# Lithium scattering on iron plate

1. May 31, 2014

### skrat

1. The problem statement, all variables and given/known data
Every second $N_{Li}=2\cdot 10^8$ atoms of $_3^7Li$ with kinetic energy $30 MeV$ is Coulomb scattering on $_{26}^{56}Fe$ plate $d_{Fe}=0.3mm$ thick with density $\rho =7800kg/m^3$. Under angle of $30^°$ and $r=0.2m$ away from the plate we have a target with surface $S_n=1cm^2$ that contains $N_n=6\cdot 10^{22}$ neutrons.

a) How many Lithium (N) falls on the neutron plate each second?
b) How many Lithium (N') per second deviate due to the scattering on neutron plate? You can say that $\sigma _n=0.2\cdot 10^{-28} m^2$ for one of the Li.

2. Relevant equations

3. The attempt at a solution

Can I ask before I show my work?

Is it ok if I calculate how much of the original Lithium flux overcomes the Fe plate, where it Coulomb scatters? This should than be the answer to part a) or is it not?

All I am trying to say is that I don't get it why the angle of neutron plate and the distance is important? (not even for part b))

2. Jun 2, 2014

### BvU

Would that be in the forward direction ? None of those reach te neutron plate ! What relevant equations do you have available to tackle this problem ?

3. Jun 2, 2014

### skrat

Well I somehow have to find out how many are scattered in $d\omega =\frac{dS}{r^2}$ I just don't know how.

Equations:

Coulomb

$\frac{\mathrm{d} \sigma }{\mathrm{d} \Omega }=(\frac{e_1e_2}{16\pi \epsilon _0T})^2\frac{1}{sin^4(\theta /2)}$

For scattering on one particle: $d \Omega (\frac{\mathrm{d} \sigma }{\mathrm{d} \Omega })=\frac{N}{jdt}$ where N is number of particles scattered in $d\theta$ and $j$ initial flux.

also

$N=\frac{N_pN_t\sigma}{S}$ where N is number of reactions, $N_p$ number of projectiles and $N_t$ number of targets.

That's more or less all I have.

4. Jun 3, 2014

### BvU

Coulomb expression: Well, there is a theta in there ! So the angle is needed. Now about the distance. Where is that hiding ? Or is it absent ? Once that is answered, your conditions for showing your work are fulfilled!

5. Jun 3, 2014

### skrat

$d\Omega =\frac{dS}{r^2}$

BUT

$N_td\Omega (\frac{d\sigma }{d\Omega })=\frac{N_{on sensor}}{jdt}$

My question: How do I find out the number of Fe in a plate, $N_t$ ??? I know nothing about area size of flux.

6. Jun 4, 2014

### BvU

What about looking around in the given data ? $\rho$ for example ? Area size of flux divides out (you may assume the beam is like a pencil ray). I don't see it in your worked out thingy anyway (?)

7. Jun 4, 2014

### skrat

I haven't shown anything yet because I don't think that size of flux divides out.

Let's take a look:

Let me use notation $N_x$ instead of $N_{on sensor}$.

$N_td\Omega (\frac{d\sigma }{d\Omega })=\frac{N_{x}}{jdt}$

$\frac{N_x}{dt}=N_tjd\Omega (\frac{d\sigma }{d\Omega })=N_tj\frac{dS}{r^2} (\frac{e_1e_2}{16\pi \epsilon _0T})^2\frac{1}{sin^4(\theta /2)}$

Now $N_t=\frac{mN_a}{M}=\frac{\rho S d N_a}{M}$. Now here is where it all stops. What am I supposed to do with that S in the numerator?

8. Jun 4, 2014

### skrat

Ahhh ok, I see it now. Why does this happen to me? Why do i see it now and not days ago?

$j=N/tS$

So $\frac{N_x}{dt}=N_tjd\Omega (\frac{d\sigma }{d\Omega })=(\frac{\rho S d N_a}{M})\frac{N_0}{tS}\frac{dS}{r^2} (\frac{e_1e_2}{16\pi \epsilon _0T})^2\frac{1}{sin^4(\theta /2)}$

$\frac{N_x}{dt}=\frac{\rho d N_a}{M} \frac{N_0}{t}\frac{dS}{r^2} (\frac{e_1e_2}{16\pi \epsilon _0T})^2\frac{1}{sin^4(\theta /2)}$