BOX17-1Why Is There a Logarithmic Relation Between Transmittance and Concentration?51OP4Beer's law, Equation 17-6, states that absorbance is proportional[dpdpto the concentration ofthe absorbingspecies.The fraction of lightdxβcdx-βc==PPpassing through a sample (the transmitance) is related logarith-10Rmically, not linearly,to the sample concentration.Why shouldThe limits of integration are P = Po at x = 0 and P=Palx=bthis be?Imagine light of irradiance P passing through an infinitesimallyPo→βcbIn P = (In Po) = βcbInthin layer of solution whose thickness is dx.A physical model oftheabsorption process suggests that, within the infinitesimally thinFinally, converting n into log, using the relation In z = (ln 10)(log z),layer, decrease in power (dP) ought to be proportional to the inci-gives Beer's law:dent power (P),to the concentration ofabsorbing species (c),and tothe thickness of the section (dx):βlogcb4A =ecb(In10)dP=-βPcdx(A)AbsorbanceConstant = ewhere β is a constant of proportionality and the minus sign indi-The logarithmicrelationof Poptocates a decrease in P as x increases. The rationale for saying thatconcentrationarisesthe decrease in power is proportional to the incident power maybecause,in each infinitesimalportion of the total volume,thebe understood from a numerical example.If 1 photon out of 1000decreasein powerisproportional to thepowerincidentupon tharincident photons is absorbed in a thin layer, we would expect thatsection, As light travels through the sample, the power loss in each2 out of 2 000 incident photons would be absorbed. The decreasesucceeding layer decreases, because the magnitude of the incidentin photons (power)is proportional to the incident flux of photonspower that reaches each layer is decreasing. Molar absorptivity(power).ranges from o (if the probability for photon absorption is O) toy10°MEquation A can berearranged and integrated to find an expressioncm-1approximately(when theprobability forphotonforP:absorption approachesunity).bIncidentAbsorbingEmerging.P-OlightlightsolutionX=0X=bd1
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EXAMPLEAbsorbance,Transmittance,andBeer'sLawFindtheabsorbanceandtransmittanceofa0.00240Msolutionof a substancewithamolarabsorptivityof313M-cm-inacellwitha2.00-cmpathlengthSolution Equation17-6givesus theabsorbance.A=εbc=(313M-cm-)(2.00cm)(0.00240M)=1.50Transmittanceis obtained fromEquation 17-5byraising 10 to thepower equal totheexpression on each side of the equation:log T=-AT = 10log 7 = 10A = 10-1.50 = 0.031 6Just3.16%oftheincidentlightemergesfromthissolutionTest YourselfThe transmittance of a 0.010 M solution of a compound in a 0.100-cm-pathlength cell is T=8.23%.Findtheabsorbance(A)and themolarabsorptivity ()(Answer:1.08,1.08×10°M-cm-)2
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TABLE17-1Colorsof visible lightColorColorWavelength of maximumabsorbedobservedabsorption (nm)Violet380-420Green-yellow420-440Violet-blueYellow440-470BlueOrangeRed470-500Blue-greenPurple500-520GreenViolet520-550Yellow-green550-580YellowViolet-blue580-620OrangeBlueRed620-680Blue-greenRed680-780Green3
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StandardMicrocells1-cmpathCylindrical20-mmpath5-mm1-mmThermalpathpathFlowFIGURE17-5Commoncuvetsforvisibleandultravioletspectroscopy.Flowcellspermitcontinuousflowofsolutionthroughthecell.Inthethermal cell,liquidfromaconstant-temperaturebathflowsthrough the cell jackettomaintain a desired temperature.[CourtesyA.H.Thomas Co, Philadelphia,PA.]
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第五章紫外可见吸收光谱分析1、特征谱基础（1）分子总能量与能级结构p6;（2）分子轨道与电子能级p6~7分子能级-电子能级（成键和反键轨道）一紫外可见光光谱p2、电子辐射与材料相互作用（1）分子的散射p25、（2）紫外、可见（吸收）光谱p29（3）分子荧光、磷光光谱p30~31p3、紫外可见吸收光谱法之基本原理p204~2095
5 第五章 紫外可见吸收光谱分析 p 1、特征谱基础 （1）分子总能量与能级结构p6； （2）分子轨道与电子能级p6~7 分子能级-电子能级（成键和反键轨道）-紫外可见光光谱 p 2、电子辐射与材料相互作用 （1）分子的散射p25、（2）紫外、可见（吸收）光谱p29 、（3）分子荧光、磷光光谱 p30~31 p 3、紫外可见吸收光谱法 之 基本原理p204~209