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

1. Mar 16, 2010

### BBoy.Goon

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.

2. Mar 16, 2010

### Char. Limit

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.

3. Mar 16, 2010

### BBoy.Goon

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