Prove Attenuation Length = Avg Photon Travel Distance

[strangequark]

Homework Statement

Show that the attenuation length, $$\Lambda$$, is just equal to the average distance a photon travels before being scattered or absorbed.

Homework Equations

my book gives:

number of photons absorbed = $$\sigma\rho I(x) dx$$

number of photons present after a thickness x = $$I(x)=I(0)e^{-\sigma \rho x}$$

attenuation length = $$\Lambda = \frac{1}{\sigma\rho}$$

The Attempt at a Solution

i'm really not sure where to go here... some idea on how to get started would be very much appreciated... thanks

 

[strangequark]

ok, nevermind, I think I got it...

$$x_{avg}=\int^{\infty}_{0}x \sigma \rho e^{\sigma \rho} dx$$

(i think)

 

[ogb p]

To prove that attenuation length is equal to the average distance a photon travels before being scattered or absorbed, we can use the equations provided.

From the equation I(x) = I(0)e^(-sigma rho x), we can see that the intensity of the photon beam decreases exponentially as it travels through a material with thickness x. This is because as the photons travel, they have a chance of being scattered or absorbed, resulting in a decrease in the number of photons present.

Using this equation, we can calculate the average distance a photon travels before being scattered or absorbed. This can be done by integrating I(x) from 0 to infinity and dividing by the initial intensity I(0):

Avg photon travel distance = \int_0^\infty x I(x) dx / \int_0^\infty I(x) dx

Substituting in the equation for I(x), we get:

Avg photon travel distance = \int_0^\infty x I(0)e^{-\sigma \rho x} dx / \int_0^\infty I(0)e^{-\sigma \rho x} dx

Simplifying, we get:

Avg photon travel distance = \frac{1}{\sigma\rho}

Which is the same as the equation for attenuation length. This shows that the attenuation length is indeed equal to the average distance a photon travels before being scattered or absorbed. This makes sense because the attenuation length represents the distance at which the intensity of the photon beam decreases by a factor of 1/e, which is equivalent to the average distance a photon travels before being scattered or absorbed.