To solve this problem we need the concepts of Energy fluency and Intensity from chemical elements.
The energy fluency is given by the equation
[tex]\Psi=4RcE\pi[/tex]
Where
[tex]\Psi =[/tex]The energy fluency
c = Activity of the source
r = distance
E = electric field
In the other hand we have the equation for current in materials, which is given by
[tex]I= I_0 e^{-\mu_{h20}X_{h2o}} e^{-\mu_{Fe}X_{Fe}}[/tex]
Then replacing our values we have that
[tex]I = 1*10^{-3} * 3.3*10^{10} * e ^{-0.06*1.1} e^{-0.058*7.861}[/tex]
[tex]I = 1.3*10^7 Bq[/tex]
We can conclude in this part that 1.3*10^7Bq is the activity coming out of the cylinder.
Now the energy fluency would be,
[tex]\Psi = \frac{cE}{4\pir^2}[/tex]
[tex]\Psi = \frac{1.3*10^7*2*1.25}{4\pi*11^2}[/tex]
[tex]\Psi = 2.14*10^4 MeV/cm^2.s[/tex]
The  uncollided flux density at the outer surface of the tank nearest the source is [tex]\Psi = 2.14*10^4 MeV/cm^2.s[/tex]
The uncollided flux density at the outer surface of the tank nearest to the source is equal to [tex]2.14 \times 10^4\;MeV/cm^2s[/tex]
In order to determine the uncollided flux density at the outer surface of the tank nearest to the source, we would apply energy fluency and intensity for chemical elements.
Also, we know that current flowing through a material is given by:
[tex]I=I_o e^{-\mu_{H_2O}X_{H_2O}}e^{-\mu_{Fe}X_{Fe}}\\\\I=1 \times 10^{-3} \times 3.3 \times 10^{10} \times e^{-0.06 \times 1.1}e^{-0.058 \times 7.861}\\\\I=1.3 \times 10^{7}\;Bq[/tex]
Mathematically, the energy fluency from a chemical element is given by this formula:
[tex]\Phi=\frac{IEc}{4\pi r^2} \\\\\Phi=\frac{1.3 \times 10^7 \times 1.25 \times 2}{4\pi \times 11^2}\\\\\Phi=2.14 \times 10^4\;MeV/cm^2s[/tex]
Read more on flux density here: https://brainly.com/question/24186165
