quiz3_key_f08 - CHEM 111A TIME ALLOWED: 15 MINUTES NAME:...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
CHEM 111A NAME: _________KEY ___________ TIME ALLOWED: 15 MINUTES ID# ____________________________ Quiz 3.a 9/25/08 All Quizzes are closed book and closed notes. *** Show your work/reasoning *** General Information: p = mv E = ½ mv 2 r n = (n 2 /Z)a o a o = 52.9 pm h = 6.626 x 10 -34 Js m electron = 9.11 x 10 -31 kg m proton = 1.67 x 10 -27 kg N A = 6.022 x 10 23 1.6022 x 10 -19 J = 1 eV 10 10 Å = 1 m 10 9 nm = 1 m 10 12 pm = 1 m E = h ν c = νλ c = 2.998 x 10 8 m/s E n = (–2.18 x 10 -18 J)Z 2 /n 2 = (–13.6 eV)Z 2 /n 2 1. The Paschen series in the atomic hydrogen spectrum begins at 1874.5 nm and ends at 820.1 nm. What is the wavelength (in nm) for the line in the series corresponding to the fifth longest wavelength? Please circle your answer. E p = () () m 10 20 . 8 m/s 10 2.998 Js 10 6.626 hc 7 8 34 × × × = λ = 2.42 x 10 -19 J –2.42 x 10 -19 J = –2.18 x 10 -18 J(1) 2 (1/n f 2 ) – (–2.18 x 10 -18 J(1) 2 (1/ 2 ) –2.42 x 10 -19 J = –2.18 x 10 -18 J(1) 2 (1/n f 2 ) – 0 n f 2 = 9 n f = 3 The longest wavelength corresponds to the 4 Æ 3 transition so the fifth longest wavelength will be the 8 Æ 3 transition. Δ
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 10/20/2008 for the course CHEM 111A taught by Professor Hockings during the Fall '08 term at Washington University in St. Louis.

Page1 / 3

quiz3_key_f08 - CHEM 111A TIME ALLOWED: 15 MINUTES NAME:...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online