Laserspt2 - Wave nature is most evident with particles that...

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Wave nature is most evident with particles that are confined to a small region of space, for ex. e - bound to a nucleus. deBroglie relation; p = h l KE of particle = E = 1 2 mv 2 = p 2 /2 m Thus, λ = h p = h nmE = h 2 2 m 1 E ; h 2 2 m =150 eV 2 Wavelength of E = 10eV is: = 150 eV ( ) 1 10 eV ~ 4 ; approx. size of atom = /2 p ~ 0.6 In order for an atom to exist it much have some KE (be orbiting the nucleus) otherwise it would collapse onto the proton. E total = E = E k + E p ; E k = 1 2 mv 2 = L 2 2 mr 2 ; E p = - 1 4 pe o e 2 r Where L = mvr angular momentum ; minus sign a attractive force ; E o = 8.85 10 - 12 C 2 Nm 2 Orbits of the e - are made up of integral # of ’s. Circumference: 2 pr = nl = n h p n 2 p h p ; angular momentum: L = mvr = pr = n h 2 p Quantized in units of h! Only certain modes or states of an e - can exist; boundary conditions will only permit certain ’s to exist and for bound e - only discrete energy levels exist. H atoms; min energy we set dE dr = 0 ; dE dr = - L 2 mr 3 + 1 4 pe o e 2 r 2 = 0 ; - 1 r 2 E k + E p ( ) = 0 E k = - 1 2 E p so that = ε o h 2 pme 2 n 2 Radius has discrete values & orbits. E min = 1 2 E p = - 1 8 pe o e 2 r min = - me 4 8 e o h 2 1 n 2 ; For n=1 E min = - me 4 8 e o h 2 = - 2.12 10 - 18 J = - 13.6 eV (ionization energy for H) r min = e o h 2 pme 2 = 5.3 10 - 11 m = 0.53 = a o Bohr radius; E(in eV) = - 13.6 eV n 2 n=1,2,3… If we apply UV light such that hv>13.6eV ( = 91 nm ) we can free the e - from the nucleus (ionization). E n = E initial - E final =13.6 eV 1 n 2 inital - 1 n 2 final Thermal Equilibrium and Population Inversion Boltzmann Distribution N i = N o e - Ei - Eo ( ) kbT i=1,2,3… N i =# atoms/mol per unit volume in the i th state, N o =# atoms/mol per unit volume in the ground state & k b =1.4 10 - 23 J / K = 8.7 10 - 5 eV / K is Boltzmann’s constant. k
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This note was uploaded on 03/03/2008 for the course A&EP 110 taught by Professor Gaeta during the Fall '05 term at Cornell.

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