# I need help with spectrophotometry ( The Beer-Lambert Law )

## Main Question or Discussion Point

I'm doing a science project about the antioxidant activity of red wines from my country. This days, I was going through the theoretical part ( the basics of spectrophotometry, photometric analysis ect. ) and I'm currently at the Beer-Lambert Law. I'm would like to know the whole mathematical path that is used so that starting from this two equations: A = 2 - log10 %T and A = $$\epsilon$$bc,

where:
A is absorbance, ( no units )
T is transmittance, ( no units )
$$\epsilon$$ is molar absorbtivity, ( L mol-1cm-1 )
b is the length of the sample ( path length of the cuvette ) ( cm ) and
c is the concentration of the compound in solution ( mol L-1 )

the equation %T = e -$$\epsilon$$bc is deduced.

note: the molar absorbtivity is represented by the letter epsilon from the Greek alphabet and it's not superscripted as it looks like.

Char. Limit
Gold Member
Well, let's see...

Start by putting the two equations equal to each other...

$$2-log_{10}(T)=\epsilon b c$$

$$2=\epsilon b c + log_{10}(T)$$

Subtract epsilon b c from both sides...

$$2- \epsilon b c = log_{10}(T)$$

And raise both sides to the power of 10.

$$T=10^2 * 10^{-\epsilon b c}$$

I got a different answer than that... I'm not sure if that's right.

I started just like you ( putting the two equations equal to each other ) but I think that e, from the equation %T = e-$$\epsilon$$bc stands for the so called Neper's number ( e = 2,72 ) and after using the rule that:

logab = logcb / logca

and several trasnformations I got stuck here:

loge%T = 2loge10 - $$\epsilon$$bc loge10