Chapter7-1

# Chapter7-1

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: f orbitals: l=0 l=1 l=2 l=3 n=1 n=2 n=3 n=4 1s 2s 3s 4s 2p 3p 4p Total # of orbitals 1 1+3=4 3d 4d 1+3+5=9 4f 1+3+5+7=16 ANSWER: 16 54 Chem 6A Michael J. Sailor, UC San Diego Weds, Oct 27 55 Chem 6A Michael J. Sailor, UC San Diego Weds, Oct 27 • Quiz 6 on Friday Oct 29 covers material through Fri Oct 22 lecture, and problem 7.35 in the homework 55 Chem 6A Michael J. Sailor, UC San Diego Weds, Oct 27 • Quiz 6 on Friday Oct 29 covers material through Fri Oct 22 lecture, and problem 7.35 in the homework • Special bonus if you come in costume to class on Oct 29 55 Chem 6A Michael J. Sailor, UC San Diego Problem: Energy of electronic transitions One of the emission lines from the star σ Ori AB in the constellation Orion occurs at a wavelength of 486.3 nm. The line arises from atomic hydrogen 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 56 Chem 6A Michael J. Sailor, UC San Diego 57 Chem 6A Michael J. Sailor, UC San Diego Solution: Energy of electronic transitions Another way to solve this problem: Assume n1 = 1, calculate the value of n2 given that λ= 486.3 nm 1 1 1 = RH 2 − 2 λ n1 n 2 1 1 1 1 = R H − 2 = R H 1 − 2 λ 1 n2 n2 € Rearranging, 1 1 1− =2 λ ⋅ RH n2 Chem 6A Michael J. Sailor, UC San Diego 58 Solution: Energy of electronic transitions 1 1 1− =2 λ ⋅ RH n2 Solve for n2: −1/ 2 1 n2 = = 1 − 1 λ ⋅ RH € 1 − λ ⋅ RH −1/ 2 1 = 1.1 n 2 = 1 − −2 486.3 ⋅ 1.097 × 10 1 59 Chem 6A Michael J. Sailor, UC San Diego Solution: Energy of electronic transitions 1.1 is not a valid quantum number! Assume that n1 = 2 1 1 1 1 1 = RH 2 − 2 = RH − 2 λ n2 2 4 n2 Solve for n2: −1/ 2 1 1 = − n2 = 1 4 λ ⋅ RH 1 − 4 λ ⋅ RH 1 −1/ 2 1 n 2 = 0.25 − −2 486.3 ⋅ 1.097 × 10 Chem 6A Michael J. Sailor, UC San Diego = 3.998 = 4 60 Solution: Energy of electronic transitions So the transition is n = 4 to n = 2; the answer is (d): 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 61 Chem 6A Michael J. Sailor, UC San Diego Quantum dots: Artiﬁcial atoms http://www.invitrogen.co...
View Full Document

## This note was uploaded on 08/31/2011 for the course CHEM 6A taught by Professor Pomeroy during the Winter '08 term at UCSD.

Ask a homework question - tutors are online