Q.60 The percentage of light that would pass through a sample with an absorbance
of 2 would be ______% .
(Round off to the nearest integer)
Beer-Lambert Law Basics
Absorbance (A) equals the negative base-10 logarithm of transmittance (T), where T is the ratio of transmitted light intensity (I) to incident light (I₀), or T = I/I₀. The percentage transmittance (%T) is then T × 100%, so A = -log₁₀(T). This logarithmic relationship means higher absorbance sharply reduces transmitted light, common in solutions with high analyte concentrations like proteins or pigments in biochemical assays.
Calculation for Absorbance 2
To find %T for A = 2: T = 10-A = 10-2 = 0.01, so %T = 0.01 × 100% = 1%. Rounded to the nearest integer, the answer is 1%. This matches standard tables: absorbance 2 corresponds exactly to 1% transmittance, while absorbance 1 is 10% and 0 is 100%.
Common Misconceptions and Options
Multiple-choice questions often test this with distractors like 20%, 10%, 50%, or 0%—here’s why they’re incorrect:
- 20% implies A ≈ 0.7 (-log(0.2) ≈ 0.699), too low for A=2.
- 10% matches A=1 exactly (-log(0.1)=1), confusing linear absorption thinkers.
- 50% gives A=0.301, typical for dilute samples.
- 0% suggests total absorption (A → ∞), not finite A=2. The key error is treating absorbance as linear (% absorbed = A × 100%), ignoring the logarithmic scale.
Practical Applications in Biology
In genetics and biochemistry labs, absorbance 2 signals overly concentrated samples—dilute 10-fold to reach A=0.2 (≈ 63% T) for accurate readings in the linear range (A=0.1-1.0). Tools like spectrophotometers rely on this for DNA quantification (A260) or enzyme kinetics, ensuring precise plant biology or microbiology experiments.
