Analytical Chem istry - DePauw University

Chapter 10 Spectroscopic Methods6631 1ν = =λ 656.3×10−91 m× = 1.524×10m 100 cmcm4 −1Click here to return to the chapter.Practice Exercise 10.2The photon’s energy is−hc ( 6.626× 10 J⋅ s)(3.00×10E = =−9λ656.3×10 mClick here to return to the chapter.34 8m/s)= 303 . × 10−19JPractice Exercise 10.3To find the transmittance, T, we begin by noting thatSolving for TA = 1.27 = –logT–1.27 = logT10 –1.27 = Tgives a transmittance of 0.054, or a %T of 5.4%.Click here to return to the chapter.Practice Exercise 10.4Making appropriate substitutions into Beer’s lawA = 0.228 = ebC = (676 M –1 cm –1 )(1 cm)Cand solving for C gives a concentration of 3.3710 -4 M.Click here to return to the chapter.Practice Exercise 10.5For this standard addition we can write the following equations relatingabsorbance to the concentration of Cu 2+ in the sample. First, for thesample, we have0. 118 =εbC Cuand for the standard addition we have⎛ 20. 00 100 .0.162 = mg Cu mL ⎞εb C Cu+ ×⎝⎜L 10.00 mL⎠⎟The value of eb is the same in both equation. Solving each equation foreb and equating