BIOCHEMICAL ENGINEERING INSTRUMENTATION - EXAMPLE 29.1 : In the application of T

BIOCHEMICAL ENGINEERING INSTRUMENTATION - EXAMPLE 29.1 : In the application of Theory of Spectrometry in spectrophotometer, let n = N x C x V, V = A x t, e = a x N where n = number of molecules, N = Avogadro's number, V = volume of cuvette, A = area of cuvette, t = thickness of cuvette, C = concentration of fluid in the cuvette, e = extinction coefficient, a = effective area of molecule. (a) By using calculus in dI = -I x a x N x C x dt, prove that ln (I / Io) = -a x N x C x t, where dI is the small difference in I and dt is the small difference in t. I = intensity of light. Io = initial intensity of light. (b) Show by calculations that ln (Io / I) = e x C x t based on the answer in the previous question (a). (c) Find the equation of log (Io / I) as a function of e, C and t based on the answer in the previous question (b).

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BIOCHEMICAL ENGINEERING INSTRUMENTATION - EXAMPLE 29.1 : In the application of Theory of Spectrometr..

Answer / kangchuentat

BIOCHEMICAL ENGINEERING INSTRUMENTATION - ANSWER 29.1 : (a) Let dI = -I x a x N x C x dt, then dI / I = -a x N x C x dt. Integrate both sides of the equation gives ln (I / Io) = -a x N x C x t. (b) When ln (I / Io) = -a x N x C x t and e = a x N, then ln (I / Io) = -e x C x t. If ln (I / Io) = - ln (Io / I), then ln (Io / I) = e x C x t. (c) Let ln (Io / I) = log (Io / I) / log E, where E = 2.718 approximately. Then log (Io / I) = log E x ln (Io / I) = log E x e x C x t. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

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