chm lab exp 5

chm lab exp 5 - CHM 132L Experiment 5 Spectroscopy of Atoms...

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Fig. 2 Transition Energy: Violet (sample calculation) E = h Δ v = (6.26x10 -34 J-s)(6.98x10 14 s -1 ) E = 4.37x10 Δ -19 J convert J to eV (6.2415x10 18 eV/J) (4.37x10 -19 J)( 6.2415x10 18 eV/J) E = 2.73 eV Δ Green: 2.32 eV Red: 1.64 eV Wavelength: Violet (sample calculation) tan = (.233m)/(1m) α = 13.11° α sin = .2268 α = D(sin ) λ α = (1895.526 nm)(.2268) = 429.9 nm λ Green: = 506.01 nm λ Red: = 713.06 nm λ Frequency: Violet (sample calculation) v = c/ λ = (3.0x10 8 m/s)/(429.9x10 -9 m) v= 6.98x10 14 s -1 Green: v= 5.93x10 14 s -1 Red: v= 4.21x10 14 s -1 Em n = RHZ21n2-1m2 Δ Find:1n2-1m2 Violet (sample calculation) - 1n2 1m2 = ( ) - ( ) 1 2 2 1 5 2 = .210 Green: .188 Red: .139 ergy: E = h Δ v =(6.26x10 -34 J-s)( 9.71x10 14 s -1 ) E = 6.08x10 Δ -19 J/atom = 6.08x10 -22 kJ/atom = 366.04 kJ/mol equency: v = c/ λ = (3.0x10 8 m/s)/(309x10 -9 m) v= 9.71x10 14 s -1 CHM 132L Experiment 5- Spectroscopy of Atoms and Molecules Introduction: In this experiment, the goal was to measure and analyze the characteristic
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This note was uploaded on 04/17/2008 for the course CHM 132L taught by Professor Krugh during the Spring '08 term at Rochester.

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chm lab exp 5 - CHM 132L Experiment 5 Spectroscopy of Atoms...

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