6 - Mid-Term #2a March. 7, 2001 .Johnny Bobby Rebecca....

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1 Mid-Term #2a CHEMISTRY 20-A SOLUTIONS March. 7, 2001 . . Student Name .Johnny Bobby Rebecca. Circle Name of TA . . Student I.D. No. PLEASE WRITE YOUR NAME & NAME OF YOUR TA and READ the DIRECTIONS on THIS PAGE PLEASE DO NOT TURN PAGE UNTIL ASKED TO DO SO . DIRECTIONS . Do all problems. All work and answers should be written on the exam. Please write legibly and clearly. Partial credit will be given; correct statements will count in your favor, incorrect ones will count against you, unintelligible or illegible ones will not count at all. Wherever applicable, show your work, justify approximations, and explain what you are doing; otherwise no partial credit can be given. In many of the problems no credfit will be given without supporting explanation. Allocate your time between problems judiciously. This is a closed book exam, but you may make use of a single 8.5" by 11" sheet of handwritten notes (both sides). This exam consists of 8 pages Problem. Max Grade Grade 1 28 2 22 3 16 4 35 --------------------------------------------------- Total 101 Some possibly useful data: μ = Qr U electrostatic Q 1 Q 2 r 12 K.E. = (1/2)mu 2 prob density = | Ψ | 2 λν = c λ mass = h/mu E n = - R H Z 2 eff 1 n 2 r n = a o n Z eff h = 6.6x10 -34 J s c = 3.0x10 8 m s -1 m e = 9.1x10 -27 kg m proton = 1.7x10 -27 kge = 1.6x10 -19 C N o = 6.0x10 -23 particles mol -1 1 eV = 1.6x10 -19 J
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2 1. What counts in this question is the reasoning by which you arrive at your answer (whatever it may be). Be sure to lay out your reasoning succinctly, as quantitatively as possible, legibly, and clearly , or else you will not get credit. The ionization energy of the H atom is about 13 eV, that of He about 23 eV. a) Making use of these numbers, estimate the effective nuclear charge, Z eff , on a 1s electron in He? Why is this reasonable (or unreasonable)? One can model the electronic energy of He in terms of shielding. The charge on the nucleus is Ze with Z = 2, but each of the 1s electrons shields the other so that Z eff lies between 1 and 2. Thus the ionization energy can be expressed as Z eff 2 R H /1 2 = 13eV(Z eff 2 ) = 23eV Z eff = (23/13) 1/2 1.33. We expect each 1s electron to be quite effective in shielding the other
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This note was uploaded on 01/31/2012 for the course CHEMISTRY 20a taught by Professor Baugh during the Spring '11 term at UCLA.

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6 - Mid-Term #2a March. 7, 2001 .Johnny Bobby Rebecca....

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