Question

Assuming the charge density distribution of a nucleus is given by the ρ(0 Saxon-Woods function(p(r) = 。)a) where the diffuseness i 0.55 fm, calculate the skin thickness of the nucleus (see lecture notes for definition). What happens to the skin thickness as the parameter d increases? What physical quantity does the parameter a correspond to?

Answer 1

d is related to skin thickness t by

$t=(4ln3)d$

Given : d = 0.557m

Then, t = (41n3) * 0.55%n (4 * 1.099) * 0.55 = 2.425m
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clearly as d increases skin thickens t also increases.

parameter a is half density radius