Beer-Lambert Law Application

This problem involves the Beer-Lambert Law, which describes how light intensity decreases as it passes through a substance. The law states that absorbance ($A$) is directly proportional to the path length ($l$) and the concentration ($c$) of the absorbing species: $A = \epsilon c l$. Transmittance ($T$) is related to absorbance by the formula $A = -\log_{10}(T)$.

Step 1: Determine Initial Transmittance and Absorbance

The problem states that 20% of incident light is absorbed in a cuvette with a path length ($l_1$) of 1.0 cm. This means the remaining 80% is transmitted.

• Initial Transmittance: $T_1 = 1 - 0.20 = 0.80$
• Path Length 1: $l_1 = 1.0$ cm
• Calculate Initial Absorbance ($A_1$): $A_1 = -\log_{10}(T_1) = -\log_{10}(0.80)$

Using a calculator, $\log_{10}(0.80) \approx -0.09691$. So,

$A_1 \approx -(-0.09691) \approx 0.09691$

Step 2: Relate Absorbance to New Path Length

For a constant concentration, absorbance is directly proportional to the path length. We can write this relationship as:

$ \frac{A_2}{A_1} = \frac{l_2}{l_1} $

We are given the new path length ($l_2$) is 3.0 cm.

• Path Length 2: $l_2 = 3.0$ cm
• Calculate Absorbance ($A_2$) for the new path length:

$A_2 = A_1 \times \frac{l_2}{l_1}$

$A_2 \approx 0.09691 \times \frac{3.0 \text{ cm}}{1.0 \text{ cm}}$

$A_2 \approx 0.09691 \times 3 \approx 0.29073$

Step 3: Calculate Final Transmittance

Now, convert the absorbance ($A_2$) back to transmittance ($T_2$) using the formula $T = 10^{-A}$:

$T_2 = 10^{-A_2}$

$T_2 \approx 10^{-0.29073}$

$T_2 \approx 0.5123$

To express this as a percentage, multiply by 100:

Transmittance Percentage $\approx 0.5123 \times 100\% \approx 51.23\%$

Rounding to one decimal place gives 51.2%.