Once the Coulomb distortion has been removed

PHYSICS · Multi-part · Fermi University Physics · Nuclear Physics · Nuclear constituents, size and binding · Nuclear constituents, size and binding

The Nucleus Resolved, Ch How big a nucleus is, and what the mass surface knows

Problem

Once the Coulomb distortion has been removed, the charge form factor of

^{208}\mathrm{Pb}

has its first zero at

q_1 = 0.665\,\mathrm{fm^{-1}}

, and at

q = 0.838\,\mathrm{fm^{-1}}

, near the top of the next lobe,

|F| = 0.0616

. A convenient stand-in for a sphere with a soft edge is a uniform sphere of radius

R

whose edge is blurred by a Gaussian of width

\sigma

F(q) = \frac{3(\sin x - x\cos x)}{x^3}\,e^{-q^2\sigma^2/2}, \qquad x = qR.

The blurring leaves the zeros where the sharp sphere puts them, at

x = 4.493

first, and only makes the envelope fall faster. For this shape

\langle r^2\rangle = \tfrac{3}{5}R^2 + 3\sigma^2

, and a Fermi shape of diffuseness

a

has nearly the same surface when

\sigma = \pi a/\sqrt{3}

. ((a)) Find

R

from the first zero. ((b)) Find

\sigma

from

|F|

q = 0.838\,\mathrm{fm^{-1}}

, and hence the Fermi diffuseness

a

and the surface thickness

t = 4a\ln 3

. ((c)) Compute

\langle r^2\rangle^{1/2}

for this shape and compare it, and your

a

, with the chapter's

5.50\,\mathrm{fm}

and

0.55\,\mathrm{fm}