the following content is provided under a Creative Commons license your support will help MIT OpenCourseWare continue to offer high quality educational resources for free to make a donation or to view additional materials from hundreds of MIT courses visit MIT opencourseware at ocw.mit.edu so today I want us to go more in depth into the photon interactions with matter and we're going to bring the theory back to something that we can actually start to use in doing shielding calculations if you want to find out how much of this material do I need to shield this much gammas we're gonna answer that question today first I want to start off again with compton scattering because I messed up a couple of the energy things from last time I got excited due to an energetic coincidence between the photo peak and the continent of our banana spectrum and one of the examples in the book so I'm going to correct that now and we'll go through in more mathematical detail why that wasn't the case and what the actual quirk of physics is because there is a constant energy thing here that I want to highlight to you guys so skipping ahead on the photoelectric effect which I think was similar to review compton scattering it's the same thing that we saw between two particles except now one of them's a photon and like I said on the next homework after quiz 1 you guys will be doing the balance between energy and momentum because the photons don't really have mass in order to figure out what's the relationship between the incoming energy of the photon the outgoing energy of the photon and the recoil energy of the electron and so this is these are the relationships we were showing last time it's it's is an interesting quirk of physics that the wavelength shift itself does not depend on the energy of the photon coming in as you can see it just depends on the angle that it scatters at and a bunch of constants where that M right there is stands for mass of the electron now that wavelength shift while that wavelength shift does not depend on the energy of the incoming photon the recoil energy does you can see it depends on both the energy and the angle and the incoming energy of the photon equals H nu and to give that quick primer on photon things I want to show you guys here why that's the case so even if you have a constant wavelength shift that might give you a non constant energy shift so even though in this Compton scattering formula the wavelength shift only matters with the angle the energy shift actually depends on the angle and the incoming energy of the photon so now let's look at a couple of limiting cases so as we have a of the photon equals H nu goes to zero what does t approach the recoil energy of the electron well let's just do out the formula right this recoil energy equals H nu which is the energy of the incoming gamma times 1 minus cosine theta over MC squared over H nu plus 1 minus cosine theta as H nu approaches 0 what happens here ya H nu goes to 0 this fraction goes to infinity and this goes to 0 hopefully that's an intuitive explanation if the incoming photon has zero energy it can transfer zero energy to the electron now the more interesting case what happens now as a gamma approaches infinity as the photon gets higher and higher in energy do you approach this almost H do I actually want to do a quick calculation without doing all the limit math let's say we had energy of the gamma was 1 GeV and extremely high energy so all we plug in and let's say we wanted to find out what's the maximum energy of this recoil electron and this is something I want to ask you guys I can't remember yesterday did we say that T is a maximum at theta equals PI or PI over 2 what did we say yesterday interesting it's PI actually the case where you have the largest energy transfer and sorry for not catching that is just like in a nuclear collision if the photon were to back scatter it transfers the maximum energy to the electron so the analogy here is like perfect between two particles hitting and between a photon and a particle hitting the maximum energy is when theta equals PI and let's actually plug that in to find out why if we say T max equals H nu depends on the electron coming in 1 minus cosine theta over MC squared over H nu plus 1 minus cosine theta at theta equals PI cosine theta goes to minus 1 and so then that also goes to minus 1 and so this becomes 2 H nu over MC squared over H nu plus 2 and so that without worrying about the numerator especially in the limit of very high-energy photons you can see that that actually maximizes the recoil energy of the electron the reason we're harping so much on this recoil energy of the electron is because that's what we measure so when we look at our banana spectrum you're not measuring the energy of the photon you're measuring the recoil energy of the electron and the ionization cascade that happens as all those electrons smash into each other creating electron hole pairs which are counted is current if you guys remember from last class in fact I'll just bring up the blackboard image because we can do that so I took a picture of the board yesterday photon interactions part one there it is we'll just use the screen as a bigger blackboard for now so if you guys remember let's say a gamma ray comes in and causes a Compton scattering nth or a photoelectric emission or whatever it doesn't matter which process and it liberates an electron either by scattering off of it or just getting absorbed and ejecting it or it doesn't really matter how but it creates this electron hole pair that electron right here has this recoil energy which depends on theta the angle that it scatters and the incoming energy and that electron is going to keep moving in this material knocking into other electrons very very efficiently so that most of the energy of that electron recoil actually gets counted as other electrons being freed we're going to go over on well next Friday electron nuclear interactions including what's the probability and energy transfer when electrons slam into each other or when ions slam into electrons or each other see so let me go back to the slides and so this maximum as you approach as this approaches infinity this actually approaches a value of H nu minus 0.25 5 MeV just to do the quick calculation to give a numerical example if I plug in theta equals PI and H nu equals 1 Giga electron volt or a thousand MeV and could anyone remind me what is mm mass of the electron C squared what's the rest mass of the electron sorry right while I'm usually against memorizing anything because that's what like books in the internet are for this is one of those quantities that I want on the tip of your tongue as nuclear engineers you should remember what the rest mass of the electron is because a lot of our cow quantities are calculated based upon it for example this ratio Emmys it was at h nu / m ec squared gives you the energy the photon in terms of the number of electron rest masses which is a useful quantity in itself so what am i plug all this stuff in so we have 0.5 1 1 over a thousand plus 2 flipped over the x-axis times 2 times H nu and we get so that T becomes nine ninety-nine point seven four five MeV interestingly enough thousand MeV are in going photon - that equals that right there so that's the interesting quirk of physics is as you as the photon increases in energy the maximum amount of energy that it can leave with or the sorry the minimum amount of energy the photon can leave with or the maximum amount it can impart to an electron approaches 0.255 MeV or the photon energy - that what do you guys notice about that number that's right it's exactly half the rest mass of the electron so as your photons hit the GeV range and above they can all leave with half the rest mass of the electron of energy which means you have more and more energy able to be transferred in a given Compton scatter as the photon gets higher and higher in energy so for the limit of low energy the photons basically bounce off without transferring much energy at all and the higher energy the photon gets the higher percentage of that maximum transfer can be so that's what I wanted to clarify from last time it was an interesting coincidence that our photo peak and our Compton edge were pretty close to that number apart but one MeV isn't quite infinity what it is pretty close that difference right there what does that come out to before I say something stupid I'll just calculate it two for one it's like 0.218 MeV so we're already most of the way there so once you reach like 10 or 100 MeV you're pretty much at that limit and so what that tells you is that the dif the distance between a photo peak and it's corresponding Compton edge for high-energy photons is going to be half the rest mass of the electron once you get to lower and lower energies that distance will start to shrink I'm sorry other way around that distance will start to grow and I have a few examples I want to show you but first in order to understand these now I want to get into the part I told you we get to yesterday which is what's the probability that a confidence got R happens at a certain angle and this polar plot explains it pretty well in the limit of really low photons like 0.01 MeV or 10 ke V photons you can see that this forward scattering to an angle of zero or back scattering to an angle of 180 there's a 180 that's blocked by an axis right there they're almost the same so forward and back scattering there's not really that much of a big difference in probability so we were going to start graphing theta of this Compton's scatter as a function of the going to introduce this new quantity this angular ly dependent cross section before we were giving you cross-sections in the form of just Sigma's like Sigma Compton now we're actually telling you what's the probability of that interaction happening in this certain angle so it's called the differential cross section and you can have all sorts of differential cross sections like energy differential cross sections angle differential cross sections whatever have you so if we try and graph what is the shape of this look like this polar plot on a more understandable graph we can see that the probability is pretty high let's just call that a relative probability of one and as we trace around this circle that value gets lower and lower until we hit 90 degrees or PI over two at which point it starts to pop back up almost to its original value so this was for a 10k evey Photon now let's take a different extreme example we have a three MeV photon right here and that's that long dashed curve so that one here in the center so we can see that the relative probability of zero degrees is the same and if we trace around to 180 we're almost at the origin which tells us that for three MeV it starts off the same and quickly drops really far down what that means is that what's called forward scattering is preferred so up on this board we were talking about what's the maximum energy that a photon can transfer which is always in the back scattering case the other part to note is that that back scattering probability gets lower and lower as the photon gets higher and higher in energy yeah exactly yeah the cross-section value will yeah goes down as you increase in energy so a total forward scatter if you had a true forward scatter where theta equals zero I'll call that a miss it means that because we saw in this formula up here when when theta equals zero there is no energy transfer nothing and so that that would be to me like a Miss if that angles ever so slightly above zero then there is some scattering but there's very little energy transfer but that smaller energy transfer becomes more likely when the photon goes higher in energy these are these sorts of like cause-and-effect relationships I want you guys to be able to reason out if I were to give you a polar plot of this differential cross-section with angle and energy I want you to be able to reproduce this and tell me what's really going on if we fill in one of the ones in between let's go with 0.2 MeV this sort of single dashed line you can see that the probability of backscattering is somewhere between the three MeV and the ten ke V yeah Luke that's right backscattering refers to the photon going back which means this situation were the difference in angle between the incoming and outgoing photon is PI 180 degrees which means it just turns around and moves the other way right so that dashed line right there would follow something like this at what did we say 0.2 MeV the form the full form of this cross section right here is referred to as the Cline Nisshin a cross section and it has been derived by quantum electrodynamics which I will not derive for you now but there are plenty of derivations online if that's your kind of thing and I had to make a trade off in this class of how deep do we go into each concept versus how many concepts do we teach and I'm going for the latter because if there's any course that's supposed to give you an overview of nuclear's 2201 there will be plenty of time for quantum in 2202 and beyond so do you want at any rate this is the general form of it and what this actually tells you is that as the energy of the photon increases the effect of that angle well you're gonna die you guys will have to work that out on a homework problem I just remembered I want to stop stealing your thunder and giving away half the homework yep yes the quantity D Sigma D Omega says let's say you have a photon coming in at our x-axis you've got an electron here what's the probability that it scatters off into some small a area D Omega so in some small D Phi or sorry D theta D Phi or some small sine theta D theta D Phi into an element of solid angle I should probably draw that smaller to be a little more differential looking so if gammas are gonna scatter off in all directions but this D Sigma D Omega tells you what's the probability that it goes through that little patch so that that Omega has some component of theta in it and some component of Phi in it since it's a solid angle it depends on both the angle of rotation and the angle of inclination which we call theta and Phi now to get from this to the regular cross-section you're used to you can integrate over all angles Omega of the differential cross-section and you'll get the regular total cross-section which is just what is the probability of Compton scattering . if you wanted to know then what's the probability of Compton scattering into this angle it sounds kind of boring right why do we care about the angle anyone bored yet you can raise your hands be honest interesting okay well I'm gonna tell you why it's not boring cuz I don't think you're honest you can actually if you know the angle at which a Compton photon scatters into actually I want to leave that stuff up there is a pretty much one-to-one relation between the energy and the angle of scattering which means that let's say you have a cargo container and inside and your arms are you don't have a cargo container you have a cargo ship full of tons and tons of these stacked up cargo containers has anyone actually ever seen one of these before I'm gonna okay in case not I'm gonna do something dangerous and go to the Internet and hopefully the search for cargo container doesn't come up with something disgusting oh look at that how about cargo container yeah okay you got one of these right and your detector goes off and it just says there's something radioactive that shouldn't be here how do you find out which container it's in without taking the ship apart interesting problem huh do you just kind of look yeah that's one way you could shield you could mask off the ship and move your detector around and then do it for the other two dimensions so that will get you there but slowly how do you do it quickly well you do it with the client ition a cross-section if you know the relationship between the energy and the angle of a Compton scatter you take two detectors detector one detector two and you form what's called a Compton camera this is awesome because with two detectors you can pinpoint the location of a radiation source by knowing let's say you had a gamma ray that entered detector one so you have your initial eegammagee a let's draw a couple of spectra I'm going to use different colors so I can make them bigger this will be intensity and this will be energy so we have our blue detector one and we have some spectrum for detector one where we get the photo peak of the gamma at Iijima and this time you don't see every possible angle you only see whatever angle you get that Compton scatter at or you might see a whole bunch of different angles nevermind so you're going to see the Compton edge and this whole continuum of things and so you know that whatever energy this corresponds to means that theta equals PI that energy corresponds to theta equals zero and you know that there's a source somewhere in here now let's say that photon scatters out of detector one and into detector two in this case you've no longer just know that you have a source of some sort you'll end up with a certain photo peak corresponding to this EE gamma prime the only energy that can Compton scatter in the direction from Tector ones a detector to because you have now determined the angles between the line between the detectors and the detector of the source so then you get a photo peak and the corresponding compton edge for your agam a prime your A game of crime that tells you the angle that it came off of for your first interaction so that is your source angle so what that means is that by using these two angles you've now pinpointed your source to lie somewhere on these two cones projected back on one of the two points were those cones at that angle intersect this is something I want you to try and think about and work out on your own but it's really cool to explain this because like with one detector you know that there's a source somewhere and you know generally where to point with two detectors and energy resolution the energy of the photo peak of the second event tells you what the angle of the first event was and this way if you know what source you're looking for from the first photo peak and you know what angle you're looking for from the Compton scattered photons photo peak because they have those then you know not only what the source is but where it is and you know which containers would take apart to start looking so this is why we care about angle because there's actual like real ways of using this to your advantage to solve some pretty insane problems like which container would that be in who's starting to get the general idea behind a Compton camera or who would like another explanation anyone I asked two questions so who would like another explanation yeah okay the idea here is with just one detector all you know is whether or not there's a source the only information you're getting is its photo peak and Compton edge and Bowl and so you know the identity of the source maybe it's cobalt 60 or something but you don't know where it is by making a second you can then determine where on the Compton spectrum of the first measurement does the photon lie so by saying okay the photo peak corresponds to this energy which corresponds to some certain angle that this had to scatter off of so then you know what this angle is you've determined yes okay angle you've then determined this angle because you know the incoming path of the photon and you now know the outgoing path which means you know the angle so if you know the line between these two detectors and your photo peak lines up with your first Compton spectrums angle you then look at that angle back you also have to sweep around in the other direction which means you end up with a cone of possible locations yep there can be only a couple of things that could happen right you could have another direct photon to shoot into detector - in which case you'll just get a little bit of photo peak but we know to expect that so we can ignore it we're specifically looking for the photo peak coming from detector one they would which gives the perfect pretext to bring up the cross sections for these processes so I'm gonna skip ahead a little bit and start getting into what are the actual cross sections look like for compton scattering photoelectric and pair production all of them have to do with the electron density of the material the more electrons in the way the more you get these events happening so yes you will get compton scatters in the air because air contains electrons and they do Compton scatter but air is not very dense so you will get comparatively less so let's say that adds a little bit of noise on the bottom which is like any detector spectrum we've ever seen there's always noise from all sorts of other processes we're not looking for so now let's start to look at what the general energy and Z ranges of each of these effects are and we're going to recreate one of the plots that I showed you at the beginning of yesterday's class so let's make a graph of energy of the photon and Z of the material and we want to try to map out where the following three processes are dominant so tau our photoelectric effect C which we'll call our Compton scattering and Kappa which we'll call pair production and these cross-sections do give us relative probabilities as a function of the energy of the photon and the medium they're going through along with the actual density that something's going to happen so let's look at the form of these the cross section for the photoelectric effect scales with Z to the fifth do you think that photoelectric effect will be more likely for lo Z or high Z materials high Z materials right so we know we're going to be in the top half area and it's scales with one over energy to the seven halves so do you think this will be most likely with low or high energy so lower energy so we're gonna fill in our photoelectric somewhat heat oh I'm sorry so we know it's going to be in this part of the Z we know it's going to be in this part of the energy curve so let's fill in that area as photoelectric let's look at the other extreme pair production it scales with Z squared so is that going to be in the low or the high Z area anyone I think still high just the bigger the Z the bigger the cross-section right so we know that pear production is going to be in the high Z area now how about the energy it scales with the log of energy but the energy is on the top so we're a low or a high energy give you more pair production high so pair production it's gonna be here leaving everything else for Compton scattering and if we jump back to the start of last class we've reproduced from the cross-sections the same sort of plot that we saw before just from looking at the relative probabilities of each effect so I think is pretty cool we can now do that with some basic physics knowledge the last thing I want to fill in for Compton scattering is that the Klein ition a cross-section which is your differential cross-section as a function of Omega combined with the probability that you scatter into a certain angle and a certain energy because there's a one-to-one relationship ends up giving you your D Sigma C over de which is your energy distribution of Compton recoil electrons and that's directly what leads to this shape right here where if you have let's say 0.51 MeV relatively low energy there's let's see what do I want to say here yeah so as you go up in higher and higher energy you get fewer and fewer Compton electrons because like Jared was saying the probability of a comput interaction does go down with increasing energy as we kind of reasoned out but in addition you get more of let's say you get relatively more of these back scattered ones which is kind of interesting I don't think about that one but this is the typical shape that you tend to see for Compton scattering for lower for high-energy photons you get a Compton peak and then a very long shallow tail and as you go lower and lower energy it starts to sort of back up yeah Luke oh yeah the photoelectric effect is an absorption followed by an ejection in compton scattering the other photon is still intact it just loses energy gains wavelength but the energy of compton scattering it's let's say four MeV photons it's on the order of hundreds of ke V plenty to eject most of the electrons from an atom so you always have to think about are you going to eject something are you above the work function the work function for most atoms so we had a nice plot of that is in the e V range so chances are yeah Compton scattering mostly is going to be ejecting electrons - that's the whole reason we can count them if there wasn't an electron ejection there would be no electron ionization cascade to count so there's some sort of empirical or experimental proof that it does happen and speaking of experimental proof you can actually see that shape in this case this is a spectrum taken from two different gamma sources together a what is it one point two eight MeV and 0.51 MeV you can see that the one point two 8 MeV first of all has a way lower number of counts showing that the cross-section does indeed decrease with higher energy and it doesn't have well you can't really even see but it does not curving back up whereas this lower energy Compton photon is scattering more often because it's much higher and it's got that bump back up at the really low energies so it's kind of neat when the math that we're looking at if I jump ahead to the cross sections you can see that you'd expect the Compton scattering to decrease with energy and you look at an experimental plot of two different energies and there you have it higher energy less total Compton scattering but different shape so any questions before I move on to how you can use them to design shielding cool so what we're getting here these cross-sections are probably better known to some of you as mass attenuation coefficients which are simpler ways of describing how many photons in a narrow beam would be undergo some sort of process and be removed from the beam and they get removed exponentially for the same sort of reason for pretty much everything in this class ends up being an exponential doesn't it where if you have some if you have some intensity of photons or some change in intensity that's proportional to a change in X in the initial let's see and some what is it constant of proportionality like the cross-section or we'll call it now mu this mass attenuation coefficient the answer to that ends up being pretty much the same and it's this simple exponential thing and the nice thing is you don't need to be able that you don't have to calculate all these different cross sections because they're tabulated for you by NIST and that's one of the links on the website on the learning modules website that I want to show to you guys now you can actually look up these total summed cross sections for gamma ray interactions as a function of energy versus their mass attenuation coefficient in centimeter squared per gram the reason it's in that unit is it just tells you what the material does it doesn't tell you how much you have in the way and that's why I've rewritten this exponential attenuation formula with a row on the bottom and a row on the top that row being the density of the material because usually you can just say it's like I naught e to the MU X but these are the things that you'll look up in tables and these rows right here is whatever density you have of your material so if you want to calculate how much better cold water is shielding than hot water because of his change in density you can then look up the value for water which I'm going to show you how to do right now back up to see the actual site so if you guys looked right here these what is it missed tables of x-ray absorption coefficients I'll show you how to read through this table now because you'll need it from everything from problem set five to the rest of your life this is like one of those places you're gonna go constantly looking for nuclear data so you can either look at elemental media or compounds and mixtures water being one of them so let's go down to water comma liquid if you notice what's actually given here is this mu over Rho so the mass D what is a density specific mass attenuation coefficient in other words it's a cross section really it's a microscopic cross section I don't know why in a lot of these courses they're introduced separately because they're the same thing their interaction probabilities and then that other row from the slides just tells you how much is in the way that Rho times X just tells you how much water's in the way in terms of how dense it is and how thick your water shield is so using these tables if you know let's say you're sending in one MeV photon so we look up 10 to the 0 MeV you then have this value this nice round value of 10 to the minus 1 centimeter squared per gram you then multiply the die the density of the water that you have multiplied by the thickness of your water shielding and you get the change in intensity of the photons and let's do an example calculation just to make it a little more real so let's say we have a beam of photons of intensity I not and we're sending it through a tank of water and the question is do you want to keep this water at 0 Celsius or at a hundred Celsius what's the difference in shielding between freezing and boiling liquid water well we can look that up and let's say we have to specify an energy of the photons we'll call it 1 MeV photons so we can look up and say at 10 to the 0 MeV we go over we get just about 0.1 mu over Rho so mu over Rho equals zero point one centimeters squared per gram and now we can set up two equations one for zero Celsius and one for 100 Celsius so we'll say I at zero C equals I naught e to the minus 0.1 times Rho at zero Celsius times X let's say we have I don't know 10 centimeters of water so we'll just put a 10 in there that works out pretty well so and then are I at 100 C is the same I naught times e to the minus 0.1 times Rho of water at 100 times 10 centimeters and keep in mind here I made sure that since my new over row units are in centimeters squared per gram I'm putting in X as 10 centimeters because whatever's up here in the exponential has to be unitless that's a good check to see why are my calculations off by a factor of a billion let's check the unit's in the exponential yeah it will change the value of 0.1 will change depending on the energy of the incoming photons I know density changes and that's what we account for here with this row so next up we have to look up the densities of water at zero and 100 Celsius so I don't actually know them so density of water at zero C I'll surprise surprise it's we'll call it one gram per centimeter cubed now what is it at a hundred C too close to tell actually that's interesting at zero it's a little lower no I think that's it sites wrong yeah let's look at some steam tables because this is a real place to look for them so water is a hundred Celsius atmosphere is okay we'll just see that I think they're down here oh the chalks not letting me use the touchpad that's kind of cool oh there we go yep I think I got it so what's the density of saturated liquid zero point nine five eight zero point nine five eight grams per centimeter cubed well actually use the last corner today awesome so you could actually then go ahead and calculate because I love how this worked out right zero point one and ten cancel of zero point one and ten cancel so we'll just do e to the minus one and we get that this I is about thirty-six point seven nine percent of the gammas will be transmitted through ten centimeters of water and here e to the power of negative point nine five eight we get thirty eight point three seven percent transmission so actually about two percent more of the gammas will be transmitted if the water is hot it's a neat little calculation that you can do now we've looked at a really fine or very very small magnitude example folks came to me yesterday saying we want to design a new type of medical x-ray apron because we're worried that people carrying around all this lead are their backs are hurting and it's like making surgeons lives difficult if they're doing radioactive procedures can we do any better can they do any better what do you guys think can you beat physics when it comes to mass attenuation it's going to be awfully difficult and the best weapon you have is these mass attenuation coefficients to look at their relative values now these again are in centimeters squared per gram so this actually ranks aluminum to uranium in a sort of like per atom basis it's nut has nothing to do with their higher densities which only help things this just tells you how effective each of these elements is relatively at blocking gammas of different energies then to get the total amount of attenuation you multiply by the density aluminum is pretty sparse LED and uranium are pretty dense there's not too many ways around this problem in fact I wouldn't say that there's a way around this problem the best thing you can do is look at the really only interesting features on these curves does anybody know why there is those jagged edges there well let's take a look at some trends for uranium the jagged edge is at about 110 kV for lead it's like 80 K evie for tin it's probably more like 50 K evey it's decreasing with Z anyone remember what sort of magnitude we've looked at things like this before and if we're talking about photon and electron interactions what could be responsible for those sudden jagged edges well we have talked before about all sorts of different decay methods including those that can eject electrons from different energy shells you're looking at the same electron energy shells if you have a let's say photoelectric capable photon entering a calcium nucleus and let's go look at calcium as an example so I'll go to the tables of coefficients I'm gonna back up to elemental media and I'm gonna go to calcium for a simple example calcium has got this jagged edge right here and if we draw a line down it is precisely for ke V for ke V i bet is going to be the K edge energy the energy of the most inner bound calcium electron to find out we can go to the other NIST page that I linked you guys to the next X NIST x-ray transition energy table and let's look at calcium well this really doesn't work with chalk on your fingers and let's look at the K edge to check lo and behold four point zero five ke V so what you're seeing here is the photoelectric peak K edge absorption what this says is that energy is below 4 ke V you can't eject the innermost electron you just don't have enough energy as soon as you hit 4 ke V those inertial electrons become accessible to you so the cross-section suddenly jumps up because you have more electrons that you can eject photoelectric Li beyond the 100 ke V or so there's no more jagged edges because any photon above 100 kV can access pretty much energy of any electron in any element except maybe the super heavy ones and we don't have this data for them yet so then you might ask well there's going to be an L edge for calcium where would that be probably off this chart but you can look up where it would be so we'll go to the NIST yeah so you had the right idea Dan it's to the left right yeah exactly so now let's look up the L edge so if I were to ask you really doesn't work with chalk oh yeah okay that's better so the l1 edge is down at 438 evey which is indeed off the scale for this graph this bottoms out at 1 ke V but if I were to ask you to draw the full mass attenuation coefficient for uranium I'd expect to see a K edge an L edge and M edge at an N edge corresponding to shell levels 1 2 3 & 4 and where do you get that data you get it from here from these NIST databases or you calculate it one at a time using that Rydberg formula where that end final goes to infinity so you can either calculate them if you don't know or if miss doesn't have them in the table and I don't think they have the N edge wow they go all the way up to fermium let's do your rhenium what do they go to yep they don't have the M or the N edge but you do know how to calculate them is with that formula and so if we were to construct any old mass coefficient good we have the little space left it's gonna look generally like this there's going to be a photoelectric region let's say that's going to correspond to our photoelectric cross-section which goes way up with lower energies there's going to be a pair production part which goes up with higher energies and there's going to be this kind of decreasing Compton cross-section and if you couldn't dance these curves you end up with a shape like that which is just like all the other mass attenuation coefficients that you see so this is why they take the shape that they do if you add up the cross sections for photoelectric effect Compton scattering and pair production and you just kind of bounce on top of those you end up with the mass attenuation coefficient the part that's not shown here is that this photoelectric effect we'll have some jagged edges whenever you hit an electron energy transition level so it's 505 of I want to see if you guys have any questions on photon interactions with matter you know it's a lot to throw at you at once but I'm gonna be giving you guys lots to calculate to try it out and to learn what's going on from a more hands-on point of view yeah they're a way to slow down gammas yeah so the question is can you slow down gammas without putting stuff in the way well then what are you doing you got a vacuum right there so hmm that's probably a deeper question that I think so what else so gammas for example do have indices of refraction and materials yeah no gammas are just photons they're just really high energy and they do have indices of refraction that are usually around one part per million or like 1.000000 one or so so you can refract or bend gammas just not very well so the question is could you do something to stop the gammas that were maybe ten feet away the answer is physics not much you can do but if they are like planetary levels away it's possible that you could bend them away from an object just like you can bend visible light away from something closer up because it's got a much higher index of refraction pretty crazy stuff did you ever think a gamma rays having indices of refraction and behaving like regular light it's just regular light it's just really high-energy light so any other questions on the photon interactions would matter cool you