This lecture, in 2 parts, covers interaction of radiation with matter and includes fundamental interactions, particle ranges, dosimetry, and hot atom chemistry. Interaction of radiation with matter covers energy loss and reactions with charged particles and photons. The stopping power of charged particles in different material is covered, including calculations on energy loss with thickness. Electron backscattering is introduced with examples on different behavior with varied elements. Discussion on photon interaction includes photoelectric effect, Compton effect, and pair production. Units of dosimetry are described. Dosimetry measurements are discussed and quality factors based on particle mass and charge are introduced. Introductory dose calculations are supplied. Radiation protection regulations and the definition of terms (ALI, DAC) are given. A review of hot atom chemistry is given.
Another very interesting lecture. My only question is in the example given for calculating detriment from maximum occupation dose (for 30yrs), Where did the 50E-3 Sv/y come from?
ReplyDeleteThe values come from data evaluation performed by the International Commission on Radiation Protection (http://www.icrp.org/)
DeleteI enjoyed the lecture especially the effects of radiation dose on health.
ReplyDeleteMy question regarding the quiz comes from question 1. I based my work off the example you provided and the ASTAR (helium) program on nist website. I found the stopping power to be 3.762 MeVcm^2g^-1. I divided 10MeV/stopping power and received .0266 g cm^-2. I found the density of aluminum to be 2.7 gcm^-3. I divided my previous answer by the density and received a thickness of .001 cm which seems really thin to me.
It seems you used the Total Stopping Power of Aluminum in your calculation, which gives you a very small answer as it is on the order or 10^2. I thought we were supposed to use the nuclear stopping power since the question said "Helium Ions", and this smaller value yields a value of 14cm. I'm not entirely sure if this is correct, however, and it could be the case the 10^-3cm answer was supposed to show us how easily alpha particles are stopped.
DeleteI found the beginning of the second part of the lecture interesting where it described the Radiolysis of water. Up to this point, I had only heard about the effect of ionizing radiation on the generation of double stranded breaks in DNA and was curious as to other immediate biological effects of exposure to ionizing rays. The various products of the interaction between water and radiation, along with the more specific examples listed on slide 4 helped me understand the effects of radiation more clearly.
ReplyDeletethanks for the comments
DeleteInteresting lecture. I enjoyed the information from the tables provided for evaluating nuclides that would be used in experiments. The examples shown after the tables were helpful in understanding how the tables would be used and gives an idea of the purpose for the tables.
ReplyDeletethanks for the comments. the presented information is regularly used in experiment preparation.
DeleteThank you for the lecture. I especially enjoyed the second half and the biological effects/ lab sections. I'm not really clear on how to read the stopping power graph (slide 10-6) Should I have been able to get the 3.762MeVcm^2/g value for aluminum from the graph to use for solving problem 1? I went to the ASTAR program instead.
ReplyDeleteEither way would work. The astar program will provide a specific value. I should reinforce that the equation provided in the lecture is just truly an estimate. There are program and some detailed publications available for providing more accurate range
DeleteHi, I finished the Quiz. For question 1, I used the ASTAR program for He ions and got 3.762MeVcm^2/g value for the stopping power at 10 MeV for aluminum. I calculated it and kept getting .00985 cm Al as my answer. I'm not sure how to get the correct answer for this question.
ReplyDeleteThe answer and example is posted. I would like to emphazise this is an estimate. I will post another example of range
DeleteAnother interesting lecture. For the part in the lecture that states the types of matter that can interact with one another are "positive ions" meant to be protons or positively charged atoms of elements? Thank you for the clarification and great lecture.
ReplyDeleteThis lecture reminded me of all the radiation surveys i did while in the navy. However, the Sievert is a unit we never discussed/used, all our radiation surveys were reported in Rads/s or Rems/s. I still find it funny that we described one erg as the amount of energy required to raise a mosquito 1 cm.
ReplyDeleteThe lecture was very interesting. I am only confused about the first question on the pdf quiz. Which value of stopping power are we supposed to use for the He ions?
ReplyDeleteI believe "Helium ions" refers to the Helium nuclease and not any electrons. This would mean we should use the Nuclear Stopping Power for Aluminum bombarded by helium ions at 10MeV given by the ASTAR database here:
Deletehttp://physics.nist.gov/PhysRefData/Star/Text/ASTAR.html
Super interesting lecture! I am applying to medical school right now, and it is really interesting to learn about the dose and how to calculate them!!!!!!!!
ReplyDeleteTerrific to hear about the overlap.
DeleteGreat lecture. I am having problems with the first question on the pdf quiz as well. I am getting half the thickness of Al of the posted answer. I am not sure which stopping power we should use in the calculation.
ReplyDeletethanks for the comment. I will need to sharpen this portion of the lecture.
ReplyDeleteI thought this was an interesting lecture, thanks for the help with the blog it helped when I did it.
ReplyDeletethanks for comment Alex
ReplyDeleteLike many others, I had issues with the first question of this quiz. Nonetheless, the rest of the material was clear and easier to follow than others (maybe because of my physics background, not sure).
ReplyDeleteI enjoyed this lecture. I remember a bit of it (particularly regarding dosimetry) from my radiation safety training, but it was helpful to review it.
ReplyDeleteThe lecture was great as always and has been most informative on the dose and the calculations in order to find them.
ReplyDelete