STUDY GUIDE 10, SELF TEST 1

© Department of Physics, University of Guelph.

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1. Seventy five percent of the light striking a chlorophyll sample is absorbed. What is the percent transmittance and the absorbance of the sample?

2. Convert an absorbance of 0.37 to % transmittance.

3. A solution yields a %T of 48 in a sample cell with a 1.0 cm path length. What would the %T value be if the same solution were placed in a sample cell with a path length of 3.0 cm?

4. A sample is irradiated at its absorption lmax (500 nm) with 1.00X1018 photons. The absorbance of the sample was found to be 0.680. Calorimetric analysis indicated that 0.100 joules of the incident light energy were converted into heat in the sample. What is the approximate value of  lmax of the fluorescent light assuming a quantum yield for fluorescence of 1.00? In this context, a quantum yield of 1.00 means that for every photon absorbed one photon is fluoresced.

5. Sketch:
(a) a graph of percent transmittance versus path length.
(b) a graph of log %T versus path length.

(Hint: Obtain expressions relating %T and pathlength in both exponential and logarithmic form.)

6.(a) You have equal volumes of two solutions in two glass cells. Extinction coefficients and concentrations are: 4200 litre.mole-1.cm-1 and 3.00X10-4 moles/litre for solution one and 2800 litre.mole-1.cm-1 and 2.60X10-4 mole/litre for solution two. The two solutions each having a path length of 1.00 cm are set one behind the other in a light beam. Find the total absorbance of the two solutions.

(b) Equal volumes of the two solutions are mixed and some of the mixture is added to a new cell having a path length of 1.00 cm. Calculate the expected percent transmittance of the mixture. (Remember that, when you mix equal volumes, concentrations are halved.