Feynman: Apparent area of a nucleus

[Aleoa]

In paragraph 5.7 of this lecture, Feynman explains how to calculate the apparent area of the nucleus, in a sheet of unspecified material.
I have two questions about the formula used by Feynman.

1) Although the sheet has a thickness, the formula considers only the superficial area of the sheet. Maybe because the material is a crystal, so the every nucleus in the superficial layer is aligned with the nuclei of below layers. What do you think ?

2) In the note Feynman says:

"This equation is right only if the area covered by the nuclei is a small fraction of the total, i.e., if (n1−n2)/n1 is much less than 1" role="presentation">1. Otherwise we must make a correction for the fact that some nuclei will be partly obscured by the nuclei in front of them"

Do you have any idea how to apply this correction factor to the previous formula?

 

[BvU]

Hi,

You can answer this yourself: what is the probability a nucleus of radius ##r_0## is covered by another nucleus if you do assume ##{n_1-n_2\over n_1} \ll 1## ??

 

[Aleoa]

Hi,

You can answer this yourself: what is the probability a nucleus of radius ##r_0## is covered by another nucleus if you do assume ##{n_1-n_2\over n_1} \ll 1## ??

If ##{n_1-n_2\over n_1} \ll 1## i know that the area covered by the nuclei is very small, but I'm not able to obtain information about the disposition of the atoms in the volume of the material...

 

[BvU]

But it says so in the text:

the fraction of the area “covered” by the nuclei is Nσ/A

in other words: the probability that a nucleus is hidden by another is that self-same ##n_1-n_2\over n_1## and we get a correction term of ##{A\over N} \left( n_1-n_2\over n_1 \right ) ^2##

 

[Aleoa]

But it says so in the text: in other words: the probability that a nucleus is hidden by another is that self-same ##n_1-n_2\over n_1## and we get a correction term of ##{A\over N} \left( n_1-n_2\over n_1 \right ) ^2##

Can you explain me why ?

I thought that ## \left( n_1-n_2\over n_1 \right ) ## simply represents the area covered by all the nuclei, but it doesn't given information about the arrangement of the nuclei ( that is, If many of them are covered with each other) ; maybe Feynman assumes that the material is a perfect crystal; however if it were so Feynman would have no reason to ask for "a correction for the fact that some nuclei will be partially obscured by the nuclei in front of them " . I'm a bit confused...

 

[BvU]

There is an area ##A## of which a fraction ##n_1-n_2\over n_1## is covered by ##N## nuclei with each an area ##\sigma##. So the area covered is ##N\sigma = A{n_1-n_2\over n_1}##. The probability that a nucleus is hidden is therefore ##n_1-n_2\over n_1##.

Don't waste too much time on this: you have seen the values: nuclei are really small in relation to atoms.

[edit]Ah, I see:

not able to obtain information about the disposition of the atoms in the volume of the material

Do some calculations: density, atomic weight, Avogadro number, and see if fig 5-10 is exaggerated or an 'understated' picture

 

[Aleoa]

There is an area ##A## of which a fraction ##n_1-n_2\over n_1## is covered by ##N## nuclei with each an area ##\sigma##. So the area covered is ##N\sigma = A{n_1-n_2\over n_1}##. The probability that a nucleus is hidden is therefore ##n_1-n_2\over n_1##.

Don't waste too much time on this: you have seen the values: nuclei are really small in relation to atoms.

[edit]Ah, I see:Do some calculations: density, atomic weight, Avogadro number, and see if fig 5-10 is exaggerated or an 'understated' picture

I'm very sorry, but i am not still able to understand where the correction factor comes from. I have understood the formula##N\sigma = A{n_1-n_2\over n_1}##, but not the intuition behind the correction factor