Please view the attached file to see the questions. There are diagrams and charts associated with the questions so I was unable to type them out.© BrainMass Inc. brainmass.com March 5, 2021, 12:28 am ad1c9bdddf
See the attached file.
Here I will explain the necessary physics and concepts that are to be used to solve the problems. Obviously, I can not do them for you as it is against policy and would be detrimental to your learning.
For problem 2:
Question statement: The problem is asking for the number of (compton, photoelectric, coherent, and pair production) interactions that take place in the 3mm^2 x 1cm box of water.
Step 1: To solve this problem we first have to find out how many photons are actually incident on the pill box cross-sectional area. There is a "mean free path" thickness of water in between the incident photons and the pill box that will interact and absorb photons thus altering the number that actually reach the pill box. To find this number you have to use the exponentially decaying formula, Eq. 1 (please look at the attached file for the equation). Here the thickness is the mean free path.
To review, the mean path length is equal to the inverse of the linear attenuation coefficient.
So to solve you would have to look up a radiological properties data table for the linear attenuation coefficient of water at 5 MeV and plug into mfp = 1/mu, then you can plug it into Eq. 1 and get the number of photons incident on the water phantom at a depth of mean free path in units of number of photons per sec per mm^2. But that still doesn't give us the number of photons incident on the pill box alone. Thus you need to multiply the cross-sectional area, A, by the number of photons at depth = mfp (which is the number we just calculated using the exponentially decaying formula). This will give you number of photons per sec incident on the pill box (which is what we want).
Step 2: Now knowing the correct number of photons incident upon the pill box we are now ready to find the number of interactions from the different processes. Each process will be done in the same way, the only difference between them is the use of each processes' own distinct interaction coefficients. I will explain how to do it step by step for the photoelectric effect, the ...
The solution discusses problems for medical radiation. The diagram and charts associated with the questions are provided.