pre test in the CHEM 110 FA15 Grades Conflict Signup Course on ANGEL due Mon 9 7 by 11 55PM Take the Intro quiz on ANGEL due Thurs 9 10 by 11 55PM eBook 2 Electronic transitions and emission lines All electronic transitions that go down to nf 1 will be in the UV range of the EM spectrum All electronic transitions that go down to nf 2 will be n 2 1 2 3 4 Energy This is due to the fact that there is a very large energy gap between n 1 and n 2 Electrons that undergo a transition through this large energy gap will release photons with high energy n 6 n 5 n 4 n 3 in the visible range of the EM spectrum All electronic transitions that go down to nf 3 will be in the IR range of the EM spectrum If we look at the Hydrogen emission spectrum in the visible range all wavelengths observed are only due to transitions that involve nf 2 High E Low E Transition 1 2 3 4 n 1 3 The Bohr model only works for one electron atoms Orbits do not really paint an accurate picture of the location of electrons In 1924 Louis de Broglie proposed that matter has dual wave particle behavior just like light h Wavelength of matter waves mv The de Broglie wavelength is only significant for extremely small masses like e de Broglie For an e the wavelength is on the order of the atom s size For a baseball and bacteria is too small to observe The wave like property of e s was proved in 1927 Scanning Electron Microscopy images http www clemson edu glimpse wp content uploads 2012 09 Broglie Louis Victor de jpg https tysontrepidations files wordpress com 2011 07 pollen under scanning electron microscope jpg http archives starbulletin com 1999 04 02 features story2 html 4 Consequence for electrons in atoms The wave like nature of matter is important when dealing with subatomic matter The Heisenberg Uncertainty Principle It is impossible to simultaneously know the exact position and momentum mv of a particle The mass of the electron is well known and the velocity can be determined therefore we can only calculate the probability of finding an e The Schr dinger equation can be used to determine the probability of finding an electron in a region of space the electron density These clouds of electron density are called orbitals https s media cache ak0 pinimg com 236x 80 d4 12 80d41204b641b13ac66f40814fb4f702 jpg 5 Orbitals are described by three quantum numbers 1 Principal quantum number n size information same from Bohr model 2 Angular momentum quantum number shape information 3 Magnetic quantum number m orientation information The orbitals only exist if we have at least one electron in them Orbitals only tell us where we are likely to find an electron 6 The principal quantum number n The principal quantum number must be an integer greater than zero n 1 2 3 The principal quantum number n gives information about The size of the orbital as n increases the size of orbital increases The energy associated with an electron in the orbital as n increases the energy increases as e are further away from nucleus become less stable higher E n 1 n 2 n 3 7 The angular momentum quantum number The values of depend on the principal quantum number n For a given n the possible values of include all integers from 0 to n 1 Examples If n 1 If n 3 l 0 n 1 1 1 0 l 1 2 max is n 1 3 1 2 shape The angular momentum quantum number gives us information about We use symbols instead of numbers for 0 1 2 Symbol s p d 3 f spherical dumkbell 4 leaf clover 8 Magnetic quantum number m The values of m depend on the angular momentum quantum number For a given the possible values of m include all integers from to Examples If 1 If 2 m l 1 0 1 three p orbitals m l 2 1 0 1 2 five d orbitals of orbital The number of m values tells you the with the given value orientation The magnetic quantum number gives information about p orbitals l 1 d orbitals l 2 n 2 OF ORBITALS 2l 1 9 Shells subshells orbitals one n A shell is described by quantum number 3rd shell n 3 two n and l A subshell is described by quantum numbers Example 3s subshell is defined by n 3 l 0 three n l and m l An orbital is described by quantum numbers Examples n 1 n 0 1 m 0 1 Subshell 2s 2p 0 m 0 Subshell one orbital 1s 1s 2 one orbital 2s 0 1 three orbitals 2p 2px 2py 2pz 10 You should understand how to determine all these numbers and what they mean 11 Clicker question Which of the following set of quantum numbers is not notpermissible permissible A n 2 1 m 1 B n 3 2 m 2 C n 2 0 m 0 D n 4 3 m 3 E n 3 3 m 0 because n 1 l l cant be same as n 12 Summary Orbitals Are allowed energy states for electrons in an atom Describe spatial distribution of electrons in these energy states Orbital Name Number of orbitals Shape s 1 spherical p 3 Dumb bell d 5 Clover leaf f 7 Quantum number defines principal n size angular momentum shape magnetic m orientation 13 Clicker question Give the n and quantum numbers that describe the following orbital Assume it is in the third shell A n 1 2 B n 2 3 C n 3 3 D n 3 1 E n 3 2 Orbital modified from http en wikibooks org wiki High School Chemistry Shapes of Atomic Orbitals media File D orbitals png 14 Quantum numbers and nodes A node is where there is zero electron density The electron will never be found there wavefunction No physical meaning electron density The number of nodes in an orbital n 1 Tells us where the electron can be found Where is the e most likely to be found for an s orbital the e is mostly found …