The first part of the nuclear reactions lecture is posted. Reaction energetics, threshold energy, and Coulomb barriers are described. The use and determination of cross section data are provided. Be prepared to discuss this lecture during meeting on 23 September.
JD: Lecture 3 Part I complete
ReplyDeleteJN: Lecture 3, Part I complete
ReplyDeleteQuestions:
1) On slide 3-7 how did we get the constant of 0.96? What does the e stand for in the e^2 term?
2) On slide 3-9 is the n term in units of /cm^2 or /cm^3?
The value is for the factor used to determine the radius in the coulomb barrier. The radius is calculated by
DeleteR=roA^1/3, with ro from 1.1 to 1.6 fm. For this equation the ro term is 1.5 fm and the factor in the Coulomb barrier is 0.96. The term e is elementary charge (http://en.wikipedia.org/wiki/Elementary_charge).
The value is nuclei/cm^3. Multiplied by target thickness (cm) and cross section (cm^2), the cm term is eliminated in the rate.
IA: Lecture 3 Part 1 Complete
ReplyDeletethanks for the note!
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