# Amount of scattering calculated from scattering cross section

1. Mar 23, 2013

### wc2351

I was reading Wikipedia article on Rayleigh scattering and came upon this:

"...the major constituent of the atmosphere, nitrogen, has a Rayleigh cross section of 5.1×10^(−31) m^2 at a wavelength of 532 nm (green light). This means that at atmospheric pressure, about a fraction 10^(−5) of light will be scattered for every meter of travel."

I am embarrassed to say that I am confused how the fraction of scattering per meter has been arrived at, starting from the scattering cross section.

I tried the following: dI/ I = -(nitrogen density)*(cross section)*(path length)

and you get an exponential decay form but when I calculated the decay constant it was way too small. What am I doing wrong here?

2. Mar 23, 2013

### Staff: Mentor

We cannot tell what you did wrong unless you show your work.

3. Mar 26, 2013

### wc2351

Well you see my work above, a differential equation.

Anyway I found out that it was simply an error of plugging in wrong numbers. Here is the answer I got.

total cross section = 5. 1 x 10 ^(-31) m
density of nitrogen = 1.251 kg / m^(3) (I don't know the source) / (mass of nitrogen molecule) ~ 2.7 x 10^(25) / m^3

decay rate = density * cross section ~ 1.4 x 10^(-5) / m

So I = I_0 * exp (-x/ l) ~ I_0 (1- x/l) => for x = 1m, the reduction in intensity is 1.4 x 10^(-5) * I_0, around what Wikipedia says.

4. Mar 27, 2013

### Staff: Mentor

I expected that, and it is impossible to find out if you don't post the actual calculation.

The density of nitrogen can be calculated with the ideal gas law, and room temperature and pressure.

5. Mar 27, 2013

### wc2351

Ah yes, I should have posted the actual (wrong) numbers in the first place. I'll behave better next time.

Thanks for pointing out the ideal gas law; 1 atm * 1 / (k*300K) gives me 2.45 X 10^(25) particles per cubic meter, which is pretty close to the value I found above.