Chapter7-1

3 nm the line arises from atomic hydrogen contained

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Unformatted text preview: contained in that star. This line probably corresponds to the following electronic transition: a) n=5 to n=1 b) n=4 to n=1 c) n=3 to n=1 d) n=4 to n=2 e) n=5 to n=2 25 Chem 6A Michael J. Sailor, UC San Diego Solution: Energy of electronic transitions First need to convert λ = 486.3 nm to J: −34 8 9 hc 6.626 × 10 J ⋅ s 3.00 × 10 m 10 nm E= = × × λ 486.3nm s m = 4.09 x 10-19 J 26 Chem 6A Michael J. Sailor, UC San Diego Solution: Energy of electronic transitions Energy of each shell can be obtained from Bohr’s expression, which is the Rydberg equation for an individual energy level (chapter 7, p225): n Energy, J Z 2 hcR E = 2 n1 1 2 3 4 5 -2.180 x10-18 -0.545 x10-18 -0.242 x10-18 -0.136 x10-18 -0.087 x10-18 Note h.c.R = (6.626x10-34J.s)(3x108m/s)(1.0967x107m-1) = 2.18x10-18J 27 Chem 6A Michael J. Sailor, UC San Diego Solution: Energy of electronic transitions Now calculate the energy of each transition: n 1 2 3 4 5 Energy, J -2.180 x10-18 -0.545 x10-18 -0.242 x10-18 -0.136 x10-18 -0.087 x10-18 transition a) n=5 to n=1 b) n=4 to n=1 c) n=3 to n=1 d) n=4 to n=2 e) n=5 to n=2 Energy, Jx1018 -(0.087-2.180) = 2.093 -(0.136-2.180) = 2.044 -(0.242-2.180) = 1.938 -(0.136-0.545) = 0.409 -(0.087-0.545) = 0.458 So the answer is (d) 4.09 x 10-19 J 28 Chem 6A Michael J. Sailor, UC San Diego Problem: Ionization energy (a) Using the Schroedinger expression for energy of a 1-electron atom, calculate the ionization energy of an electron in the 3s orbital of a sodium atom, in kJ/mol. 29 Chem 6A Michael J. Sailor, UC San Diego Solution: E = -1310(Z2/n2) kJ/mol Z2 Z2 E = -1310 2 − 2 n final ninitial 112 112 E = - 1310 2 − 2 3 ∞ 121 E = - 1310 − = 17, 612 9 kJ/mol * Note I mistakenly used the value of +23 on the board during the lecture. The positive charge on the Na 30 nucleus is +11, so Z = 11 in this example Chem 6A Michael J. Sailor, UC San Diego Problem: Ionization energy (cont) (b) The measured ionization energy of an electron in the 3s orbital of a sodium atom is IE = 495 kJ/ mol. Why is this so far off from the value calculate...
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This note was uploaded on 08/31/2011 for the course CHEM 6A taught by Professor Pomeroy during the Winter '08 term at UCSD.

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