# Absorption length

1. Jan 3, 2016

### says

1. The problem statement, all variables and given/known data
Aluminium (Ar = 26.98, ρ = 2700 kg/m3) has cross-section of thermal neutron absorption 0.23 barn. What is the absorption length of thermal neutrons in the aluminium? What value decreases neutron intensity in 0.6 mm, 8 mm and 80 mm aluminium?

2. Relevant equations
Beer-Lambert Law

I'm not sure what x and lambda are in this equation.

3. The attempt at a solution
I'm not sure how to calculate the absorption length from the information provided.

Absorption length
[PLAIN]http://gisaxs.com/images/math/c/5/0/c50b9e82e318d4c163e4b1b060f7daf5.png, [Broken] the distance over which the intensity falls to 1/e.

Last edited by a moderator: May 7, 2017
2. Jan 4, 2016

### BvU

is strange when you then state
. Or are you worried about the relation between $P$ and intensity ?
Is it really that hard to imagine what $x$ could be ?
Did you Google some examples ?

Last edited by a moderator: May 7, 2017
3. Jan 5, 2016

### says

The intensity falls to 1/e = absorption length. But what is the original intensity?

4. Jan 5, 2016

### BvU

Why not use 1 for the original intensity ? Or $I_0$ and gamble that $P = {I\over I_0}$ ? At least then you can move on !

5. Jan 5, 2016

### says

$P = {I\over I_0}$ = 1

1/e = .367
1-.367879 = 0.633

The intensity has fallen, I'd assume x is the distance the thermal neutron has traveled through the atom, maybe. not sure about lambda though.

6. Jan 5, 2016

### BvU

We're not making much headway like this. Can you imagine an aluminum plate with a certain thickness, for example 0.6 mm, at which a beam of thermal neutrons is directed. The plate is a void with an awful lot of aluminum atoms. For a neutron in the beam this looks like a volume with a certain density of targets with an area of 0.23 barn per atom, the neutron cross section.
Thicker plate, more targets. Your $\lambda$ has something to do with the density and the area of these targets.

You have a textbook or lecture notes to introduce you to the matter at hand ? This exercise doesn't come out of the blue, I should think?

7. Jan 5, 2016

### says

You'd be surprised...

λ = density * area of sphere ? (Assuming nucleus has a sphere shape)
= density * 4*pi*radius of atomic nucleus

8. Jan 5, 2016

### says

If I take the value of kinetic energy from that link you posted. 0.025 eV I can calculated the scattering cross section. From there I can calculate the absorption cross section.

9. Jan 5, 2016

### BvU

My impression is the 0.23 nb is the absorption cross section. The exercise is a lot simpler than you now make it look.

10. Jan 5, 2016

### says

Yes, that makes sense. 0.23 is the absorption cross section. I have absolutely nothing in my lecture slides relating to neutron cross section or neutron absorption. No mathematical equation in which I can understand what happens. Is λ the mean free path?

11. Jan 5, 2016

### says

Σ=attenuation coefficient cm-1
N= Atom density
σ = cross section

Σ=Nσ
Σ=621 cm-1

λ=1/Σ= 1/621=0.00161

12. Jan 5, 2016

### BvU

What do you get for N ? And how come your units end up as cm-1 ?

13. Jan 5, 2016

### says

Sorry, N = density = ρ = 2700 kg/m3

I can see this isn't correct now though through dimensional analysis. density * cross section = kg/m3 * m2 ≠ cm-1

14. Jan 5, 2016

### BvU

You want the number of Al atoms per volume, not the density in kilograms per volume !

15. Jan 6, 2016

### says

oh ok, so number density = number of objects / volume = 100.74

100.74*.23 = 23.017 cm-1
λ=1/Σ= 1/23.017=0.043446

0.367 = e^-x/lambda
-1.00239 = -x/0.043446 =
x= -0.04355

This answer is still incorrect. The solution is 0.639 m according to my textbook but I don't see how I can get that.

16. Jan 6, 2016

### BvU

Do you seriously believe there are 100 Al atoms in a 1 m3 block of Aluminum ? Or do you work in some other system of units ?