Q.41. A solution absorbs 𝟐𝟎% of the incident light in a cuvette of path length 1.0 cm .
The amount of light transmitted by the same solution in a cuvette of 3.0 cm path length is ____ % (rounded
off to one decimal place).

The correct answer is 51.2%. This spectroscopy problem tests the Beer-Lambert Law, where transmittance scales exponentially with path length.

Core Concept

Beer-Lambert Law states absorbance A = ε c l = -log₁₀ T, where T is transmittance (fraction of light transmitted). For fixed concentration c, A &propro; l, so T₂ = T₁^(l₂/l₁).

Here, 20% absorption means T₁ = 0.80 for l₁ = 1.0 cm. For l₂ = 3.0 cm, T₂ = 0.80³ = 0.512 or 51.2% (rounded to one decimal).

Detailed Derivation

Initial absorbance A₁ = -log₁₀(0.80) ≈ 0.0969. New absorbance A₂ = A₁ × 3 = 0.2908. Thus, T₂ = 10^-0.2908 = 0.512, since 0.8³ = 64/125 = 0.512.

Physically, tripling path length is like stacking three 1 cm cuvettes: each transmits 80%, total 0.8 × 0.8 × 0.8 = 51.2%.

Common Mistakes (Pseudo-Options)

• Linear absorption (wrong: 40%): Assuming 20% per cm so 60% absorbed in 3 cm. Ignores exponential decay.
• Natural log misuse (51.2%, correct numerically): T = e^{(-κ l)}, but yields same T₂ ≈ 0.512 as constants adjust.
• Absorbance linear for T (wrong: 60%): Thinking transmission = 100% – 20% × 3. Violates Beer-Lambert.
• No scaling (wrong: 80%): Ignoring path length change[web:1].

Exam Tips

Practice with logarithms and exponents for quick calc: memorize T₂ = T₁^(ratio).