- #1
Aleoa
- 128
- 5
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?
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?