In spectrophotometric measurements, absorbance (A) is related to percent transmittance (%T) by which true statement?

In spectrophotometric measurements, absorbance (A) is related to percent transmittance (%T) by which true statement?

• A. A and T are directly proportional
• B. A increases with increasing %T
• C. A increases as %T decreases
• D. A decreases as %T decreases

Answer: (C)

Absorbance is defined from the amount of light that passes through a sample on a logarithmic scale. Specifically, A = -log10(T), where T is the transmitted light fraction (I/I0). Since percent transmittance %T = 100 × T, you can write A = -log10(%T/100) = 2 - log10(%T). This means that as percent transmittance decreases, the logarithm term increases in magnitude in the negative direction, so A increases. Conversely, as %T increases, A decreases. So the true statement is that absorbance increases as percent transmittance decreases. For example, %T of 80% gives a small A (~0.097), while %T of 40% gives a larger A (~0.398), and %T of 4% gives an even larger A (~1.398).

Absorbance is defined from the amount of light that passes through a sample on a logarithmic scale. Specifically, A = -log10(T), where T is the transmitted light fraction (I/I0). Since percent transmittance %T = 100 × T, you can write A = -log10(%T/100) = 2 - log10(%T). This means that as percent transmittance decreases, the logarithm term increases in magnitude in the negative direction, so A increases. Conversely, as %T increases, A decreases.

So the true statement is that absorbance increases as percent transmittance decreases. For example, %T of 80% gives a small A (0.097), while %T of 40% gives a larger A (0.398), and %T of 4% gives an even larger A (~1.398).