Neutrons scattering on Silver plate

[skrat]

Homework Statement

A 2 mm thick plate of natural Silver absorbs 11% of neutron flux with kinetic energy 1 eV. What is the total scattering cross section for neutrons?
$\rho (Ag)=10500 kg/m^3$ and $M(Ag)=107.9kg/kmol$
What is the ratio between the calculated cross section and geometric cross section of nucleus?

Homework Equations

The Attempt at a Solution

No problems with the first part, but also no idea what to do with the second part of the problem?

If I am not mistaken, than I should be able to get the geometric cross section from $\rho (Ag)=10500 kg/m^3$ and $M(Ag)=107.9kg/kmol$; while calculated cross section should be in some relation to scattering?

Any hints?

 

[skrat]

Actually, here is what brings me to wrong solution:

From the first part $\sigma _{tot}=9,945\cdot 10^{-28}m^{2}=10^{-27}m^{2}$

Now I said that $\sigma _{tot}=\pi R^2$. From where $R_c=(\frac{\sigma _{tot}}{\pi })^{1/2}$. This should be calculated cross section.

Geometric cross section:

$\rho =\frac{m}{V}=\frac{m}{\frac{4}{3}\pi R^3}=\frac{M(Ag)}{\frac{4}{3}\pi R^3N_a}$ from where $R_g=(\frac{3M}{\rho N_a4\pi})^{1/3}$

Than $\frac{R_c}{R_g}=0.00011$ instead of 11.6.

 

[SteamKing]

Is M supposed to be the molar mass of mercury? Mercury has an atomic weight of about 200.6.

 

[skrat]

Ooops, sorry, I made I mistake when posting my question. The element is Silver, not Mercury.

I will edit my original post.

 

[haruspex]

I very much doubt the two measures should disagree so violently.
It's hard to say much more because there's so little detail in your post. I've no idea how you compute σ_tot, so I'll just accept that. For the geometric radius I get roughly 10^-10m from your equation. From what I see in Wikipedia, that's about right. Is that what you got? You didn't mix up kmol with mol, right?

 

[skrat]

I very much doubt the two measures should disagree so violently.
It's hard to say much more because there's so little detail in your post. I've no idea how you compute σ_tot, so I'll just accept that.

Ok. My result for this part is the same as the "official".

For the geometric radius I get roughly 10^-10m from your equation. From what I see in Wikipedia, that's about right. Is that what you got? You didn't mix up kmol with mol, right?

Yes, $R_g=(\frac{3M}{\rho N_a4\pi})^{1/3}=1.6\cdot 10^{-10} m$

Of course, maybe I understood "geometric" wrong. Maybe something else is meant by geometric and not something you get out of this formula. :/

 

[haruspex]

Ok. My result for this part is the same as the "official".



Yes, $R_g=(\frac{3M}{\rho N_a4\pi})^{1/3}=1.6\cdot 10^{-10} m$

Of course, maybe I understood "geometric" wrong. Maybe something else is meant by geometric and not something you get out of this formula. :/

Just noticed a crucial word... Geometric cross section of NUCLEUS. I believe you computed the cross section of the whole atom.

 

[SteamKing]

Ooops, sorry, I made I mistake when posting my question. The element is Silver, not Mercury.

I will edit my original post.

The thread title has now been edited to change 'Mercury plate' to 'Silver plate' to avoid further confusion.

SteamKing