Problem Set 2_Sol

Problem Set 2_Sol - Answer Key Side 1 of 9 Ch 1a Problem Set Two Problem One EM Radiation 15 Points 5 Each hc hc"= a(5 Points E = h = E Sig fig

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Answer Key Ch 1a, Problem Set Two Side 1 of 9 Problem One – EM Radiation – 15 Points, 5 Each a. (5 Points) E hc hc h E = ! = = " # = 6.626 # 10 $ 34 J % s ( ) 2.998 # 10 8 m s ( ) 10 9 nm 1 m & ( ) * + 3.61 # 10 $ 19 J = 550.27nm Final Answer: 550 nm, Yellow b. (5 Points) ! ! " # $ $ % & ( = = ) 2 1 2 1 1 1 * hc hc hc E " E = 6.626 # 10 $ 34 J % s ( ) 2.998 # 10 8 m s ( ) 1 365.0 nm $ 1 600.5 nm & ( ) * + 10 9 nm m & ( ) * + 1 eV 1.602 # 10 $ 19 J & ( ) * + = 1.332 eV Final Answer: 3.546 eV c. (5 Points) "# = c $ = c = 2.998 % 10 8 m s 89.3 % 10 6 s & 1 = 3.3572 meters E = h ν = 6.626 " 10 # 34 J $ s " 89.3 " 10 6 s # 1 = 7.06994x10 # 26 J Final Answer: KROQ’s wavelength is 3.36 m with E = 5.92 x 10 -26 J 2 pts 2 pts for correct wavelength 1 point for correct color (orange ok) 3 pts Sig fig = -1/2 Calculation error = 1-2 Wrong units = -1 2 points -1/2 for significant figures 3 pts 2 points, 1 each -1/2 each for significant figures
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Answer Key Ch 1a, Problem Set Two Side 2 of 9 Problem Two – Bohr Model – 15 Points Write Rydberg Equation in terms of wavelength difference we’ll call " fi : For Hydrogen atom E = h = hc # fi = hcR H Z 2 1 n f 2 $ 1 n i 2 % & ( ) * * fi = 1 R H 1 n f 2 # 1 n i 2 $ % & ( ) # 1 For Helium E = h = hc fi = hcR H Z 2 1 n f 2 $ 1 n i 2 % & ( ) * * fi = 1 4R H 1 n f 2 # 1 n i 2 $ % & & ( ) ) # 1 The leading coefficient of this equation is a constant that is easily calculated: 1 4R H = 22.8 nm Now we must move systematically through each n i value of each n f until we have exhausted the possibility of finding visible light…for simplicity, use the series discussed in class… n f n i Lyman Lines: n f =1, n i 2 Location 1 2 91.2 nm ( ) " 1 1 2 # 1 2 2 $ % & ( ) # 1 = 30.5 nm UV 1 3 91.2 nm ( ) " 1 1 2 # 1 3 2 $ % & ( ) # 1 = 25.75 nm UV 1 4 91.2 nm ( ) " 1 1 2 # 1 4 2 $ % & ( ) # 1 = 24.33 nm UV 1 5 91.2 nm ( ) " 1 1 2 # 1 5 2 $ % & ( ) # 1 = 23.75 nm UV n f n i Balmer Lines: n f =2, n i 3 Location 2 3 91.2 nm ( ) " 1 2 2 # 1 3 2 $ % & ( ) # 1 = 164 nm UV 2 4 91.2 nm ( ) " 1 2 2 # 1 4 2 $ % & ( ) # 1 = 122 nm UV 2 5 91.2 nm ( ) " 1 2 2 # 1 5 2 $ % & ( ) # 1 = 109 nm UV n f n i Paschen Lines: n f =3, n i 4 Location 3 pts for correct derivation 3 pts for correct derivation
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Answer Key Ch 1a, Problem Set Two Side 3 of 9 3 4 91.2 nm ( ) " 1 3 2 # 1 4 2 $ % & ( ) # 1 = 469 nm Visible -Blue 3 5 91.2 nm ( ) " 1 3 2 # 1 5 2 $ % & ( ) # 1 = 321 nm UV All other transitions will therefore be in the visible or higher wavelengths.
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Answer Key Ch 1a, Problem Set Two Side 4 of 9 Problem Three – Range of Colors – 5 Points Final Answer: For Helium ion, the
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This note was uploaded on 03/30/2010 for the course CH 1a taught by Professor Lewis during the Fall '08 term at Caltech.