More Solved Problems in Quantum nkurur/2013-14/IIsem/cyl100/mo... · More Solved Problems in Quantum…

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More Solved Problems in Quantum ChemistryNarayanan KururFebruary 5, 20141. Determine the radii for the nodal surfaces of the 3s orbital of the hydrogen atom.300 = 3s =2813a30(27 18 ra0 + 2(ra0))exp( ra0)Solution: For a 3s orbital, the factor(27 18 ra0 + 2(ra0)2)vanishes at r = 181084 a0which gives the radii of the nodal surfaces.2. Confirm that Ylm =( 38)sin exp i is an eigen function of the L2 and Lz operators.Solution: From the form of the function we can identify that l = 1 and m = 1.L2 = 1sinsin 1sin2 22L2Ylm =38(1sinsin cos +1sin2 sin)exp i= 38sin expi(cos2 sin2 sin2 +1sin2 )= 2( 38sin exp i)As expected, we find that L2Ylm = l(l + 1)Ylm = 2Ylm in this case. The Lz operator isih().LzYlm =ih()( 38sin exp i)= 1h( 38sin exp i)Once again, as expected, we find that LzYlm = mhYlm.3. Show that r2s 6= r2pz. You are given that R20 = 12(1a0)3/2(1 r2a0)er2a0 and R21 =126(1a0)3/2ra0er2a0 .Solution: From the expectation value postulate we know that Op=Opd, whichfor this case is r=RrRr2 dr.4. Consider a system whose state is given as =33 1 +23 2 +23 3, where 1, 2, and 3are orthonormal. (a) Calculate the probability of finding the system in any of the states1, 2, or 3. (b) Consider an ensemble of 810 systems on which measurements are made.How many systems will be found in each one of the states 1, 2, or 3?Solution: (a) We first verify whether the state is normalized, which it is - 3/9 + 4/9 +2/9 = 1. The probability of finding the system in any of the states i is |ci|2. In this caseit is 1/3, 4/9, and 2/9 respectively. (b) If a large number of measurements, (N say, aremade, the number of systems being found in a state i is piN.5. An electron is moving freely in a box which extends from 0 to a. The electron is in theground state of the box. If the wall at a is suddenly moved to 4a, what is the probabilityof finding the electron in the (a) ground state and (b) first excited state of the new box?Solution: The electron is initially in the state =2a sin (xa ). The ground and first ex-cited states of the new box are 1 =24a sin (x4a ) and 2 =24a sin( 2x4a). The probabilityof finding the electron in the state i is given by | a0 idx|2.6. An electron in hydrogen atom is in the energy eigenstate Nrer2a0 sinei. (a) Find N(b) If L2 and Lz are measured, what will be the results? (c) And if Lx is measured? (d)What is the probability per unit radial interval (dr) of finding the electron at r = 2a0?Solution: (a) N2 = 1 r=0 =0 2=0 r2e ra0 sin2 r2 dr sin d d(b) The l and m values are, respec-tively, 1 and 1. If L2 is measured, one would get l(l + 1)h2 = 2h2. Measurement of Lzyields mh = 1h. (c) The state is not an eigenfunction of Lx (you can verify this by findingthe operator form of Lx. The expectation value of Lx in this state is zero. (d) The desiredprobability is found by integrating 2 over and and evaluating the resulting functionof r at r = 2a0.7. If A is the operator i(x2 + 1) ddx + ix, find the state (x) for which A(x) = 0. Normalize(x). Calculate the probability of being in the region 1 x 1 if the particle in thestate (x).Solution: The state (x) is the solution of the differential equation ddx =xx2+1 , which isNx2+1. The function is normalized as always, N2dx = 1 implying that N = 1.The probability of being found in the specified region is 1 111x2+1 dx = 1/2.


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