Solutions – PHYS 352 Design Assignment #2 QUESTION ONE a) At 30 keV: I_1 = I_0 exp(–mu_tissue*10) = 0.026 I_0 I_2 = I_0 exp(–mu_tissue*7) exp(–mu_bone*3) = I_o exp(–mu_tissue*7 – mu_bone*3) I_2 = 0.000863 I_0 The contrast is 2(I_1–I_2)/(I_1+I_2) = 1.871 Obviously, the contrast will be worse for 1.5 MeV photons since the absorption coefficients are not as different between tissue and bone. Calculating like above, we find at 1.5 MeV: The contrast is 2(I_1–I_2)/(I_1+I_2) = 0.087 b) If the tissue were 20 cm thick, that would be like adding 10 cm of tissue on either end. This would attenuate the intensity by a factor exp(–mu_tissue*10) and that would be the same attenuation factor on I_1 and I_2. Hence, the contrast which is the ratio 2(I_1–I_2)/(I_1+I_2) isn’t affected at all since the attenuation factor shows up in both numerator and denominator. This statement is the same whether it is 30 keV or 1.5 MeV. The extra tissue thickness doesn’t affect the image contrast. QUESTION TWO Design question, several approaches. I’m looking for calculations that show the thickness of liquid xenon required to have good efficiency for detecting 30 keV x-­‐
rays. This is the main supporting calculation for your design. QUESTION THREE (from the 2010 final exam): It’s all Compton scattering in this calculation and what you want is the number of electrons since the Compton cross section is given as 0.26e–24 cm^2 per electron Argon density is 1.784 g/L at STP (just a tiny complication thrown in here) so you need the Ar density at 0.1 atm, since the counter is stated to operate at 0.1 atm. In 1 cm^3, there is 1.784e–3 g of Ar gas at STP and thus 1.784e–4 g of Ar gas at 0.1 atm In 1 cm^3 there is 1.784e–4 [g]/39.95 [g/mol] * 6.022e23 [atoms/mol] = 2.689e+18 [atoms/cm^3] and multiply by Z=18 to get 4.84e+19 [electrons/cm^3] Use the electron density to calculate the attenuation coefficient, which is the number of targets times the cross section per target… mu for Ar is 4.84e19*0.26e–24 = 1.2585e-­‐05 cm^–1 x = 5 cm I/I_0 = exp(–mu x) = 0.999937 The interaction probability is the gas is 1–0.999937 = 6.3e–5. Repeat for steel, using Fe In 1 cm^3 there is 7.85 [g]/55.85 [g/mol]*6.022e23 [atoms/mol]*26 = 2.20e+24 [electrons/cm^3] mu for Fe is 2.20e+24*0.26e-­‐24 = 0.57218 x = 0.1 cm I/I_0 = exp(–mu x) = 0.944388 The interaction probability in the 1 mm steel wall is 1–0.9444 = 0.0556 QUESTION FOUR: Detection efficiency involves the intrinsic part (probability for photon to interact in the detector medium) and the geometric part (the solid angle fraction), and the full calculation folds them both together. mu/rho for Ge at 1 MeV is 5.727e–02 cm^2/g multiply by rho = 5.3 g/cm^3 given in the problem to get mu = 0.303531 cm^–1 Now, for different angles from the source through the detector, there is a different path length and efficiency for that path length. To average a function f(θ) over solid angle we want to integrate over all angles theta. π
∫
0
π
f (θ )sin θ dθ
π
∫ sinθ dθ
=
∫
1
f (θ )sin θ dθ
0
2
=
∫ f (θ ) d(cosθ )
−1
2
0
Above: that’s how we would average over all angles (cylindrical symmetry means we don’t need to integrate over φ). !"#$
&'()*+,-.+$
%$
#$
%!$ %/$
#$
cos(theta_1) = 10/sqrt(10^2+2.5^2)=0.9701425 from angle theta=0 to theta_1, the path length is half the hypotenuse cos(theta) = 10/hypotenuse therefore, the path length is 5/cos(theta) for angles theta=0 to theta_1 next: tan(theta_1) = 2.5/10; theta_1 = 0.2449786631 and cos(theta_1) = 0.9701425 tan(theta_2) = 2.5/5; theta_2 = 0.463647609 and cos(theta_2) = 0.8944272 from angle theta=theta_1 to theta_2, the path length is the hypotenuse minus 5/cos(theta) because 5/cos(theta) is the path length before the gamma ray hits the detector. Between theta_1 and theta_2: sin(theta) = 2.5/hypotenuse; the hypotenuse is 2.5/sin(theta) and the path length through the detector is 2.5/sin(theta)–5/cos(theta) sanity check: at angle theta_1 path length is: 2.5/sin(theta_1) – 5/cos(theta_1) = 10.307764 – 5.153882 = 5.153882 or 5/cos(theta_1) = 5.15388203 For each path, the geometric efficiency is just the solid angle fraction. The probability for interaction is f(θ) = 1–exp(–mu x), where x is the path length. 1
f =
∫ f (θ ) d(cosθ )
−1
2
1
⎡ cos θ1 ⎡
⎤
⎡
5 ⎞⎞ ⎤
⎛ ⎛ 2.5
⎛ ⎛ 5 ⎞⎞ ⎤
= 0.5 ⎢ ∫ ⎢1 − exp ⎜ − µ ⎜
−
dcos
θ
+
1
−
exp
−
µ
d
cos
θ
⎥
⎜⎝ ⎜⎝ cosθ ⎟⎠ ⎟⎠ ⎥
∫ ⎢
⎝ ⎝ sin θ cosθ ⎟⎠ ⎟⎠ ⎥⎦
⎢⎣ cos θ2 ⎣
⎥⎦
⎦
cos θ1 ⎣
f = 0.5[0.0289446 + 0.023461] = 0.0262028
Calculation performed with Mathematica, see following page. Therefore, the efficiency of the Ge detector is 2.62%. Make sure you understand that the factor 0.5 isn’t because we are averaging two different formulae for calculating the efficiency (which depends on the path length). It’s a factor of 0.5 because over the whole solid angle, the integral over all values of cos(theta) is 2. Here’s the Mathematica output: In[3]:= mu = 0.303531 Out[3]= 0.303531 In[13]:= costheta1 = 10/Sqrt[10^2 + 2.5^2] Out[13]= 0.970143 In[14]:= costheta2 = Cos[ArcTan[2.5/5]] Out[14]= 0.894427 In[11]:= Integrate[1 -­‐ Exp[-­‐mu 5/x], {x, costheta1, 1}] Out[11]= 0.023461 In[15]:= NIntegrate[1 -­‐ Exp[-­‐mu (2.5/Sqrt[1 -­‐ x^2] -­‐ 5/x)], {x, costheta2, costheta1}] Out[15]= 0.0289446 In[16]:= 0.5 (%11 + %15) Out[16]= 0.0262028 In[17]:= 1 -­‐ Exp[-­‐mu 5] Out[17]= 0.780775 In[18]:= 1 -­‐ costheta1 Out[18]= 0.0298575 In[20]:= costheta1 -­‐ costheta2 Out[20]= 0.0757153 Test your understanding: -­‐ if the gamma rays were incident normally on 5 cm of Ge, the efficiency would be the answer in Out[17] or 78.1% -­‐ the solid angle fraction between cos(theta) = 1 and cos(theta) = 0.970143 is the answer in Out[18] times 0.5 or 0.5[0.0298575] -­‐ compare 0.5[0.0298575] to 0.5[0.023461], the answer in Out[11] -­‐ similarly, compare 0.5[0.0757153] to 0.5[0.0289446], the answer in Out[15]