Question

Monochromatic light of wavelength 532 nm is used to measure the absorption coefficient of a material in a UV-Visible Spectrophotometer. The measured light intensity after transmission through a 1 cm thick sample of the material is 0.414 mW/cm\(^2\). For a sample of thickness 2 cm, the measured light intensity is 0.186 mW/cm\(^2\). The absorption coefficient (in cm\(^{-1}\)) of the material is _________ (round off to two decimal places)

Hint:

The absorption coefficient can be determined using the Beer-Lambert law, which relates the ratio of intensities to the thickness of the sample and the absorption coefficient.

Answer 1

Correct Answer: 0.75

The absorption coefficient \( \alpha \) is given by the Beer-Lambert Law: \[ I = I_0 e^{-\alpha x} \] where:
- \( I_0 \) is the initial light intensity,
- \( I \) is the transmitted light intensity,
- \( \alpha \) is the absorption coefficient,
- \( x \) is the sample thickness.
We are given the following: - For \( x = 1 \, \text{cm} \), \( I = 0.414 \, \text{mW/cm}^2 \), - For \( x = 2 \, \text{cm} \), \( I = 0.186 \, \text{mW/cm}^2 \). To calculate \( \alpha \), we first take the ratio of the two intensities: \[ \frac{I_2}{I_1} = \frac{0.186}{0.414} = e^{-\alpha (2 - 1)}. \] Simplifying: \[ 0.449 = e^{-\alpha}. \] Taking the natural logarithm: \[ \ln(0.449) = -\alpha, \] \[ \alpha = -\ln(0.449) \approx 0.804 \, \text{cm}^{-1}. \] Thus, the absorption coefficient is approximately \( 0.80 \, \text{cm}^{-1} \).