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UVA CHEM 3420 - Problem Set #3. Answers

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CHEM 342. Spring 2002. Problem Set #3. Answers. Linear Momentum; Orbital Angular Momentum 1. Calculate the average linear momentum of a particle described by the wave functionikxe, where  x.    kpxikxipxikdxipdxdxeikeipdxeedxedxdiepdxdxppxxxikxikxxikxikxikxikxxxxˆˆˆˆˆˆˆ**2. Given that yzxzpypL , derive an expression for xLˆ, the orbital angular momentum operator. yzzyiLyizziyLpzpyLxxyzxˆˆˆˆˆ3. Show that the wave function   iesin is an eigenfunction of zL. What is the eigenvalue? Hint: iLzˆ         zizizizLeLieiLeiLˆsinˆsinˆsinˆThe eigenvalue is .4. What is the orbital angular momentum for electrons in 3s, 3p, and 3d orbitals (expressed in terms of )? How many radial and angular nodes do each of these orbitals have? 1CHEM 342. Spring 2002. Problem Set #3. Answers. For the 3s orbital, l = 0. 01  llL The angular momentum is zero.The number of angular nodes equals l, so there are no angular nodes.The number of total nodes is n  1 = 3  1 = 2. Therefore there are 2 radial nodes.For the 3p orbital, l = 1. 21   llLThe number of angular nodes equals l, so there is one angular node.The number of total nodes is n  1 = 3  1 = 2. Therefore there is one radial node.For the 3d orbital, l = 2. 61   llLThe number of angular nodes equals l, so there are 2 angular nodes.The number of total nodes is n  1 = 3  1 = 2. Therefore there are no radial nodes.5. For a 2p electron in a hydrogenlike atom, what is the magnitude of the orbital angular momentum and what are the possible values of Lz? Express all answers in terms of .  21   llL,0,zlzLmLLaguerre Polynomials; Radial Factors6. The radial wave function for the 2s orbital of the hydrogen atom can be written as 2/2/30,2221ρlneρaZR where 02naZrρ . Find the location of the radial nodes in this orbital in terms of a0. The radial nodes occur when the radial probability function equals zero. 022212/2/30, ρlneρaZRAt this point, 2ρ and therefore 220naZrρ. For a 2s orbital, n = 2; and for a hydrogen atom Z = 1. Therefore, we obtain the following:  0022212ararρ7. The eigenfunction for a 1s electron of a hydrogen-like atom is given by 0/ aZrkeψ, where k is a constant, ao is the radius of the first Bohr orbit for hydrogen. Show that the radius at which there is a maximum probability of finding a 1s electron (in any direction) is just Zar0max. 2CHEM 342. Spring 2002. Problem Set #3. Answers. The probability of finding an electron in unit element of volume at distance r is given by2ψ. The probability of finding the electron at distance r, irrespective of direction, is given by224 ψrP , which can be written as o/222aZrerkP, where 2k is a constant.The radius at which P is maximum occurs when 0drdP. We obtain the following:012022o/22/2o2/22000aZrrekdrdPeaZrrekdrdPaZraZraZrZaraZroo1Hamiltonians and the Schrodinger Equation8. Give an expression for the Hamiltonian of a two-electron atom. Explain what each term represents and which of the terms is the most difficult to evaluate. Hint: Ignore any magnetic interaction between the spin and the orbital motions of the electrons (i.e. the spin-orbit coupling term, which is small for light atoms). The Hamiltonian is as follows      120220210222221244422ˆrerZerZemmHee where 2hThese terms represent, in the order given, (1) kinetic energy of electron #1(2) kinetic energy of electron #2(3) Coulombic attraction of electron #1 to nucleus(4) Coulombic attraction of electron #2 to nucleus(5) electron-electron Coulombic repulsion, the most difficult term to evaluateThe last term contains r12 which is very complicated mathematically. Normalization of Hydrogenlike Orbitals9. The radial wave function for the 1s orbital of a hydrogen atom is 0/1arsAeR. Find the normalization constant A. Hint: 1!naxnandxex3CHEM 342. Spring 2002. Problem Set #3. Answers.   2/303023220/22202*241!21100aAaAAdrerAdrRrRaarOrbitals and Quantum Numbers10. A hydrogen-like wave function nlmψ is of the form  4sinsin44krψnlm. Explain what the values of n, l, and m are. The radial portion of a hydrogen-like wave function is a simple power of r (not a polynomial in r) when l has the maximum possible value, and the power is then (n1). Thus, in this case n = 5, and l = 4. The  function is a simple power of sin if m has the maxium valuefor that l, and the power is then l. Thus in this case m = 4. This is further confirmed by the function in . 11. The hydrogen-like wave functions for n = 2 are sinsin2412241cossin241cos2412/2/3o42/2/3o32/2/3o22/2/3o1eaZψeaZψeaZψeaZψwhere ao is the radius for the first Bohr orbital for hydrogen, and oaZr. For each of the above wave functions, explain if the wave function is the      zyx2por ,2p ,2p ,s2  function. Give the l and m quantum number values for each of the above wave functions. 3 is the only spherically symmetric function, and thus must be  s2 which has l = m = 0. Next, 1 is symmetric about the z-axis, and thus must be  zp2 with (l = 1 and m = 0). Then 2 and 4 must be for l = 1 and m =  1. Further, 2 has a maximum at  90, so2 must be  xp2 and 4 must be  yp2.4CHEM 342. Spring 2002. Problem Set #3. Answers. 12. The following figure shows a plot of the radial wave


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