Question

A. A spectrophotometer is known to have 0.800% stray light reach the detector for all analyses due to poor light sealing. Determine the true absorbance, apparent absorbance, and apparent transmittance for a sample with a true transmittance of 0.0657.

B. What amount of stray light produces an apparent absorbance of 1.995 with a true transmittance of 0.0100?

C. In the process of performing an analysis, you determine that you need to improve the precision by limiting the error in the absorbance to 0.00007200. You attempt to achieve this by reducing the amount of stray light that reaches the detector. What is the maximum amount of stray light that can reach the detector that will limit the error in the absorbance to 0.00007200 for a true absorbance of 1.1000000?

Answer 1

(a) True transmittance P = 0.0657
True transmittance P =  -log [True absorbance]
True absorbance

With 0.8 % stray light (S) , the apparant transmittance =

The apparent absorbance = -log {apparent transmittance} =

(b) apparent absorbance = 1.995

The apparent absorbance = –log(apparent transmittance)

Apparent transmittance

With stray light (S) , True transmittance P and Standard transmittance Po the Apparant transmittance =

Stray light percent =

(c) Error in the absorbance = 0.000072
Error = Apparent absorbance-True absorbance = 0.000072
True absorbance = 1.10
Apparent absorbance = 0.000072 + 1.100000 = 1.100072
True transmittance

Apparent transmittance

from the formula of stray light and transmittance the

Apparent transmittance =