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Beer-lambert law and UV absorption spectroscopy.

  • Thread starter Matt15
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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
 

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