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<title>Fundamentals of radiation chemistry</title></titleStmt>

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The recordings and transcriptions used in this study come from the British

Academic Spoken English (BASE) corpus, which was developed at the

Universities of Warwick and Reading under the directorship of Hilary Nesi

(Warwick) and Paul Thompson (Reading). Corpus development was assisted by

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<date>30/10/1998</date><equipment><p>video</p></equipment>

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<item n="module">"UV Spectroscopy, Photochem and Rad Chem"</item>

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<u who="nm0692"> right well yesterday i was # <pause dur="0.5"/> talking about the <pause dur="2.2"/><kinesic desc="puts on transparency" iterated="n"/> idea of a track <pause dur="0.3"/> in <pause dur="0.2"/> radiation chemistry <pause dur="0.4"/> where <pause dur="0.3"/> as the particle moves along <pause dur="0.6"/> it's losing energy we know it loses energy at a great rate we know the energy of the particle <pause dur="0.4"/> we know the range of the particle <pause dur="0.8"/> what happens to this energy how does it get transmitted to the medium <pause dur="0.5"/> and <pause dur="0.2"/> this <pause dur="0.2"/> gave an indication <pause dur="0.3"/> that the principal <pause dur="0.2"/> acts of energy deposition <pause dur="0.4"/> are in the form of ionization <pause dur="0.4"/> and also <pause dur="0.2"/> <trunc>o</trunc> of excitation as well <pause dur="0.2"/> that that's <pause dur="0.2"/> an overall picture <pause dur="0.6"/><kinesic desc="changes transparency" iterated="y" dur="16"/> now <pause dur="0.4"/> that being the case you might say well <pause dur="0.2"/> all right you've got ions in excited states <pause dur="0.3"/> or so you say <pause dur="0.5"/> # <pause dur="0.4"/> what happens immediately after <pause dur="0.4"/> this <pause dur="0.4"/> event has occurred <pause dur="0.6"/> and <pause dur="0.2"/> i suppose <pause dur="0.4"/> the first thing that happens <pause dur="0.5"/> is that the electrons <pause dur="0.7"/> which are formed <pause dur="0.6"/> in the ionization act <pause dur="0.4"/> these are very small <pause dur="0.4"/> and they're very mobile <pause dur="1.4"/> and they're certainly very small compared with the <pause dur="0.2"/> cations that are formed 'cause those are going to be molecular size <pause dur="0.4"/> so the electrons tend to diffuse away very rapidly <pause dur="0.5"/> simply because they

are so small <pause dur="0.6"/> and # <pause dur="1.4"/> they have a a very high mobility <pause dur="0.9"/><kinesic desc="reveals covered part of transparency" iterated="n"/> what about the # <pause dur="1.0"/> the cations that are formed <pause dur="0.4"/> well if you just consider a general i put R-H-plus but it could be anything <pause dur="0.4"/> but suppose it was i don't know hexane or something <pause dur="0.4"/> then <pause dur="0.2"/> the <pause dur="0.7"/> cation radical <pause dur="0.4"/> which is the thing you get by taking the electron out of the molecule <pause dur="1.0"/> # <pause dur="0.5"/> will <pause dur="1.4"/> react with electrons that are <pause dur="0.2"/> are nearby the ones that haven't got away so to speak <pause dur="0.4"/> and you get a certain amount of ion recombination <pause dur="0.7"/> and when this electron returns to that cation <pause dur="0.5"/> it's very likely <pause dur="0.4"/> to form it in an excited state <pause dur="0.2"/> so you can get additional excited states <pause dur="0.2"/> from <pause dur="0.3"/> from ion recombination <pause dur="1.4"/><kinesic desc="reveals covered part of transparency" iterated="n"/> but also what can happen <pause dur="0.9"/> is that the R-H-plus the cation radical <pause dur="0.7"/> # is a very powerful proton donor <pause dur="0.7"/> and it will give a proton to almost anything in sight <pause dur="0.7"/> # and so <pause dur="0.3"/> this will tend to happen <pause dur="0.4"/> and perhaps the best known example of this is if you imagine <pause dur="0.3"/> R-H-plus is <pause dur="0.5"/> H-two-O-plus <pause dur="0.6"/> the water cation <pause dur="0.4"/> then <pause dur="0.2"/> that will give away a proton to a nearby water molecule <pause dur="0.4"/>

to give H-three-O-plus <pause dur="1.0"/> and you're left with O-H behind <pause dur="2.2"/> so you do get these # <pause dur="1.6"/> proton transfers occurring <pause dur="1.3"/> how many ions do we get <pause dur="1.6"/> how do you measure the efficiency of a radiation chemistry <pause dur="1.0"/> process <pause dur="0.6"/> well in photochemistry it was very easy because you had the idea of a photon <pause dur="1.0"/> and the photon <pause dur="0.7"/> entered a molecule <pause dur="1.0"/> and so much no <trunc>s</trunc> a certain percentage of the molecule was converted <pause dur="0.7"/> to a product <pause dur="0.4"/> and you had the idea of a quantum yield <pause dur="0.4"/> the idea that you could destroy <pause dur="0.5"/> such a percentage of molecules <pause dur="0.7"/> you would form such a percentage of product molecules <pause dur="0.4"/> and you had a quantum yield for this and this was always less than one <pause dur="0.5"/> unless you had a chain reaction <pause dur="0.4"/> so it's a very very clear idea <pause dur="0.6"/> but with a gamma ray quantum i the point i was trying to make yesterday it's completely different <pause dur="0.4"/> because the gamma ray quantum <pause dur="0.4"/> has got enormous amounts of energy <pause dur="0.7"/> and <pause dur="0.8"/> or the alpha particle has <pause dur="0.4"/> and you don't have the notion <pause dur="0.3"/> of just one molecule <trunc>be</trunc> <trunc>be</trunc> being converted by one

quantum <pause dur="0.6"/> you've got the <trunc>n</trunc> the idea of along the track <pause dur="0.2"/> hundreds of thousands of molecules being converted <pause dur="0.6"/> so how do we cope with this idea <pause dur="0.5"/> well we cope with it <kinesic desc="reveals covered part of transparency" iterated="n"/> by saying <pause dur="1.3"/> how many molecules <pause dur="0.2"/> how or how many ions do we get <pause dur="0.7"/> for a certain amount of energy deposited <pause dur="0.7"/> and there is a definition of a thing called a G value <pause dur="0.8"/> and the G value <pause dur="0.6"/><kinesic desc="reveals covered part of transparency" iterated="n"/> which is a measure of the product yield or indeed the <trunc>react</trunc> the <pause dur="0.2"/> destruction of the reactant 'cause # that would be then the negative G value but <pause dur="0.5"/> they normally normally you look at products to measure positive G values <pause dur="1.0"/><kinesic desc="reveals covered part of transparency" iterated="n"/> they are the number of molecules <pause dur="0.2"/> formed <pause dur="0.3"/> or destroyed <pause dur="0.5"/> for every hundred electron volts <pause dur="2.7"/> and so you if you think about it a hundred electron volts <pause dur="1.5"/> ionization of a typical organic is about ten eleven or twelve let's so say ten for the sake of argument <pause dur="0.6"/> then <pause dur="0.5"/> you would say well if we got ten <pause dur="0.6"/> <trunc>prod</trunc> if we got ten ions <pause dur="0.5"/> that would be <pause dur="0.2"/> a hundred per cent efficient <pause dur="1.2"/> but in fact you don't get ten ions <pause dur="0.3"/> you get typically <pause dur="0.4"/> two or three ions <pause dur="0.6"/> maybe

slightly more than three sometimes but not much more than three <pause dur="0.7"/> and so the ionization efficiency <pause dur="0.8"/> is not enormously high <pause dur="1.4"/> but of course <pause dur="1.4"/> you also get the excited states as well <pause dur="0.7"/> and so if you're at the excited state yield which is also often about two or three <pause dur="0.5"/> but well for some systems <pause dur="0.9"/> you're using about fifty per cent of the of the <trunc>ou</trunc> energy <pause dur="1.0"/> chemically productively <pause dur="0.4"/> what happens to the rest <pause dur="0.5"/> well the rest <pause dur="0.3"/> you get ion recombination the ions don't escape at all they they <trunc>si</trunc> they're formed and they recombine instantly <pause dur="0.2"/> don't measure them <pause dur="0.3"/> 'cause they're not around to be measured any more <pause dur="1.8"/> under those conditions <pause dur="1.7"/> you simply end up by getting excited states <pause dur="0.4"/> and some of these simply end up by heating up the medium <pause dur="0.3"/> so you you you actually simply do a lot of vibration excitation of the medium <pause dur="0.3"/> so the medium will become warm <pause dur="0.3"/> as you put radiation through it <pause dur="1.1"/> so <pause dur="0.3"/> the energy is partitioned between ionization events <pause dur="0.4"/> excitation events <pause dur="0.3"/> and merely thermal warming <pause dur="0.8"/><kinesic desc="reveals covered part of transparency" iterated="n"/> and this <pause dur="0.2"/> G value <pause dur="0.6"/> as i've <pause dur="0.2"/> put down

there is the analogue of the quantum yield <pause dur="0.2"/> it's not as precise in a way <pause dur="0.4"/> you can't picture it as as nicely <pause dur="0.7"/> but you've got to start from somewhere <pause dur="0.4"/> and you say well if i've got a million electron volts <pause dur="0.3"/> how <trunc>m</trunc> what do i get for each hundred electron volts that i deposit in the medium <pause dur="0.5"/> and so that is your <pause dur="0.2"/> fundamental parameter the G value <pause dur="0.3"/> and whenever you're discussing radiation chemistry of anything <pause dur="0.4"/> you begin <pause dur="0.3"/><event desc="takes off transparency" iterated="n"/> by saying well what is what is its G value <pause dur="0.3"/> is it # <pause dur="1.0"/> one or <pause dur="1.0"/> three or five or <pause dur="0.4"/> and if it's a lot more <pause dur="0.3"/> than <pause dur="0.6"/> than ten or if it becomes hundreds again it's a you got a chain reaction <pause dur="0.4"/> running away <pause dur="1.3"/> now <pause dur="1.6"/> let's think about some of the # <pause dur="1.9"/><gap reason="inaudible" extent="1 sec"/> experiments <pause dur="0.2"/> to get some idea of what these G values are <pause dur="0.4"/> the very simplest experiment you can do <pause dur="0.3"/> is using an ionization chamber and i'll draw one of these <kinesic desc="writes on transparency" iterated="y" dur="16"/> it's a very simple idea <pause dur="0.7"/> if you imagine you've got a glass vessel <pause dur="0.8"/> with two electrodes in <pause dur="1.1"/> you got a you apply a voltage <pause dur="0.4"/> that's voltage there <pause dur="1.1"/> and you have an ammeter for measuring current <pause dur="0.4"/> there <pause dur="2.5"/> now <pause dur="0.6"/> normally of

course <pause dur="1.6"/> if you have that <pause dur="0.2"/> globe filled with <pause dur="0.5"/> argon <pause dur="0.3"/> or helium <pause dur="0.7"/> or <pause dur="0.5"/> hexane vapour or something propane <pause dur="1.0"/> if you apply a voltage <pause dur="1.0"/> that's not enormous <pause dur="1.2"/> then there's no current <pause dur="0.8"/> because the gases are insulating <pause dur="0.6"/> i know if you take it to a very very high voltage you should get diametric breakdown <pause dur="0.3"/> we're not thinking about that at all <pause dur="0.4"/> we're thinking of quite low voltages <pause dur="0.3"/> get no current <pause dur="0.5"/> but if you bring up <pause dur="0.9"/> a <pause dur="0.4"/> other a radioisotope <pause dur="0.4"/> to the outside <pause dur="0.4"/><kinesic desc="writes on transparency" iterated="y" dur="1"/><kinesic desc="indicates point on transparency" iterated="n"/> there <pause dur="0.3"/> close to the glass <pause dur="0.4"/> vessel so if we have isotope <pause dur="3.0"/><kinesic desc="writes on transparency" iterated="y" dur="7"/> this is giving out <pause dur="0.7"/> radiation <pause dur="1.0"/> like so <pause dur="0.8"/> or you could fire a very very fast electron beam <pause dur="0.7"/> into the globe <pause dur="1.6"/> if you're still applying the voltage <pause dur="1.5"/> you don't get a current <pause dur="0.6"/> I <pause dur="0.4"/> becomes greater than nought <pause dur="0.6"/><kinesic desc="writes on transparency" iterated="y" dur="13"/> with radiation <pause dur="4.1"/> I <pause dur="0.5"/> becomes greater than zero <pause dur="1.1"/> without radiation <pause dur="2.2"/> I <pause dur="0.3"/> equals zero <pause dur="0.7"/> so clearly the radiation is producing ions in the gas <pause dur="1.1"/> and you know they're there <pause dur="0.8"/> because the whole thing is carrying a current <pause dur="0.5"/> once you apply a voltage <pause dur="1.1"/> positive ions are moving to one electrode <pause dur="0.6"/> negative ions are moving

to another <pause dur="0.7"/> so immediately we know <pause dur="0.6"/> that <pause dur="0.3"/> the application <pause dur="0.8"/> of ionizing radiation to a gas <pause dur="0.6"/> produces ions <pause dur="0.7"/> and you might say well <pause dur="0.8"/> # <pause dur="1.5"/> what sort of <pause dur="0.4"/> yields do we get <pause dur="0.7"/> from gases <pause dur="0.7"/> and maybe from liquids <pause dur="0.4"/> well from gases <pause dur="0.6"/><kinesic desc="writes on transparency" iterated="y" dur="6"/> the <pause dur="0.2"/> ion yield is <pause dur="0.2"/> sort of <pause dur="0.5"/> reasonable it's <pause dur="0.2"/> say <pause dur="0.2"/> one <pause dur="0.6"/> to three <pause dur="0.3"/> depends on the gas <pause dur="0.5"/><kinesic desc="writes on transparency" iterated="y" dur="3"/> for liquids <pause dur="1.6"/> interestingly if you if you fill that globe with <pause dur="0.2"/> liquid hexane or liquid benzene <pause dur="0.8"/><kinesic desc="writes on transparency" iterated="y" dur="1"/> G liquid it's nought-point-one <pause dur="1.6"/> now why should that be <pause dur="0.7"/> well the answer again if we go back to photochemistry <pause dur="3.7"/> a lot of <pause dur="1.0"/> photochemical acts <pause dur="0.6"/> give you <pause dur="0.7"/> two species formed very close to each other <pause dur="2.2"/> if they <pause dur="0.8"/> can get away from each other <pause dur="0.4"/> you get products <pause dur="1.0"/> but if they <pause dur="0.8"/> are forced <pause dur="0.2"/> by being surrounded by solvent <pause dur="1.0"/> back onto each other <pause dur="0.4"/> they recombine <pause dur="0.7"/> and so what's happening in the liquid <pause dur="0.3"/> is if you <kinesic desc="writes on transparency" iterated="y" dur="16"/> get <pause dur="0.9"/> a positive charge and a <trunc>neg</trunc> negative charge formed <pause dur="0.4"/> and these are surrounded <pause dur="2.7"/> by solvent molecules <pause dur="3.4"/> like so <pause dur="0.4"/> and this of course is a positive solvent molecule and that's a <pause dur="0.3"/> either a negative solvent molecule or an electron <pause dur="0.5"/>

the thing is trapped <pause dur="0.6"/> and you've got <kinesic desc="writes on transparency" iterated="y" dur="8"/> basically what's called a cage effect the thing is trapped <pause dur="1.3"/> in a solvent so i'll say solvent cage effect <pause dur="0.7"/> that's very important in photochemistry <pause dur="1.7"/> it's very important in radiation chemistry as well <pause dur="0.7"/> so <pause dur="0.3"/> in liquids <pause dur="0.4"/> you've got the solvent cage effect operating <pause dur="0.4"/> in a gas phase you haven't once the ions are formed zoom <pause dur="0.5"/> they they go charging off towards the electrodes <pause dur="0.2"/> so there's quite a difference there <pause dur="0.7"/> so <pause dur="0.7"/> that was the <trunc>f</trunc> the first indication <pause dur="0.3"/> that you were getting ionization <pause dur="0.8"/> in the gases <pause dur="0.2"/> to quite high yields <pause dur="0.4"/> but in the liquids <pause dur="0.9"/> to very low yields <pause dur="0.2"/> and for a long time <pause dur="0.2"/><kinesic desc="writes on transparency" iterated="y" dur="1"/> that experiment <pause dur="0.9"/> convinced everybody <pause dur="0.7"/> that ionization <pause dur="0.7"/> was not important in liquids <pause dur="1.6"/> not at all <pause dur="0.7"/> because <pause dur="0.6"/> the conductivity experiments indicated the yields are very low <pause dur="1.0"/> and that that sort of held <pause dur="0.4"/> sway for a long time <pause dur="0.7"/> but then there were some <pause dur="0.3"/> very clever experiments <pause dur="0.4"/> that were done <pause dur="0.4"/> by a chap <pause dur="0.4"/> called Hamill <pause dur="0.7"/><kinesic desc="changes transparency" iterated="y" dur="18"/> and i'll <pause dur="0.8"/> as i've got these on a <pause dur="0.5"/> a slide i'll put the slide on <pause dur="2.7"/> and it goes as follows <pause dur="6.4"/> Hamill worked <pause dur="1.5"/>

with <pause dur="1.0"/> solutions <pause dur="0.6"/> they were quite dilute say a millimolar <pause dur="0.8"/> of things like naphthalene and biphenyl <pause dur="0.3"/> and he worked with lots of other ones as well but they're just examples <pause dur="0.6"/> in <pause dur="0.2"/> what are called glass glassy forming solvents or <pause dur="0.6"/> # <pause dur="0.2"/> glass forming solvents <pause dur="1.4"/> there were quite a few of these around i mentioned them in the photochemistry section of the course these are solvents that <kinesic desc="reveals covered part of transparency" iterated="n"/> when you freeze them down <pause dur="1.4"/> they give a perfect glass they don't crystallize <pause dur="1.3"/> they've got an irregular structure <pause dur="0.4"/> you've got the methyl group here <pause dur="0.4"/> tetrahydrofuran <pause dur="0.5"/> crystallizes when you freeze it <pause dur="0.2"/> in liquid nitrogen <pause dur="0.5"/> methyltetrahydrofuran is less symmetrical <pause dur="0.4"/> and it forms a glass <pause dur="1.6"/> pentane <pause dur="0.2"/> will crystallize if you freeze it <pause dur="0.3"/> you put a methyl group on get three-methylpentane <pause dur="1.4"/> it gives a perfect glass <pause dur="0.6"/> and if you've got a perfect glass <pause dur="0.7"/> it means you can freeze the thing in a cell a caught cell <pause dur="0.6"/> and then run an optical spectrum <pause dur="0.7"/> at liquid nitrogen temperature <pause dur="0.9"/> in an ordinary spectrophotometer <pause dur="1.8"/> and <pause dur="0.3"/> what did

Hamill do <pause dur="0.4"/> well <pause dur="0.5"/> he rotated these solutions in these glassy forming solvents and then ran U-V spectra <pause dur="0.8"/><kinesic desc="reveals covered part of transparency" iterated="n"/> and for example <pause dur="2.2"/> if you # <pause dur="0.6"/> take <pause dur="1.2"/> naphthalene <pause dur="1.4"/> C-ten-H-eight <pause dur="0.8"/> this is <trunc>co</trunc> naphthalene's colourless <pause dur="0.8"/> dissolve it up in T-H-F or three-methyl-pentane <pause dur="0.4"/> if you gamma irradiate it <pause dur="1.2"/> the thing goes deep green <pause dur="0.3"/> it <trunc>s</trunc> it goes a deep green colour it's quite a spectacular experiment to do it's a nice demonstration but of course it's <pause dur="0.3"/> not easy to do 'cause you'd need <pause dur="0.4"/> # <pause dur="0.8"/> # a gamma source to do it <pause dur="0.2"/> and we we don't have one here <pause dur="0.8"/> # <pause dur="1.6"/> to get deep green colour <pause dur="0.6"/> this is this has got a it's quite a structured spectrum it's not just a broad band <pause dur="0.4"/> and it's exactly the same spectrum as <kinesic desc="indicates point on transparency" iterated="n"/> this thing <pause dur="0.7"/> in other words if you take a sodium film <pause dur="0.3"/> a film of <trunc>so</trunc> if you purify sodium <pause dur="0.7"/> in fact you heat it up and evaporate it <pause dur="0.4"/> and then could allow the sodium vapour to get to get condense on glass <pause dur="0.3"/> you get <pause dur="0.6"/> a <pause dur="0.2"/> mirror <pause dur="0.3"/> formed <pause dur="0.2"/> a silver mirror <pause dur="0.4"/> of sodium <pause dur="0.8"/> if you take that silver mirror of sodium which is very very pure <pause dur="0.9"/> and you react it <pause dur="0.5"/>

with <pause dur="1.1"/> a dilute solution of naphthalene <pause dur="0.2"/> in an ether <pause dur="0.7"/> you end up by getting a deep green solution <pause dur="0.6"/> and this is sodium-plus <pause dur="0.3"/> <trunc>C-ten-H-et</trunc> <pause dur="0.2"/> H-eight-<pause dur="0.3"/>minus <pause dur="0.2"/> with a free radical symbol <pause dur="0.3"/> # it's a free radical <pause dur="0.5"/> and it's it's strongly paramagnetic <pause dur="0.3"/> and it shows an E-S-R spectrum quite a complex one <pause dur="0.4"/> but it can be perfectly analysed in terms of # <pause dur="0.4"/> in terms of naphthalene <pause dur="0.5"/> so <pause dur="2.5"/> you've got this formed <pause dur="0.3"/> now we know <pause dur="0.3"/> the extinction coefficient <pause dur="0.3"/> of this <pause dur="0.7"/> thing <pause dur="0.2"/> because we can <pause dur="0.5"/> measure <pause dur="0.7"/> how much sodium has been consumed or <pause dur="0.5"/> we can indeed # <pause dur="1.8"/> # <pause dur="1.0"/> we can quantitate it reasonably well <pause dur="1.0"/> and so if we know the extinction of the <pause dur="0.4"/><kinesic desc="indicates point on transparency" iterated="n"/> of this we can work out the yield <pause dur="0.5"/> of these <pause dur="1.1"/> <trunc>cat</trunc> <pause dur="0.2"/> anions <pause dur="0.3"/> which are formed in the glass <pause dur="0.8"/><kinesic desc="reveals covered part of transparency" iterated="n"/> and if you work it out <pause dur="0.6"/> it comes out to be about two <pause dur="0.5"/> for hydrocarbons <pause dur="0.3"/> and three <pause dur="0.3"/> for alcohols and ethers <pause dur="0.6"/> and this <unclear>retracts from all</unclear> people's thinking because <pause dur="0.3"/> what it meant was <pause dur="0.3"/> that if you <pause dur="0.2"/> operate at low temperature <pause dur="1.9"/> where everything was frozen down <pause dur="1.6"/> and you gamma irradiate it <pause dur="1.1"/> the electrons <pause dur="0.8"/> formed <pause dur="0.5"/> by attacking <pause dur="0.6"/> the

solvent <pause dur="0.5"/> migrated through the medium <pause dur="0.4"/> and ended up <pause dur="0.8"/> on the naphthalene <pause dur="1.1"/> now this is quite this is a very important idea here <pause dur="0.2"/> which i which i i i want to stress so i'll i'll <trunc>th</trunc> i'll go on about it a bit <pause dur="1.1"/> when you do photochemical excitation <pause dur="0.6"/> you match <pause dur="0.3"/> the quantum <pause dur="0.9"/> of the light <pause dur="1.7"/> to the energy level <pause dur="0.2"/> of the solute <pause dur="1.2"/> okay <pause dur="0.4"/> you have a solvent and a solute <pause dur="0.6"/> the solvent normally doesn't absorb any light <pause dur="0.3"/> you're looking at the solute <pause dur="0.3"/> and that <pause dur="0.4"/> has <trunc>spectra</trunc> spectral levels <pause dur="0.5"/> that correspond to the irradiation wavelength in other words you're <pause dur="0.3"/> pumping a particular state within the solute molecule <pause dur="0.8"/> that's photochemistry <pause dur="0.6"/> in radiation chemistry it's completely different <pause dur="1.0"/> as the <pause dur="1.3"/> gamma ray traverses a medium or the alpha particle whatever it is <pause dur="1.0"/> it <pause dur="0.3"/> interacts with the solvent <pause dur="0.4"/> and it ionizes the solvent <pause dur="1.6"/> you might say what happens to the solute <pause dur="0.3"/> does it ionize that as well <pause dur="0.7"/> well <pause dur="0.3"/> it's purely statistical <pause dur="0.7"/> it will <pause dur="0.5"/> excite electrons in whatever it's passing by <pause dur="1.9"/> and of course statistically <pause dur="1.4"/> if you take something like

methanol or pentane <pause dur="0.6"/> then take a litre of that and work out how many <trunc>m</trunc> how many moles there are in liquid pentane <pause dur="1.0"/> it's about ten molar <pause dur="0.4"/> liquid pentane <pause dur="0.2"/> is about ten moles per litre <pause dur="0.8"/> the naphthalene is ten-to-the-minus-three <pause dur="0.4"/> so it's a a ten-thousandfold excess of the solvent <pause dur="0.7"/> and so ninety-nine-<pause dur="0.7"/>point-nine-nine per cent <pause dur="0.3"/> of all the energy <pause dur="1.1"/> goes into the solvent <pause dur="0.3"/> and nought-point-nought-one per cent goes into solute <pause dur="0.4"/> so <pause dur="0.2"/> all of the <pause dur="0.5"/> reactions shown by the solute <pause dur="1.7"/> reflect what's happened to the solvent <pause dur="1.6"/> and with this solute <pause dur="0.2"/> it's picking up the electrons that have been formed in the solvent <pause dur="0.6"/> they've moved through this medium at low temperature <pause dur="0.5"/> and they've <trunc>al</trunc> alighted <pause dur="0.3"/> and been trapped <pause dur="0.7"/><vocal desc="cough" iterated="n"/><pause dur="0.2"/> on the solute to give this green colour <pause dur="1.3"/> and <pause dur="0.8"/> this process of having a very small amount of something there <pause dur="0.8"/> that captures <pause dur="1.0"/> the negative charge in the system <pause dur="0.9"/> so you can then look at it optically <pause dur="1.0"/> that's called a scavenger <pause dur="0.4"/> okay <pause dur="0.4"/> so it's playing the role <pause dur="0.5"/><kinesic desc="writes on transparency" iterated="y" dur="15"/> of a scavenger <pause dur="1.4"/> i'll i've actually writing on the glass at

the moment so i'll <pause dur="0.5"/> get back onto the <pause dur="2.8"/> so the naphthalene <pause dur="0.4"/> is <pause dur="0.2"/> C-ten-H-eight <pause dur="1.0"/> acts <pause dur="1.0"/> as <pause dur="0.5"/> a scavenger <pause dur="2.5"/> and the <pause dur="0.7"/> this was a <pause dur="0.8"/> the key <pause dur="0.2"/> to <pause dur="1.3"/> getting a much better understanding what was going on <pause dur="0.4"/> so we now know that in the solid organic matrices <pause dur="0.4"/> you know frozen down <pause dur="0.5"/> we had high ion yields <pause dur="1.5"/> and that result <pause dur="0.2"/> is complete opposite <pause dur="0.4"/> to the one we got <pause dur="0.5"/> using the conductivity cell <pause dur="0.7"/> and in the conductivity cell it's almost what's happening properly there <pause dur="0.6"/> is that # <pause dur="0.4"/> you're <trunc>simp</trunc> <trunc>is</trunc> you're simply <pause dur="0.7"/> not you haven't got a very efficient means of <trunc>g</trunc> of <pause dur="0.2"/> gathering the charge <pause dur="0.5"/><kinesic desc="reveals covered part of transparency" iterated="n"/> here <pause dur="0.5"/> you've got these solute molecules spread right through the system and they <trunc>c</trunc> they captured electrons before they can <pause dur="0.4"/> find positive ions to <pause dur="0.7"/> collide with <pause dur="0.5"/> or even destroy each other which # is <trunc>i</trunc> you know <pause dur="0.2"/> can also happen <pause dur="0.7"/> and there is such a reaction which i'll <pause dur="0.3"/> perhaps i'll talk about later <pause dur="1.0"/> well that was a position in <pause dur="0.6"/> frozen matrices <pause dur="0.4"/> and a question then arose <pause dur="0.3"/> well <pause dur="0.3"/> oh that's fine for frozen matrices <pause dur="0.8"/> what about liquids <pause dur="0.7"/> well <pause dur="1.1"/> the answer

to that <kinesic desc="reveals covered part of transparency" iterated="n"/> came when people developed a technique called pulse radiolysis <pause dur="1.3"/> we talked about flash photolysis before that's where you take <pause dur="0.5"/> a flash lamp <pause dur="0.8"/> or a laser that's pulsed <pause dur="0.4"/> and you administer <pause dur="0.2"/> a flash of light or a pulse of light from the laser <pause dur="0.4"/> to the sample <pause dur="0.5"/> and you look <pause dur="1.9"/> spectroscopically on a very short time base nanoseconds or even shorter than nanoseconds as to what's going on <pause dur="0.9"/> in pulse radiolysis exactly the same idea <pause dur="0.4"/> what you've got this time <pause dur="0.5"/> instead of radiolysing with a continuous beam <pause dur="0.2"/> which is what you normally do with a cobalt source or a <pause dur="0.3"/> linear accelerator <pause dur="0.9"/> with a linear accelerator you can <pause dur="0.2"/> arrange the electronics <pause dur="0.6"/> to chop the beam <pause dur="1.1"/> and you can chop it <pause dur="0.4"/> right down <pause dur="1.2"/> into nanosecond pulses <pause dur="0.7"/> and so <pause dur="0.9"/> just as in flash photolysis <pause dur="0.2"/> you had a nanosecond flash of light <pause dur="0.6"/> in pulse radiolysis <pause dur="0.7"/> you've got <pause dur="0.6"/> a nanosecond <pause dur="0.2"/> burst of electrons <pause dur="0.5"/> you can pulse an electron beam <pause dur="0.6"/> fairly easily <pause dur="0.5"/> and get <pause dur="0.3"/> a pulse of electrons <pause dur="0.4"/> so it's exactly the same type of idea <pause dur="0.9"/> and with that of course you can

then <pause dur="0.7"/> attack liquids with these electron pulses <pause dur="0.2"/> and look at <pause dur="0.8"/> ions that are formed <pause dur="0.4"/> and what you have to do <pause dur="0.4"/><kinesic desc="reveals covered part of transparency" iterated="n"/> is to put in <pause dur="0.6"/> again <pause dur="1.0"/> a scavenger <pause dur="1.0"/> most <pause dur="0.2"/> aromatics <pause dur="0.2"/> # sorry most aliphatics <pause dur="0.5"/> and ethers if you imagine that you were to ionize an ether <pause dur="0.9"/> you get an ether cation or ether radical and electron <pause dur="1.3"/> those those aliphatic radicals absorb right down in the U-V they're very hard to see <pause dur="0.3"/> so if you want to know what's going on what you again do <pause dur="0.3"/> is to put in <pause dur="0.6"/> another scavenger <pause dur="0.6"/> # maybe naphthalene again or biphenyl but this time <pause dur="0.4"/><kinesic desc="reveals covered part of transparency" iterated="n"/> you're putting it at room temperature <pause dur="0.5"/> instead of of a very low temperature <pause dur="0.9"/> so it's the same experiment in a way that Hamill did <pause dur="0.2"/> except that it's in the liquid phase <pause dur="1.0"/> to begin with <pause dur="0.3"/> so you're looking at the mobility of the of the ions which you don't get in the <pause dur="0.4"/> frozen matrix <pause dur="0.2"/> and also <pause dur="1.0"/> # <pause dur="0.2"/> you're <pause dur="0.5"/> using fast <pause dur="0.2"/> spectroscopic recording you're recording optical spectra <pause dur="0.9"/> on the nanosecond timescale <pause dur="0.5"/> and if you do that <pause dur="0.6"/><kinesic desc="reveals covered part of transparency" iterated="n"/> then you find <pause dur="0.4"/> that <pause dur="0.3"/> you get # <pause dur="1.6"/> ion yields <pause dur="0.5"/> # <pause dur="0.2"/> in water

about two-point-three <pause dur="0.2"/> this is using nanosecond <pause dur="0.6"/> N-S <pause dur="0.8"/> in methanol about one-point-three <pause dur="0.6"/> and <pause dur="0.2"/> in hexane about point-five <pause dur="0.2"/> so the ion yields are much higher <pause dur="1.2"/> than the pure conductivity experiment indicated <pause dur="0.4"/> and this is using <pause dur="0.2"/> the scavengers <pause dur="0.5"/> but this time <pause dur="1.1"/> in <pause dur="0.3"/> a <trunc>l</trunc> in the liquid phase <pause dur="0.2"/> so again <pause dur="0.4"/> we've got direct physical evidence <pause dur="0.3"/> for <pause dur="0.4"/> quite <pause dur="0.2"/> large yields of ions <pause dur="0.3"/> even in hydrocarbons <pause dur="0.3"/> from <pause dur="1.1"/> the <trunc>r</trunc> the radiolysis experiment <pause dur="1.9"/> there's another thing that you can do <pause dur="0.7"/> and which was done <pause dur="2.1"/> all of these experiments were done on a <trunc>nat</trunc> a nanosecond timescale it's ten-to-the-minus-nine of a second okay ten-to-the-minus-nine <pause dur="0.7"/> and the question raised in people's minds <pause dur="0.5"/> okay this is what you see at ten-to-the-minus-nine <pause dur="0.4"/> but what if we could actually shorten the pulse further <pause dur="0.6"/> would we see earlier events after all in the track model <pause dur="0.2"/> you had the pictures of these spurs <pause dur="0.2"/> with positive and negative ions <pause dur="0.4"/> and the question was <pause dur="0.4"/> would it be conceivable <pause dur="0.5"/> to see <pause dur="0.4"/> these # scavengers <pause dur="0.6"/> capturing electrons or maybe capturing positive charges as

well <pause dur="1.3"/> on a very very short timescale <pause dur="2.4"/> giving ion yields higher than what you see at one nanosecond because at one nanosecond <pause dur="0.6"/> some of your naphthalene minuses <pause dur="0.5"/> might have combined <pause dur="0.2"/> with solvent cations <pause dur="0.3"/> or naphthalene pluses <pause dur="0.4"/> also formed <pause dur="0.3"/><kinesic desc="changes transparency" iterated="y" dur="7"/> and so people then began <pause dur="0.3"/> to to develop <pause dur="0.5"/> shorter timescales <pause dur="0.2"/> and they developed <pause dur="0.3"/> picosecond pulse radiolysis <pause dur="0.4"/> and that's ten-to-the-minus-twelve of a second <pause dur="0.9"/> so here we go <pause dur="0.5"/> picosecond pulse radiolysis ten-to-the-minus-twelve of a second <pause dur="0.4"/> basically it get gives higher ion yields <pause dur="0.5"/> and so if you actually <pause dur="0.3"/> measure the ion yield <pause dur="0.3"/> as a function of the <pause dur="0.3"/> pulse length that you're applying <pause dur="0.3"/> you find that ten-to-the-minus-twelve of a second <pause dur="0.4"/> everything <pause dur="0.4"/> gives ion yields of three <pause dur="0.4"/> at ten-to-the-minus-nine it's dropped down to about sort of two or one-point-five <pause dur="0.5"/> microsecond <trunc>i</trunc> it's fairly stable <pause dur="0.4"/> and then gradually of course <pause dur="0.3"/> with even longer pulses it gets very low <pause dur="0.4"/> and you get to the figure <pause dur="0.2"/> that you record <pause dur="0.3"/> in a conductivity experiment <pause dur="0.5"/> so <pause dur="0.6"/> you actually find the ion yield

depends on time <pause dur="0.5"/> and it depends on that <pause dur="0.4"/> because <pause dur="0.3"/> you form these things inhomogenously <pause dur="0.6"/> <trunc>el</trunc> in the spurs <pause dur="0.8"/> they're formed <pause dur="0.6"/> they begin to migrate away <pause dur="1.0"/> on a very short timescale they they alight on the solute and give you say a solute minus that # <pause dur="0.2"/> you can see at the very short timescales <pause dur="0.6"/> but <pause dur="0.7"/> not long afterwards some of these have already recombined <pause dur="0.5"/> with positive ions in the medium <pause dur="0.3"/> to give lower yields <pause dur="1.3"/> well <pause dur="0.7"/> this <pause dur="0.5"/> was the picture with ions <pause dur="0.7"/> what about excited states <pause dur="1.0"/> well there are various ways of forming the excited states <pause dur="0.3"/> and various ways of trying to see them <pause dur="0.5"/> let's let's so let's look now <pause dur="0.4"/><kinesic desc="reveals covered part of transparency" iterated="n"/> at <pause dur="0.5"/> excited states <pause dur="0.6"/> measurement of <pause dur="1.0"/> excitation yields in radiolysis <pause dur="1.4"/> what we think happens <pause dur="0.3"/> and what i've done to start with <pause dur="0.3"/><kinesic desc="reveals covered part of transparency" iterated="n"/> is simply take a very simple system <pause dur="0.6"/> and i've taken cyclohexane <pause dur="0.8"/> now you might wonder <pause dur="0.3"/> why cyclohexane <pause dur="0.6"/> and the answer is <pause dur="0.4"/> that <pause dur="0.2"/> if you're looking at free radical chemistry <pause dur="0.6"/> cyclohexane's quite a good thing to work with because <pause dur="0.3"/><kinesic desc="writes on transparency" iterated="y" dur="7"/> if you look at cyclohexane <pause dur="1.7"/> you've got <trunc>hy</trunc> <trunc>the</trunc> two

pairs of hydrogens all the way around <pause dur="1.6"/> it doesn't matter which of those bonds you break you always end up <pause dur="1.0"/><kinesic desc="writes on transparency" iterated="y" dur="4"/> with the same radical C-six-H-eleven <pause dur="1.4"/> if you add hexane <pause dur="0.7"/> you can take a hydrogen off <pause dur="0.5"/> the methyl group at the end <pause dur="0.2"/> or the next C-H-two in <pause dur="0.2"/> or the next C-H-two after that so immediately you got three different radicals <pause dur="0.4"/> and the chemistry becomes much more complicated <pause dur="0.4"/> so cyclohexane is very much a preferred solvent to work with <pause dur="0.4"/> because you can get it very very pure <pause dur="0.9"/> and it gives just one radical <pause dur="0.5"/> there's only one C-six-H-eleven <pause dur="0.8"/> so whichever that one <trunc>w</trunc> whichever those bonds you break you always have the same radical <pause dur="0.7"/> now if you <pause dur="0.4"/> take cyclohexane and irradiate it and i've just put <pause dur="0.2"/><kinesic desc="indicates point on transparency" iterated="n"/> this is the symbol by the way that sort of # <pause dur="0.7"/> lightning looking symbol <pause dur="0.2"/> # with the little <trunc>gam</trunc> gamma above <pause dur="0.4"/> that's the # <pause dur="1.3"/> symbol for <pause dur="0.4"/> high energy radiation ionizing radiation <pause dur="0.4"/> # i put gamma <pause dur="0.2"/> i could have put alpha if it was alpha radiation <pause dur="0.5"/> and you get <pause dur="0.7"/> three things happening you get the cation <pause dur="1.1"/> and again

that's a <pause dur="0.2"/> a radical cation <pause dur="0.2"/> initially so i'll <kinesic desc="writes on transparency" iterated="n"/> perhaps i'll put a dot there just to remind you <pause dur="0.6"/> the electron <pause dur="0.8"/> and the C-six-H-twelve-star <pause dur="0.7"/> now we can measure <kinesic desc="indicates point on transparency" iterated="n"/> this by scavenging <pause dur="0.2"/> at low temperatures <pause dur="0.3"/> we can measure <kinesic desc="indicates point on transparency" iterated="n"/> that as well by scavenging at low temperatures <pause dur="0.6"/><kinesic desc="reveals covered part of transparency" iterated="n"/> what if you put anthracene in the system <pause dur="0.8"/> what <pause dur="0.4"/> happens if anthracene is there <pause dur="1.2"/> right <pause dur="0.8"/> well i if i've got anthracene A <pause dur="0.7"/> what happens is this <pause dur="0.5"/> the electron goes onto the anthracene to give anthracene-minus <pause dur="2.0"/>

the hole <pause dur="0.9"/> present in the C-six-H-twelve-plus-dot <pause dur="0.4"/> goes onto the anthracene <pause dur="0.3"/> to give anthracene-plus <pause dur="1.3"/> it's much <pause dur="0.2"/> easier to ionize anthracene than it is to <trunc>anthrace</trunc> to ionize cyclohexane <pause dur="0.3"/> so thermodynamically <pause dur="0.4"/> the positive charge will migrate <pause dur="0.3"/> from <kinesic desc="indicates point on transparency" iterated="n"/> this molecule <pause dur="0.2"/> onto the anthracene <pause dur="0.3"/> to give anthracene-plus <pause dur="0.7"/> any excited states <pause dur="0.8"/> will be of very high energy 'cause they're going to be sigma sigma star states you remember there are no pi-electrons there <pause dur="0.4"/> but there are excited states formed <pause dur="0.7"/> and they give <pause dur="0.4"/> the excitation to the anthracene <pause dur="0.4"/> to give singlet anthracene-plus <pause dur="0.6"/> and triplet anthracene <pause dur="0.9"/> meanwhile <pause dur="0.6"/> # <kinesic desc="reveals covered part of transparency" iterated="n"/> what can also happen in the system <pause dur="0.2"/> is the anthracene-plus <pause dur="0.7"/> and the anthracene-minus <pause dur="0.3"/> if you don't track them <pause dur="0.5"/> at a low temperature but if you have them at room temperature <pause dur="0.8"/><kinesic desc="indicates point on transparency" iterated="n"/> these will find each other <pause dur="1.1"/> and annihilate each other in other words the plus will destroy the minus to give excited states <pause dur="0.6"/> so you'll get yet more singlet anthracene <pause dur="0.2"/> and triplet

anthracene <pause dur="0.8"/> the singlet anthracene <pause dur="0.3"/> will emit <pause dur="0.5"/> fluorescence <pause dur="0.2"/> so you can see fluorescence <pause dur="0.3"/> coming out of the system <pause dur="0.4"/> and you can measure it <pause dur="2.1"/> it is the case <pause dur="1.5"/> that if you take <pause dur="1.6"/> a solvent such as cyclohexane <pause dur="0.3"/> or benzene or toluene <pause dur="0.3"/> or <pause dur="0.2"/> some other solvents <pause dur="0.6"/> and put into the solvents <pause dur="0.8"/> a molecule that's a good fluorescing agent <pause dur="1.0"/> and you expose it to radiation <pause dur="0.8"/> it will emit <pause dur="0.4"/> light it emits fluorescence <pause dur="0.4"/> so you're not putting any light in <pause dur="0.6"/> you're only putting ionizing radiation into this system <pause dur="0.7"/> and you're getting fluorescence out <pause dur="0.6"/> and this is the basis of many detectors <pause dur="0.4"/> and this is a a a a a very slight digression <pause dur="0.4"/> but <pause dur="0.7"/> very many of the most important experiments that have been done in <pause dur="0.5"/> <trunc>cos</trunc> in radiochemistry <pause dur="0.3"/> but also astrophysics have relied on this <kinesic desc="indicates point on transparency" iterated="n"/><pause dur="0.6"/> because <pause dur="1.5"/> if you <pause dur="0.4"/> think about things like cosmic rays that are constantly pulsing through the earth <pause dur="0.6"/> they're pulsing through you as well <pause dur="1.5"/> are there ways of detecting them <pause dur="0.4"/> well if you take <pause dur="0.4"/> # <pause dur="0.2"/> either a plastic block <pause dur="0.3"/> loaded with anthracene <pause dur="0.5"/> or maybe a solution

of toluene <pause dur="0.3"/> with some anthracene in it <pause dur="0.6"/> and you take it down to the bottom of a coal shaft <pause dur="1.2"/> you can actually see it fluorescing <pause dur="0.4"/> and it's fluorescing <pause dur="0.3"/> because of cosmic rays <pause dur="0.6"/> passing right through the earth <pause dur="0.2"/> and passing through that <pause dur="1.2"/> sample <pause dur="0.5"/> and <trunc>f</trunc> giving you light <pause dur="0.4"/> so we are all the time <pause dur="0.4"/> under constant irradiation <pause dur="0.4"/> from <pause dur="0.4"/> outside the solar system <pause dur="0.7"/> and you can detect that <pause dur="0.4"/> using these <pause dur="0.3"/> systems <pause dur="1.4"/> and these <trunc>th</trunc> these systems <pause dur="0.3"/> # <pause dur="0.3"/> they have a special name in in that technology <pause dur="0.5"/> they're called liquid scintillators <pause dur="0.5"/> so if i write if i put in brackets under here <kinesic desc="writes on transparency" iterated="y" dur="10"/> liquid scintillators <pause dur="9.1"/> they're a a very good way of measuring <pause dur="0.3"/> ionizing radiation <pause dur="1.2"/> you can use liquid ones and also you could there are <trunc>s</trunc> there are crystal ones as well <kinesic desc="reveals covered part of transparency" iterated="n"/> there are there's some some very nice crystalline scintillators that you can make solid state ones <pause dur="0.4"/> and they are used again for measuring radiation <pause dur="0.6"/> if there's been a leakage or something you go you can either go round with a gigacounter <pause dur="0.2"/> which is one possibility <pause dur="0.4"/> or you can

go round with a solid state scintillator <pause dur="0.2"/> and you measure <pause dur="0.6"/> the # <pause dur="2.3"/> the scintillation spectrum <pause dur="0.6"/> okay <pause dur="0.3"/> well if you <trunc>de</trunc> # <pause dur="0.2"/> that's that was a bit of an aside but i thought i'd mention it in case # <pause dur="0.4"/> you sort of read about it sometime and wondered how it all linked up with this <pause dur="1.0"/> the <pause dur="0.3"/> G value that's a hundred electron volt yield of singlet anthracene is about two <pause dur="0.9"/> and the yield of triplet anthracene <pause dur="0.4"/> measured by pulse radiolysis this time <pause dur="0.8"/> is also about two <pause dur="0.3"/> so you get quite high excited state yields <pause dur="0.4"/> from <pause dur="0.2"/> radiolysis <pause dur="0.4"/> of something like <pause dur="0.7"/> cyclohexane <pause dur="0.5"/> hexane <pause dur="0.6"/> # dioxan <pause dur="0.2"/> benzene and toluene <pause dur="0.2"/> and all those sorts of solvents <pause dur="0.8"/> high yields from organic solvents <event desc="takes off transparency" iterated="n"/> particularly <pause dur="1.4"/> if there are no <pause dur="0.4"/> # <pause dur="0.3"/> organic radicals there sorry if there are <trunc>n</trunc> sorry if there are no hydroxy groups there they they that disturbs things but i i'll talk about that later as well <pause dur="2.8"/> that's using <pause dur="0.8"/> spectroscopic methods <pause dur="1.7"/> this detection of excited states <pause dur="0.6"/> we've <pause dur="0.9"/> looked at the fluorescence emission <pause dur="0.6"/> from the anthracene say or whatever else you were

using <pause dur="0.8"/> you can look at the triplet states 'cause they won't emit at room temperature 'cause you know about that anyway from photochemistry <pause dur="0.3"/> but you can see them in <trunc>p</trunc> # by you doing triplet triplet <trunc>abdorp</trunc> triplet triplet absorption <pause dur="0.5"/> as in pulse radiolysis <pause dur="0.5"/> what about chemical effects <pause dur="0.4"/> well you can also <kinesic desc="puts on transparency" iterated="n"/> induce <pause dur="1.1"/> cis-trans isomerization <pause dur="0.8"/> i mentioned in <trunc>t</trunc> <pause dur="0.2"/> # <trunc>w</trunc> in the photochemistry lectures that # <pause dur="0.4"/> one of the things that # you can do <pause dur="0.2"/> is to make molecules <pause dur="0.6"/> isomerize by shining light on them <pause dur="0.7"/> but you can also do it with radiation <pause dur="0.5"/> and if you take <pause dur="0.2"/> liquid benzene <pause dur="0.9"/> and add <pause dur="0.4"/> a small amount of <pause dur="0.3"/> cis-butene <pause dur="0.4"/> which is what that is <pause dur="0.8"/> and gamma irradiate it <pause dur="1.2"/><kinesic desc="reveals covered part of transparency" iterated="n"/> it turns into a mixture of cis and trans-butene <pause dur="0.5"/> so you can bring about isomerization <pause dur="1.0"/> what how does this happen <pause dur="0.4"/> well <pause dur="0.5"/> <trunc>i</trunc> <pause dur="0.6"/> from the previous overhead you ought to have a good idea <pause dur="0.6"/> but i'll i'll write it all i'll put it down again <pause dur="0.3"/><kinesic desc="reveals covered part of transparency" iterated="n"/> to indicate <pause dur="1.1"/> got <pause dur="0.5"/> this is benzene <pause dur="0.9"/><kinesic desc="indicates point on transparency" iterated="n"/> benzene <pause dur="0.4"/> undergoes radiolysis <pause dur="0.9"/> and this is a perfectly general equation you see you've got it for

cyclohexane you've now got it for benzene <pause dur="0.6"/> it applies to every molecule there is <pause dur="0.4"/> so <pause dur="0.8"/> you know if you were given <pause dur="0.2"/> say a question <pause dur="0.4"/> in which you were given some molecule you hadn't come across before <pause dur="0.4"/> and asked what happens if you radiolyse this <pause dur="1.0"/> the first thing to write down is M-plus <pause dur="0.5"/> minus an M-star where the molecule is M <pause dur="0.5"/> then you've got everything that can happen <pause dur="1.1"/> in the spur <pause dur="0.2"/> <trunc>i</trunc> along the track <pause dur="1.5"/> well this is <kinesic desc="indicates point on transparency" iterated="n"/> benzene-plus minus benzene excited state benzene <trunc>exci</trunc> excited state is singlet and triplet <pause dur="0.5"/> you get them both formed <pause dur="0.5"/> and what happens in particular <pause dur="0.7"/><kinesic desc="reveals covered part of transparency" iterated="n"/> is that triplet benzene <pause dur="0.8"/> will react <pause dur="0.3"/> with this <pause dur="0.5"/><kinesic desc="indicates point on transparency" iterated="n"/> and it will transfer its energy <pause dur="0.3"/> so we've got energy transfer <pause dur="0.4"/> to give excited <pause dur="0.7"/> cis-butene <pause dur="0.5"/> plus ground state benzene so i could maybe <pause dur="0.3"/> just <kinesic desc="writes on transparency" iterated="y" dur="8"/> write in ground state benzene just to remind you that the <pause dur="0.7"/> you've got energy transfer so i if i put E-N <pause dur="0.9"/> energy transfer <pause dur="0.4"/> there <pause dur="0.9"/> and this excited state falls back to the <kinesic desc="reveals covered part of transparency" iterated="n"/> ground state <pause dur="0.5"/> and it falls back into a mixture of cis and trans <pause dur="0.8"/> and of course <pause dur="0.8"/> # <pause dur="0.9"/>

<kinesic desc="indicates point on transparency" iterated="n"/> this has another go then this can go round the circuit again and get converted into trans <pause dur="0.5"/> the isomerization yield <pause dur="1.0"/><kinesic desc="reveals covered part of transparency" iterated="n"/> is two-point-three <pause dur="0.5"/> you convert two-point-three molecules of cis-butene <pause dur="0.3"/> for every <pause dur="0.3"/> hundred electron volts <pause dur="0.5"/> of <pause dur="0.7"/> energy that falls on there <pause dur="0.5"/> and if that <pause dur="0.7"/> if it's a case there's a fifty-fifty chance of falling <kinesic desc="indicates point on transparency" iterated="n"/> this way <kinesic desc="indicates point on transparency" iterated="n"/> or that way <pause dur="0.5"/> it means <pause dur="0.4"/><kinesic desc="reveals covered part of transparency" iterated="n"/> the excited state yield <pause dur="0.6"/> the triplet state yield <pause dur="0.4"/> of the cis-butene is in fact twice that it's four-point-six <pause dur="0.5"/> 'cause half of them <pause dur="0.4"/> go <kinesic desc="indicates point on transparency" iterated="n"/> this way <pause dur="0.4"/> and <kinesic desc="indicates point on transparency" iterated="n"/> half of them go that way <pause dur="0.7"/> all right <pause dur="0.9"/> so <pause dur="0.3"/> that <pause dur="0.3"/> is <pause dur="0.6"/> an indication <pause dur="0.2"/> of how you can use a scavenger <pause dur="0.8"/> and this time it's the cis-butene <pause dur="0.3"/> to register <pause dur="0.4"/> what's going on in the benzene <pause dur="0.4"/> you rely on the scavenger <pause dur="0.4"/> picking up <pause dur="0.5"/> the intermediates <pause dur="1.5"/> of course if you <pause dur="0.6"/> were as a person were undergoing <pause dur="0.4"/> gamma radiolysis <pause dur="1.7"/> in the water in your body you would get <trunc>positi</trunc> you'd get <pause dur="0.4"/> <trunc>ion</trunc> ionization <pause dur="1.8"/> and you get the intermediates formed from that <pause dur="0.2"/> and then your D-N-A in your proteins acts as the scavengers <pause dur="0.3"/> so <pause dur="0.2"/><kinesic desc="indicates point on transparency" iterated="n"/> this picture i've

given you <pause dur="0.3"/> of scavengers <pause dur="0.2"/> small amounts of <trunc>s</trunc> materials <pause dur="0.4"/> in a <unclear>bulk</unclear> solvent system <pause dur="0.5"/> is relevant to radiobiology <pause dur="1.2"/> because in radiobiology <pause dur="0.4"/> eighty per cent of you is water eighty per cent <pause dur="0.3"/> and so if you <pause dur="0.4"/> get <pause dur="0.3"/> radiation effects in water <pause dur="0.3"/> these will <pause dur="0.2"/> transmit themselves <pause dur="0.5"/> to the building blocks <pause dur="0.3"/> of your cells <pause dur="0.2"/> okay but <pause dur="0.3"/> that's for a later lecture but i mention it now <pause dur="0.5"/> this picture <pause dur="0.2"/> carries all the way through this notion of scavenging <pause dur="1.1"/> now <pause dur="0.2"/> i've talked about the <pause dur="1.1"/> capturing the <trunc>n</trunc> the <pause dur="0.3"/> electrons <pause dur="0.8"/> i've talked about capturing the excited states <pause dur="0.7"/> the last thing i want to talk about is capturing the positive ions <pause dur="0.3"/> how do we know <pause dur="0.5"/> well well we there must be positive ions there 'cause we see the negative ions <pause dur="0.4"/> and you can't have negative ions without positive ions but it would be nice to see them as well <pause dur="0.5"/><kinesic desc="reveals covered part of transparency" iterated="n"/> and what you can do <pause dur="0.5"/> is this <pause dur="1.8"/> you can take <pause dur="1.5"/> a <pause dur="1.2"/> a molecule <pause dur="1.9"/> which we know from electrochemistry <pause dur="1.4"/> is very easily oxidized <pause dur="0.8"/> to give <pause dur="2.1"/> a highly coloured <pause dur="0.3"/> species <pause dur="0.8"/> and a nice example <pause dur="0.6"/> is <pause dur="1.2"/> triphenylamine <pause dur="0.3"/> nitrogen

with three phenol groups on <pause dur="0.7"/> this is a pale <pause dur="0.4"/> pale yellow material <pause dur="0.5"/> crystalline <pause dur="0.5"/> if you dissolve it in <pause dur="0.5"/> cyclohexane <pause dur="0.4"/> and submit it to pulse radiolysis <pause dur="0.5"/> you get a brilliant colour <pause dur="0.8"/> just for a short time <pause dur="0.5"/> but you can measure that <pause dur="0.3"/> and it's known <pause dur="0.5"/> from electrochemical oxidation <pause dur="0.2"/> of triphenylamine <pause dur="0.7"/> so <pause dur="0.2"/> # <trunc>i</trunc> # it's a known species <pause dur="0.3"/> the extinction coefficient's known as well <pause dur="0.5"/><kinesic desc="reveals covered part of transparency" iterated="n"/> and what's happening <pause dur="0.4"/> is you're getting this this # <pause dur="0.5"/> you're getting C-six-H-twelve-plus <pause dur="0.4"/> with an odd little <pause dur="0.2"/> little radical dot <pause dur="0.8"/> formed <kinesic desc="indicates point on transparency" iterated="n"/> that's your solvent cation radical <pause dur="0.8"/><kinesic desc="indicates point on transparency" iterated="n"/> this is your <pause dur="0.3"/> amine <pause dur="0.9"/><kinesic desc="indicates point on transparency" iterated="n"/> this gives up an electron far more readily than <kinesic desc="indicates point on transparency" iterated="n"/> this does <pause dur="0.2"/> so the electron <pause dur="0.3"/> passes from <kinesic desc="indicates point on transparency" iterated="n"/> here <pause dur="0.4"/> onto <kinesic desc="indicates point on transparency" iterated="n"/> there to neutralize it and give you back the solvent <pause dur="0.6"/><kinesic desc="indicates point on transparency" iterated="n"/> this becomes <pause dur="0.8"/> the cation <pause dur="0.8"/> and you can measure these cations <pause dur="0.6"/> # <pause dur="0.7"/> by pulse radiolysis that's what # P-R stands for <pause dur="0.5"/> at room temperature <pause dur="0.4"/> or you can do it at liquid nitrogen temperature as well <pause dur="0.3"/> and you get the same coloration there it's a blue-green colour <pause dur="0.9"/> so this is called a positive ion scavenger <pause dur="0.4"/> and

it's been chosen <pause dur="0.3"/> because it's got a very low ionization energy <pause dur="0.3"/> and it's marvellous <pause dur="0.3"/> at # picking up positive ions in that sort of way <pause dur="1.2"/> okay <pause dur="0.4"/> so those are <pause dur="0.9"/> all <pause dur="1.4"/> experiments <pause dur="0.7"/> either at low temperature <pause dur="0.7"/> in some cases <pause dur="0.5"/> matrix isolation as i've called it before <pause dur="0.5"/> or room temperature <pause dur="0.8"/> to measure <pause dur="0.6"/> cations <pause dur="0.4"/> here <pause dur="0.5"/> excited states there <pause dur="0.2"/> and electrons on the previous overhead <pause dur="0.6"/><event desc="takes off transparency" iterated="n"/> so that is how <pause dur="0.3"/> we actually get at <pause dur="0.5"/> the <pause dur="0.2"/> the yields <trunc>o</trunc> of materials <pause dur="2.1"/> now <pause dur="0.5"/> # <pause dur="2.5"/> i'm going to skip <pause dur="0.2"/> the scavenger equation 'cause i'm going to come back to that later on <pause dur="1.3"/> what i'd like to do now <pause dur="0.2"/> is to say a little bit about water <pause dur="1.1"/> water is probably the most important system looked at <pause dur="0.5"/> mainly because as i said <pause dur="0.4"/> we are <pause dur="0.6"/> we consist largely of water <pause dur="0.8"/> so there's huge interest <pause dur="0.3"/> there must must be over a thousand papers on the radiation chemistry of water and probably half a dozen books as well <pause dur="1.0"/> # <pause dur="0.3"/> what happens <pause dur="0.2"/> with water <pause dur="1.2"/> well <pause dur="0.2"/> in a way it's quite interesting <pause dur="0.5"/> if you take <pause dur="0.4"/> a vessel <pause dur="1.1"/> and you <trunc>fi</trunc> you fill it almost to the top <pause dur="1.4"/> put a stopper in <pause dur="2.6"/> # <pause dur="0.3"/>

take this vessel of pure water completely full <pause dur="2.5"/> and if you gamma irradiate it <pause dur="1.8"/> for days <pause dur="1.9"/> and then you do a a big analysis of it <pause dur="1.2"/> what happens <pause dur="0.4"/> and the answer is <pause dur="1.3"/> almost nothing <pause dur="1.1"/> water appears <pause dur="0.2"/> to survive <pause dur="1.3"/> radiolysis <pause dur="0.6"/> well that's a very strange sort of thing with hydrocarbons <pause dur="0.6"/> you get all sorts of products you get dimers you get hydrogen you get goodness knows what <pause dur="0.4"/> with methanol you get ethylene glycol <pause dur="0.6"/> plus hydrogen <pause dur="0.5"/> but water <pause dur="0.6"/> is extraordinary <pause dur="0.5"/> it's <trunc>a</trunc> it appears to be amazingly radiation resistant <pause dur="1.1"/> and so <pause dur="0.2"/> for quite a while <pause dur="0.8"/> certainly <pause dur="0.2"/> in in the earlier parts in the earlier <trunc>y</trunc> <trunc>y</trunc> years of this century <pause dur="0.5"/> people thought water was very boring <pause dur="0.5"/> because it didn't seem to do anything <pause dur="0.7"/> # <pause dur="0.9"/> but then <pause dur="0.7"/> people got used to got used to this this idea of using scavengers and they began to look at water <pause dur="0.6"/><kinesic desc="puts on transparency" iterated="n"/> with <pause dur="0.2"/> scavengers <pause dur="1.2"/> so they they took water <pause dur="0.7"/> and added scavengers <pause dur="0.5"/> so the first point to note is water on its own <pause dur="0.4"/> appears <pause dur="0.5"/> not to do anything <pause dur="0.9"/> but if you put in scavengers <pause dur="0.4"/> and the two i've chosen

are iron-two and cerium-four <pause dur="0.6"/> iron-two is a reducing agent <pause dur="0.7"/> cerium-four is an oxidizing agent <pause dur="1.2"/> what do we think we get in <pause dur="1.8"/><kinesic desc="reveals covered part of transparency" iterated="n"/> well let us suppose <pause dur="0.4"/> and course i i know i know the answer to this but let us suppose <pause dur="0.2"/> that water does indeed undergo radiolysis <pause dur="0.5"/> to give an oxidizing species <pause dur="0.5"/> and a reducing species <pause dur="0.5"/> but let's also suppose <pause dur="0.3"/> that <kinesic desc="writes on transparency" iterated="y" dur="1"/> they are extremely good <pause dur="0.5"/> at going back again to water <pause dur="0.9"/> right <pause dur="1.6"/> let's <pause dur="0.4"/><kinesic desc="reveals covered part of transparency" iterated="n"/> look at it in more detail now what <pause dur="0.2"/> what <pause dur="0.5"/> from the experience with cyclohexane and benzene what <trunc>mi</trunc> might we expect to see <pause dur="1.0"/> well water <pause dur="0.4"/> we might have thought would ionize <pause dur="0.3"/> so we write H-two-O-plus and E-minus <pause dur="0.7"/> and H-two-O-star <pause dur="0.8"/> that would be a very good starting point <pause dur="0.4"/> what happens to those things <pause dur="0.8"/> well the first thing that happens <pause dur="0.9"/><kinesic desc="reveals covered part of transparency" iterated="n"/> # <pause dur="0.8"/> and i want to concentrate on <kinesic desc="indicates point on transparency" iterated="n"/> this side of the picture <pause dur="0.4"/> so perhaps i'll <pause dur="0.5"/><event desc="covers part of transparency" iterated="n"/> cover that up for the moment <pause dur="1.1"/> H-two-O-plus is known in mass spectrometry you can you can ionize water in a mass spectrometer <pause dur="0.5"/> and what's known is a very very fast reaction <pause dur="0.3"/> between H-two-O-plus <pause dur="0.2"/> and

water <pause dur="0.3"/> to give H-three-O-plus and O-H radical <pause dur="1.4"/> so <pause dur="0.2"/> we might have thought well <pause dur="0.2"/> perhaps we're getting some O-H radical <pause dur="2.5"/> what about the electrons <pause dur="0.6"/><kinesic desc="reveals covered part of transparency" iterated="n"/> well <pause dur="0.8"/> # <pause dur="0.7"/> an electron <pause dur="1.3"/> the kind of electron that goes along a wire in a torch or in a T-V set <pause dur="0.5"/> or to this overhead projector <pause dur="0.5"/> if you put an electron in water <pause dur="1.2"/> it becomes solvated <pause dur="2.7"/> i don't know what you know about <pause dur="0.2"/> solutions of sodium in ammonia or potassium in ammonia but if you take ammonia <pause dur="0.5"/> and you dissolve potassium or sodium <pause dur="0.4"/> have you done that have you done an experiment in the lab <pause dur="1.1"/> with <pause dur="0.2"/> sodium in ammonia have you ever had liquid ammonia as a reagent <pause dur="1.9"/> maybe not <pause dur="0.8"/> well if you take ammonia <pause dur="0.8"/> it's a liquid <pause dur="0.6"/> it # boils at minus-thirty-three but it's got quite a high <pause dur="1.0"/> latent heat of vaporization so it <trunc>s</trunc> hangs around for quite a while before it all evaporates <pause dur="0.3"/> it's around for really quite a long time <pause dur="0.6"/> if you put sodium in ammonia <pause dur="0.2"/> it dissolves and gives you a deep blue solution <pause dur="1.1"/> and these deep blue solutions are <trunc>p</trunc> are <trunc>p</trunc> highly magnetic <pause dur="0.7"/> and basically you've got N-A-plus <pause dur="1.0"/>

and free electrons <pause dur="0.4"/> in the ammonia <pause dur="0.9"/> and those free electrons are solvated by the ammonia molecule which has got an electron <pause dur="0.2"/> with say six ammonias round it <pause dur="0.4"/> well here <pause dur="0.3"/><kinesic desc="indicates point on transparency" iterated="n"/> you got an electron with six waters round it that's called the aquated electron or the hydrated electron <pause dur="0.6"/><kinesic desc="reveals covered part of transparency" iterated="n"/> okay <pause dur="0.5"/> and you can <trunc>s</trunc> you can see that that is there <pause dur="0.2"/> in pulse radiolysis <pause dur="0.6"/><kinesic desc="reveals covered part of transparency" iterated="n"/> if you just pulse radiolyse pure water <pause dur="0.5"/> you get <pause dur="0.2"/> a band at seven-hundred <pause dur="0.5"/> and this thing <pause dur="0.5"/> of course <pause dur="0.4"/> i mentioned before <pause dur="0.5"/> in the photochemistry section of the course <pause dur="0.2"/> if you flash photolyse <pause dur="0.8"/> iodide ion <pause dur="0.8"/> or ferrocyanide ion <pause dur="0.3"/> again you get <kinesic desc="indicates point on transparency" iterated="n"/> this broad band <pause dur="0.4"/> peaking at seven-hundred <pause dur="0.2"/> that's right out in the red so it's a blue colour <pause dur="1.2"/> you see this <pause dur="0.3"/> and you do indeed get electrons <pause dur="0.5"/> the O-H radicals <pause dur="0.7"/> well <pause dur="0.7"/><event desc="covers part of transparency" iterated="n"/> i mentioned cerium-four and iron-three <pause dur="1.4"/> you take water <pause dur="0.4"/> and radiolyse it with iron-two there <pause dur="1.0"/> the iron-two becomes iron-three <pause dur="1.4"/> if you take water with cerium-four there <pause dur="0.8"/> the cerium-four becomes reduced to cerium-three <pause dur="1.5"/> if we put <kinesic desc="indicates point on transparency" iterated="n"/> these two

scavengers in <pause dur="1.6"/> we get oxidation <pause dur="0.5"/> of this <pause dur="0.2"/><kinesic desc="indicates point on transparency" iterated="n"/> and <kinesic desc="indicates point on transparency" iterated="n"/> reduction of that <pause dur="0.6"/> that was the first hint <pause dur="0.3"/> that there were oxidizing species in water <pause dur="0.7"/> these are very classical type experiments using those things <pause dur="0.8"/> <trunc>e</trunc> but <kinesic desc="reveals covered part of transparency" iterated="n"/> in the pulse radiolysis <pause dur="0.5"/> we can do this <pause dur="0.3"/> and we can see the O-H radicals <pause dur="0.5"/><kinesic desc="indicates point on transparency" iterated="n"/> here <pause dur="0.3"/> which are formed in <kinesic desc="indicates point on transparency" iterated="n"/> that step there <pause dur="0.5"/> they attack various things <pause dur="0.3"/> they attack <pause dur="0.5"/> benzene <pause dur="0.2"/> <trunc>s</trunc> <pause dur="0.2"/> add on to benzene to give <kinesic desc="indicates point on transparency" iterated="n"/> this thing <pause dur="0.5"/> and that has got a known optical absorption spectrum this is you can <trunc>s</trunc> you can make this thing photochemically <pause dur="0.6"/> so we we know the <trunc>o</trunc> O-H is adding to benzene <pause dur="0.4"/> to give <kinesic desc="indicates point on transparency" iterated="n"/> this <pause dur="0.5"/> but we also know that O-H if we have a different experiment <pause dur="0.6"/> it attacks <pause dur="0.2"/> various things <pause dur="0.4"/> and if we have <pause dur="0.2"/> sodium thiocyanate <pause dur="0.8"/> # <pause dur="0.4"/> N-A-<pause dur="0.2"/>S-C-N-<pause dur="0.5"/>minus it's a scavenger <pause dur="1.0"/> then <pause dur="0.4"/> we get electron transfer reaction electron transfer <pause dur="0.4"/> we get S-C-N-dot in other words the O-H pulls the electron off there <pause dur="0.3"/><kinesic desc="indicates point on transparency" iterated="n"/> to give O-H-minus hydroxide <pause dur="0.4"/> which then reproteinates to water <pause dur="0.7"/> meanwhile <kinesic desc="indicates point on transparency" iterated="n"/> this has formed S-C-N <pause dur="0.4"/> and this actually loves <pause dur="0.2"/> to add to an S-C-N-minus

to give <kinesic desc="reveals covered part of transparency" iterated="n"/>this thing <pause dur="0.5"/> and that is very highly coloured <pause dur="0.3"/> and you can see it in pulse radiolysis <pause dur="0.9"/> so in the pulse radiolysis experiment <pause dur="0.2"/> without the scavenger <pause dur="0.2"/> you get this <pause dur="0.7"/><kinesic desc="indicates point on transparency" iterated="n"/> with the scavenger <pause dur="0.3"/> you see this <pause dur="0.5"/><kinesic desc="indicates point on transparency" iterated="n"/> or you see <kinesic desc="indicates point on transparency" iterated="n"/> that <pause dur="0.8"/> and so <pause dur="0.5"/> although water <pause dur="0.4"/> appears <pause dur="0.6"/> not to give any <pause dur="0.2"/> reaction to radiation <pause dur="0.4"/> what's happening is you are getting <pause dur="0.4"/> yields of oxidizing species and reducing species <pause dur="0.5"/> the oxidizing species <pause dur="1.2"/> is in fact O-H <pause dur="0.6"/> and the reducing species <pause dur="0.4"/> is in fact E-minus <pause dur="0.5"/> and so <trunc>thi</trunc> this is basically <pause dur="0.3"/> the guts of water radiolysis <pause dur="0.4"/> i put H-two-O-star there and excited state of water <pause dur="0.6"/> are they very important <pause dur="0.7"/> well <pause dur="0.5"/> surprisingly <pause dur="0.2"/> perhaps <pause dur="0.5"/> they're not <pause dur="0.6"/> # you might have thought there might be a reasonable yield <pause dur="0.7"/> but in fact <pause dur="2.3"/> when you <pause dur="0.5"/> radiolyse water <pause dur="1.5"/> you <pause dur="0.7"/> form <kinesic desc="indicates point on transparency" iterated="n"/> these two in the spur <pause dur="0.9"/> and there's a big difference between water and hexane <pause dur="0.5"/> in hexane <pause dur="0.2"/> if you've got two charges a few angstroms away <pause dur="1.6"/> they have an enormous coulombic attraction <pause dur="0.7"/> because the <trunc>cou</trunc> the attraction between two charges is given by Coulomb's law <pause dur="0.3"/>

Q-one-Q-two <pause dur="0.3"/> over epsilon-R <pause dur="0.3"/> squared <pause dur="1.8"/> and <pause dur="1.3"/> <trunc>Q</trunc> <trunc>Q</trunc> Q-one and Q-two are both one <pause dur="1.0"/> it's the epsilon that counts the dielectric constant <pause dur="0.8"/> the dielectric constant or electric permittivity <pause dur="0.5"/> of hexane is about two <pause dur="1.0"/> has <trunc>any</trunc> does anyone have any idea what it is for water <pause dur="0.4"/> the dielectric constant of water or electric permittivity of water <pause dur="1.0"/> any <pause dur="0.3"/> any feel for that <pause dur="1.1"/> well it water's a highly dielectric medium <pause dur="0.3"/> and it the figure's about eighty <pause dur="0.8"/> so the <pause dur="0.2"/> force between the positive and <trunc>neg</trunc> negative charges <pause dur="1.0"/> in hexane <pause dur="0.3"/> is enormous <pause dur="0.3"/> because <pause dur="0.4"/> on the bottom line you've got epsilon being two <pause dur="0.3"/> but in water <pause dur="0.3"/> epsilon's eighty <pause dur="0.4"/> so the coulombic attraction is forty times weaker <pause dur="0.3"/> so there's a much better chance they'll get away <pause dur="0.5"/> and not recombine <pause dur="0.5"/> to give excited states <pause dur="0.6"/> and indeed <pause dur="0.5"/> the excited states of water <pause dur="0.6"/> # <pause dur="0.4"/> are purely dissociative if you excite water <pause dur="0.8"/> # with a very deep U-V quantum <pause dur="1.8"/> you get excited water <pause dur="0.3"/> but it breaks into <pause dur="0.3"/> hydrogen atoms and O-H radicals <pause dur="0.2"/> and so <pause dur="0.2"/> you do not have stable excited states of water <pause dur="0.3"/> if they are

formed at all <pause dur="0.2"/> they dissociate <pause dur="0.2"/> to O-H <pause dur="0.2"/> and hydrogen <pause dur="0.3"/> and immediately the hydrogen atom <pause dur="0.5"/> will form an electron <pause dur="0.3"/> because if you think about it <pause dur="4.1"/><kinesic desc="writes on board" iterated="y" dur="8"/> the solvated electron <pause dur="0.7"/> will with react with a proton <pause dur="0.5"/> to give you <pause dur="2.0"/> a hydrogen atom <pause dur="0.9"/> so you can convert electrons to hydrogen atoms <kinesic desc="writes on board" iterated="y" dur="9"/> that's the H atom <pause dur="2.2"/> that's the proton <pause dur="1.1"/> and that's the solvated electron <pause dur="2.6"/> and so <pause dur="0.3"/> these are the three very very simple species <pause dur="0.3"/> but you always knew <pause dur="0.5"/><kinesic desc="writes on board" iterated="y" dur="1"/><kinesic desc="indicates point on transparency" iterated="n"/> that the hydrogen atom <pause dur="0.2"/> was a proton plus an electron <pause dur="0.3"/> so it's not surprising <pause dur="0.4"/> that <kinesic desc="writes on board" iterated="y" dur="1"/> in water <pause dur="0.9"/> you actually get the hydrogen atom dissociating into <kinesic desc="indicates point on transparency" iterated="n"/> those two bits <pause dur="0.6"/> or they can recombine to give that <pause dur="0.2"/><kinesic desc="indicates point on transparency" iterated="n"/> and you've actually got <pause dur="0.3"/><kinesic desc="writes on board" iterated="y" dur="2"/> there is a P-K-A <pause dur="0.4"/> the hydrogen atom <pause dur="0.2"/> it's a <trunc>lit</trunc> it's a little tiny acid all in its own right <pause dur="0.8"/> well on that amazing thought <pause dur="0.5"/> i'll <pause dur="0.2"/> <shift feature="voice" new="laugh"/>i'll <shift feature="voice" new="normal"/>bring things to an end today <pause dur="0.5"/> so we've talked about <pause dur="1.2"/> radiolysis <pause dur="0.3"/> using scavengers of organic systems <pause dur="0.3"/> and i've finished up by talking a bit about water <pause dur="0.5"/> the big difference is the organic systems <pause dur="0.2"/> have excited states that's a very important part of their radiolysis <pause dur="0.8"/> water <pause dur="0.5"/> is entirely <pause dur="0.4"/> ionization <pause dur="0.3"/> to give <pause dur="0.2"/> O-H-dot <pause dur="0.4"/> and E-minus those are the two principal species <pause dur="0.3"/> and they dominate radiation biology <pause dur="0.9"/> okay

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