Question
a) Find the full width at half maximum* (in nm) of the attenuated projection of a solid gold
(density = 19.3 g/cm3) spheres with diameters of 10 nm, 20 nm, and 30 nm in vacuum. Also, find the
maximum attenuation fraction through the sphere. Assume that the soft x-rays (energy = 520 eV) incident
on the sphere have uniform intensity and are parallel to one another. Assume, also, that the detector has
uniform sensitivity, has infinitely fine pixels sufficient to cover the area of the projection, and is located
perpendicular to the incoming x-rays. Neglect any refraction that may occur. The following website can
be used as a resource about x-ray properties of materials: http://www.cxro.lbl.gov/.
b) Repeat the calculations for x-rays at 520 eV through solid carbon spheres (density 2.27 g/cm3) with
diameters of 10 nm, 20 nm, and 30 nm.
c) Repeat the calculations for x-rays at 726 eV through solid iron spheres (density 7.87 g/cm3) with
diameters of 10 nm, 20 nm, and 30 nm.
* The full width at half maximum is a common parameter used to describe the width of a “bump” on a curve
or function. It is the distance between points on a curve at which the function reaches half its maximum
value. Reference: http://mathworld.wolfram.com/FullWidthatHalfMaximum.html
Solution
a)
Au, 520 eV
FWHM (nm)
Au sphere, d = 10 nm
Au sphere, d = 20 nm
Au sphere, d = 30 nm
8.8 nm
17.9 nm
27.3 nm
C, 520 eV
FWHM (nm)
C sphere, d = 10 nm
C sphere, d = 20 nm
C sphere, d = 30 nm
8.7 nm
17.4 nm
26.2 nm
Fe, 726 eV
FWHM (nm)
Fe sphere, d = 10 nm
Fe sphere, d = 20 nm
Fe sphere, d = 30 nm
8.8 nm
17.7 nm
26.8 nm
Max Fraction
Attenuated
.208
.373
.504
b)
Max Fraction
Attenuated
.0278
.0548
.0810
c)
Max Fraction
Attenuated
.126
.235
.331
The sphere of radius R can be described as a uniform 3-D distribution of the photoabsorption cross
section μ.
f (x , y , z) = μ for (x 2 + y 2 + z 2 ) 1/2 ≤ R2
0 for (x 2 + y 2 + z 2 ) 1/2 > R2
A line integral through this 3-D distribution, then, represents the total attenuation of a beam of x-rays
traveling straight through the object. The attenuated projection of the sphere at each point (x , y) is
e-μ ρ d (x , y) where ρ is the density of the material and d (x , y) is the distance of the chord between (x , y) and
(x , -y). A plot of the attenuated projection is shown below (for gold sphere with diameter of 10 nm):
A center slice of the projection can be described as:
2 μ (R2-d2)1/2 for |d| ≤ R
0
for |d| > R
A plot of the center slice of the projection is shown below (for gold sphere with diameter of 10 nm):
The FWHM of the center slice (or equivalently, any other slice) is shown above. The maximum percentage
of attenuation is found at the diameter of the sphere and is represented on the center slice as the value given
by the maximum of the curve.