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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 function ikx e , where x . pˆ x pˆ dx dx d e e dx i dx e e dx e ik e dx i dx ikdx i * x * pˆ x ikx ikx ikx ikx pˆ x ikx ikx pˆ x x ikx pˆ x i x pˆ x k 2. Given that L x yp z zp y , derive an expression for L̂ x , the orbital angular momentum operator. Lˆ x ypˆ z zpˆ y Lˆ x y z i z i y Lˆ x y z i z y 3. Show that the wave function sin e i is an eigenfunction of L z . What is the eigenvalue? Hint: Lˆ z i 1 CHEM 342. Spring 2002. Problem Set #3. Answers. sin e i Lˆ z i Lˆ z sin e i i i Lˆ z sin e i Lˆ z The 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? For the 3s orbital, l = 0. L l l 1 0 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. L l l 1 2 The 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. L l l 1 6 The 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 . L l l 1 2 L z ml L z ,0, Laguerre Polynomials; Radial Factors 6. The radial wave function for the 2s orbital of the hydrogen atom can be written as 3/ 2 2Zr 1 Z 2 ρ e ρ / 2 where ρ . Find the location of the radial nodes in this R n ,l na 0 2 2 a0 orbital in terms of a0. The radial nodes occur when the radial probability function equals zero. R n ,l 1 Z 2 2 a 0 3/ 2 2 ρe ρ / 2 0 2 CHEM 342. Spring 2002. Problem Set #3. Answers. 2 Zr 2 . For a 2s orbital, n = 2; and for a na 0 hydrogen atom Z = 1. Therefore, we obtain the following: At this point, ρ 2 and therefore ρ ρ 21r 2 2a 0 r 2a 0 7. The eigenfunction for a 1s electron of a hydrogen-like atom is given by ψ ke Zr / a0 , 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 a rmax 0 . Z The probability of finding an electron in unit element of volume at distance r is given by 2 ψ . The probability of finding the electron at distance r, irrespective of direction, is given by P 4r 2 ψ 2 , which can be written as P k 2 r 2 e 2 Zr / ao , where k 2 is a constant. dP 0 . We obtain the following: The radius at which P is maximum occurs when dr Z dP k 2 2re 2 Zr / a0 r 2 2 e 2 Zr / a0 0 dr ao Zr dP k 2 2re 2 Zr / a0 1 0 dr ao Zr 1 ao r ao Z Hamiltonians and the Schrodinger Equation 8. 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 2 2 Ze 2 Ze 2 e2 2 2 ˆ H 1 2 40 r1 40 r2 40 r12 2m e 2m e h where 2 These terms represent, in the order given, (1) kinetic energy of electron #1 (2) kinetic energy of electron #2 3 CHEM 342. Spring 2002. Problem Set #3. Answers. (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 evaluate The last term contains r12 which is very complicated mathematically. Normalization of Hydrogenlike Orbitals 9. The radial wave function for the 1s orbital of a hydrogen atom is R1s Ae r / a0 . Find n! the normalization constant A. Hint: x n e ax dx n1 a R Rr dr 1 A r e dr 1 * 2 0 2 2 2 r / a0 0 A 2 A2 A 2! 2 3 a0 1 4 a 03 2 a 03 / 2 Orbitals and Quantum Numbers 10. A hydrogen-like wave function ψ nlm is of the form ψ nlm kr 4 sin 4 sin 4 . 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 (n1). Thus, in this case n = 5, and l = 4. The function is a simple power of sin if m has the maxium value for 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 4 CHEM 342. Spring 2002. Problem Set #3. Answers. 3/ 2 1 Z ψ1 4 2 a o Z ψ2 4 2 a o 3/ 2 1 Z ψ3 4 2 a o 3/ 2 Z ψ4 4 2 a o 3/ 2 1 1 e / 2 cos e / 2 sin cos 2 e / 2 e / 2 sin sin Zr . For each of the above ao wave functions, explain if the wave function is the 2s , 2p x , 2p y , or 2p z function. where ao is the radius for the first Bohr orbital for hydrogen, and 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 2s which has l = m = 0. Next, 1 is symmetric about the z-axis, and thus must be 2 p z 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 , so 2 must be 2 p x and 4 must be 2 p y . 12. The following figure shows a plot of the radial wave function R and a polar plot of the wave function, in the xy plane, for a particular hydrogen-like orbital. Explain what the values of n, l, and m must be, or at least what can definitely be said about them. y R x r The R function has no nodes. Therefore, n l 1 = 0 so that n = l +1. The function must be sin because there is no maximum at 0 ; in addition, there is a maximum at 45 indicating that m = 2. If the maximum extension is in the xy plane, then the l value is the maximum possible for the m value, or l = 2. Then, n = 3. 5