Beer-lambert law and UV absorption spectroscopy.

[Matt15]

In a flash photolysis experiment, the time dependant concentration of [OH] was measured by UV absorption spectroscopy at around 308 nm, where the effective molecular absorption cross section σ = 4.13*10^-16 cm^2.

Using the beer lamber law calculate the concentration (number density) of OH radicals if 10% of the initial UV light is absorped at a path length of 10cm

Convert the OH conentration into mol m^-3, what is the corresponding partial pressure in Pa at room temperature

Homework Equations

I = I₀exp(-σ[OH] l)

A = ln(I₀/I) = σ[OH] l

The Attempt at a Solution

I'm pretty much clueless how to go about solving this one. If I₀ is the incident intensity, would that be 100% ? and if so would that make I = 90%. If that were the case then [OH] = ln(100/90)/σ * l




Thanks in advance

 

[Nino MMP]

.Yes, I₀ would be 100% and I would be 90%. You are correct that the concentration of OH radicals can be calculated using the Beer-Lambert Law as follows: [OH] = ln(I₀/I) / (σ*l), where l is the path length (10 cm in this case). The units of [OH] will be mol m^-3. To calculate the partial pressure of OH at room temperature, you need to use the ideal gas law: P = nRT/V, where n is the number of moles of OH, R is the gas constant (8.314 J mol^-1 K^-1), T is the temperature (in Kelvin) and V is the volume.