# Solve Scattering Problem: Find Wall Thickness for Half Incident Particles

• Anro
In summary, in order to scatter particles without them going through a wall of a specific thickness, half the particles must reach the other side without scattering.f

## Homework Statement

A beam of particles strikes a wall containing 2 × 10^29 atoms/m^3. Each atom behaves as a hard sphere of radius 3 × 10^–15 m.

Find the thickness of the wall such that exactly half the incident particles go through without scattering.

## Homework Equations

N(sc) = N(inc) × n(tar) × σ ---> (1)
N(sc) = the number of scattered particles
N(inc) = the number of incident particles
n(tar) = the target density
σ = cross sectional area of the target = πR^2
n(tar) = ρt/m ---> (2)
ρ = density
t = thickness of the wall
m = mass

## The Attempt at a Solution

Hello everyone; this is a straightforward scattering problem but I seem to be missing something here. This is what I did so far:
From the given information I have:
N(sc) = ½ N(inc) ---> 2N(sc) = N(inc)
R = 3 × 10^–15 m ---> σ = πR^2 = 9π × 10^–30 m^2
Plugging these values in equation (1), I can get the value of n(tar) which turned out to be 1.77 × 10^28 m^–2
Having this value of n(tar), I can substitute into equation (2) and solve for t. This is where I’m stuck, as I’m supposed to find numerical values for ρ and m in order for me to find t. These values can be obtained from the given information that the wall contains 2 × 10^29 atoms/m^3, and I think that I can get my values for ρ and m from the following equation:
The number of atoms per unit volume = (Avogadro’s number × density)/atomic mass
That is: 2 × 10^29 atoms/m^3 = (6.022 × 10^23 mol^–1 × density)/atomic mass
This is where I’m not sure how to proceed, as I need to find the density and atomic mass (and convert it to kg as well) from this equation; any help would be appreciated.

Be careful here.

We know that the macroscopic cross-section, Σ, for an interaction is given by Σ = nσ, where n is the atomic density and σ is the microscopic cross section.

So Σs = nσs.

Now we also know that N(x) = No exp (-Σs x), where x is the distance traveled, and No is the initial particle intensity (we could use I(x) and Io).

All we need to do is find the distance (thickness) t at which half of the particles reach without scattering, and so they leave the other side (of slab thickness t), without having scattered.

So what is N(t)/No?

Now it makes sense using your equations; still straightforward, but different way of looking at things. Thank you very much for your help Astronuc, I really appreciate it.