Absorption of light by molecules is the first step of all photochemical reactions. The Beer – Lamber — Analytical Chemistry Chemistry Question
Absorption of Light by Molecules
Absorption of light by molecules is the first step of all photochemical reactions. The Beer – Lambert law relates the absorbance A of a solution containing an absorbing species of molar concentration c with the optical path length d:
A = log(P0/P) = ε c d
ε is the molar absorptivity (also called extinction coefficient).
Light can be considered as a stream of photons, each carrying an energy of
E = h c / λ
h is Planck’s constant, λ is the wavelength and c the speed of light.
A solution with a dye concentration of c = 4×10^–6 mol dm^–3 has a molar absorptivity of ε = 1.5×10^5 dm^3 mol^–1 cm^–1. It is illuminated with green laser light at a wavelength of 514.5 nm and with a power of P0 = 10 nW.
What is the percentage of light that is absorbed by the sample after a path length of 1 m?
Model Answer
A = ε c d = 1.5×10^5 mol^–1 dm^–3 cm^–1 × 4×10^–6 mol dm^–3 × 1×10^–4 cm = 6 × 10^–5
Since A = log(P0/P), the ratio P/P0 is 0.99994. The percentage of photons absorbed by the solution is: (P0 – P) / P0 = 1 – P/P0 = 1.38×10^–4 or 0.0138 %.
Calculate the number of photons per second absorbed by the sample.
Model Answer
According to our previous result, 0.0138 % of the 10 nW laser light entering the sample solution are absorbed:
Pabs = 1.38×10^–4 × 10 nW = 1.38×10^–3 nW = 1.38×10^–12 J s^–1
The energy of one photon is:
E = hc / λ = 6.626×10^–34 J s × 3.00×10^8 m s^–1 / 514.5×10^–9 m = 3.86×10^–19 J
The number of photons absorbed by the solution per second is:
Nabs = 1.38×10^–12 J s^–1 / 3.86×10^–19 J = 3.58×10^6 s^–1.
The absorption cross section of a molecule is the effective area that captures all incoming photons under low illumination conditions (like an idealized solar cell that would capture all light photons hitting its surface). At room temperature, this corresponds roughly to the molecular area exposed to the light beam. If you calculate it from the molar absorptivity, imagine that all molecules interacting with the light are arranged periodically in a plane perpendicular to the incoming light beam.
What area is occupied by each molecule?
Model Answer
Let’s imagine that the laser illuminates an area of 1 cm^2 of the dye solution. The light beam passes through a volume of V = 1 cm^2 × 1 m = 1×10^–7 dm^3. The number of illuminated molecules is:
N = c V NA = 4×10^–6 mol dm^–3 × 1×10^–7 dm^3 × 6.022×10^23 mol^–1 = 2.409×10^11
Each molecule would therefore occupy an area of
Smol = 1 cm^2 / 2.409×10^11 = 4.15×10^–12 cm^2 or 415 nm^2, if it was projected onto a plane.
Calculate the molecular absorption cross section in units of Å^2.
Model Answer
The molecular absorption cross section σ is the area of one molecule that captures all incoming photons. Under the experimental conditions, only 0.0138 % of the light interacting with one molecule is absorbed, so that σ is:
σ = 1.38×10^–4 × 415 nm^2 = 0.057 nm^2 = 5.7 Å^2
A crucial photochemical reaction for life on our planet is photosynthesis, which converts the absorbed light energy into chemical energy. One photon of 680 nm is necessary to produce one molecule of ATP. Under physiological conditions, the reaction requires an energy of 59 kJ per mol of ATP.
What is the energy efficiency of photosynthesis?
Model Answer
The energy of one 680 nm photon is:
E = h c / λ = (6.626×10^–34 J s × 3.00×10^8 m s^–1) / 680×10^–9 m = 2.92×10^–19 J
Photosynthesis requires 59 kJ per mol of ATP, which corresponds to
EATP = 59×10^3 J mol^–1 / 6.022×10^23 mol^–1 = 9.80×10^–20 J per ATP molecule.
The energy efficiency of photosynthesis is:
η = 9.80×10^–20 J / 2.92×10^–19 J = 0.34 or 34 %.