ME501HW2SolnF2011

ME501HW2SolnF2011 - 1 ME 501 Homework #2 Due Monday,...

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1 ME 501 Homework #2 Due Monday, October 3, 2011 Prof. Lucht (E-mail address: [email protected]) 1. NML, problem 3.5, p. 148.     2 2 2 . () - 2 : , 0, Free particle of mass m moving on circle of radius a a Schrodinger equation r V r r steady state SWE m For free particle V r t r a operator in cylindrical coord       : 1 inates r rr r     0 22 2 1 rz  0 2 2 2 2 2 1 2 0 2 2 2 0 2 A 0 in in in Hence But I ma ma d so Id b Solution to Schrodinger equation dI Ae d Substitution gives I ne A e             2 2 2 * 0 12 2 Normalization gives d =1 e e d =1 1 So 2 1 2 Therefore the normalized solution becomes 2 e = 2I / in in in I n A AA n        
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2     11 2 22 2 (c) Allowed discrete energy levels Boundary condition for wave function arises from the need for a continuous function 2 2 2 1 in in in Thus so e e en        * 0, 1, 2,. .. Hence from 2 / 0, 1, 2,. .. 2 (d) An orthonormal set requires For , 1 via normalization as shown in part (b) For , we m n mn mn n nI n I dm n mn        mn - 00 require =0. On this basis, 1 cos - sin( - ) These two trigonometric integrals vanish becau inm im in eed e d i nm d       se they cover complete cycles of cosive and side, respectively Hence, for 0, 1, 2,. .. 0 (e) Free-electron model for benzene Applying the Pauli exclusion principle the fi mn n m and n m n  rst election occupies 0, the second 0, the third and fourth 1, the fifth and sixth 1. hence two electrons (or opporite spin) occupy each energy level. The frist electronic transition nn   2 2 is then 1 to 2. Frequency: 8 3 For 2 and 8 h hc n m cI h nm cI  
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3 2. The time-independent part of the wavefunction for the 3, 0 nm  state of the hydrogen atom is spherically symmetric and is given by    2 00 0 300 3/2 0 27 18 2 exp 3 ,, 81 3 1 rr r aa a R r a           The Bohr radius   22 1 0 4 / 0.52918 10 e am e m  and the electron spin is neglected. (a) Calculate the expectation value of the potential energy of the electron Vr , where 2 0 /4 e r  . [ Note: In these formulas the factor 0 is the dielectric permittivity, not an energy,   12 2 0 8.854 10 CJ m  ] . Recall that for a spherically symmetric function d  4 r 2 dr . For parts a and b: 1 0 !
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This note was uploaded on 12/26/2011 for the course ME 501 taught by Professor Na during the Fall '10 term at Purdue University.

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ME501HW2SolnF2011 - 1 ME 501 Homework #2 Due Monday,...

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