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<title>Radiation and Photochemistry</title></titleStmt>

<publicationStmt><distributor>BASE and Oxford Text Archive</distributor>


<availability><p>The British Academic Spoken English (BASE) corpus was developed at the

Universities of Warwick and Reading, under the directorship of Hilary Nesi

(Centre for English Language Teacher Education, Warwick) and Paul Thompson

(Department of Applied Linguistics, Reading), with funding from BALEAP,

EURALEX, the British Academy and the Arts and Humanities Research Board. The

original recordings are held at the Universities of Warwick and Reading, and

at the Oxford Text Archive and may be consulted by bona fide researchers

upon written application to any of the holding bodies.

The BASE corpus is freely available to researchers who agree to the

following conditions:</p>

<p>1. The recordings and transcriptions should not be modified in any


<p>2. The recordings and transcriptions should be used for research purposes

only; they should not be reproduced in teaching materials</p>

<p>3. The recordings and transcriptions should not be reproduced in full for

a wider audience/readership, although researchers are free to quote short

passages of text (up to 200 running words from any given speech event)</p>

<p>4. The corpus developers should be informed of all presentations or

publications arising from analysis of the corpus</p><p>

Researchers should acknowledge their use of the corpus using the following

form of words:

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

funding from the Universities of Warwick and Reading, BALEAP, EURALEX, the

British Academy and the Arts and Humanities Research Board. </p></availability>




<recording dur="00:55:35" n="8640">


<respStmt><name>BASE team</name>



<langUsage><language id="en">English</language>



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<personGrp role="speakers" size="36"><p>number of speakers: 36</p></personGrp>





<item n="speechevent">Seminar</item>

<item n="acaddept">Chemistry</item>

<item n="acaddiv">ps</item>

<item n="partlevel">UG</item>

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





<u who="nm5033"> well thank you for coming along this morning # i put out a # another question sheet on Friday this was typed in a great hurry by <gap reason="name" extent="1 word"/> and # # she brought it into the lab and i checked it but i have very little budget for copy you see a couple of couple of embarrassing typos there but i'm sure you'll manage to overlook that # the idea of course is to help you i think prepare # in a way for the exams and also to test where you've got to in your understanding of some of the material that we've covered in in C-H-three-eight-seven and i thought the the the routine we used last term went quite well people seem to have made quite a serious shot of # doing some of these things so i thought we'd # do do this this this kind of thing again # in in the groups that we had last time so that you all can remember each other # also i should indicate that today we have a guest here this is this is <gap reason="name" extent="2 words"/> from the Centre of English Language Teaching and he will video part of this because he'll probably use

out of the whole hour we're here about two minutes as part of a educational film and it's really for students overseas to get some idea of what it's like to come to British university and what the classes are like and a year ago he videoed two lectures in this <trunc>pr</trunc> in this course # and out of the two hours of lecturing i think he actually got <trunc>fe</trunc> five minutes of different things so that # # please don't be shy you know <trunc>i</trunc> almost certainly anything you say won't be used and if it's outstandingly good then it will be used and you'll be beautifully happy okay if if you sort of you slip on something don't don't bother it it just won't be used at all # okay if anybody's got any real problems with that could they # they can indicate if they like okay anyone got any real problems with # with being videoed maybe and # your # efforts being recorded for students overseas it's they're not looking for accuracy of chemistry by the way it's just how how teaching is done in the U-K all right well if we maybe the best thing is to perhaps go to the sheet that i put out on Friday

and # </u><u who="sf5034"> this morning </u><u who="nm5033"> no it went it went out on Friday but you might have picked it up this morning but <gap reason="name" extent="1 word"/> put it out on Friday </u><u who="sm5035"> yesterday </u><u who="sf5034"> it wasn't in the pigeonholes on Friday or yesterday </u><u who="sm5035"> it was yesterday <gap reason="inaudible due to overlap" extent="1 sec"/> </u><u who="nm5033"> was it yesterday </u><u who="sm5035"> yesterday evening it was yesterday evening </u><u who="sf5036"> it <trunc>w</trunc> it was in <gap reason="inaudible" extent="1 sec"/> all nighter 'cause i picked on up <gap reason="inaudible" extent="1 sec"/> on Friday evening </u><u who="nm5033"> # ah well i'm sorry about that # </u><u who="sf5036"> so it must have gone on late <gap reason="inaudible" extent="1 sec"/></u><u who="nm5033"> yeah okay well never mind # you you you <gap reason="inaudible" extent="1 sec"/> you have had it and some of you have had a go at it i believe i've been told <gap reason="inaudible" extent="1 sec"/> groups have been working away # so # i notice that group A is here so who who who is the kind of spokesman for group A </u><u who="sf5037"> ooh yeah <gap reason="name" extent="1 word"/> </u><u who="nm5033"> isn't this <gap reason="inaudible" extent="1 sec"/> </u><u who="sm5038"> # no no that's not fair </u><u who="nm5033"> <gap reason="inaudible" extent="1 sec"/> i'm sorry </u><u who="sm5038"> # no don't <gap reason="inaudible" extent="1 sec"/> that's not fair they're going to pick on me now </u><u who="ss"> aah </u><u who="sm5039"> as if </u><u who="nm5033"> right why why why don't you share it with somebody else who else is in group A </u><u who="sm5040"> <gap reason="inaudible" extent="1 sec"/> </u><u who="sm5040"> ah you <trunc>s</trunc> come on </u><u who="sm5041"> can tell by the <gap reason="inaudible" extent="1 sec"/></u><u who="sf5042"> i think <gap reason="name" extent="1 word"/> should<gap reason="inaudible" extent="1 sec"/></u><u who="ss"> yeah </u><u who="nm5033"> okay </u><u who="sf5042"> 'cause at least he's looked at it </u><u who="nm5033"> okay so </u><u who="sm5038"> yeah </u><u who="sm5043"> well <gap reason="name" extent="1 word"/>'s <gap reason="inaudible" extent="3 secs"/> </u><u who="nm5033"> you haven't done anything at all oh okay do you do you want to have a crack at what you've done i'll <trunc>g</trunc> i'll i'll <trunc>g</trunc> i'll i'll give you a <gap reason="inaudible" extent="1 sec"/> here put one on there and # come on i'll give you some useful props okay as you go along do you do you want to come along <gap reason="inaudible" extent="1 sec"/> on the overhead and then <gap reason="inaudible" extent="1 sec"/> okay <gap reason="inaudible" extent="1 sec"/> start with that one there # right

if we could all simmer down a bit # yes in there's a much greater <gap reason="inaudible" extent="1 sec"/> how could you make the following conversions and you were given a a series of good reactions and the idea was there might be a chemical reagent would convert A to B or there might be a particular type of stimulation either laser light or light or U-V light or gamma rays or whatever # but # if if you <trunc>ha</trunc> have a crack at # one or two of these i'll i'll kind of # indicate how you're <trunc>ge</trunc> how you're getting along okay </u><u who="sm5038"> all right </u><u who="nm5033"> <gap reason="inaudible" extent="1 sec"/> well you there's we we seem to be slightly over the top i know let's let's bring that down a bit let's bring that okay all right let's E where's E-minus then these lights are very bright actually aren't they i shall i shall i turn the lights down slightly </u><u who="sm5038">

yeah </u><u who="nm5033"> # # okay # anyone got any <trunc>ob</trunc> any observations on that converting E-minus to O-H any any thoughts from anyone </u><u who="sm5044"> just add water </u><u who="nm5033"> sorry </u><u who="sm5044"> just add water </u><u who="nm5033"> yeah you you need <trunc>basi</trunc> yeah you certainly need water there # but <gap reason="inaudible" extent="1 sec"/> you generate E-minus with pulse radiolysis of water that that's absolutely true # if you actually allow the electrons to react with H-two-O-two so i think the H-two-O-two bit is right but # there there is another way of doing it as well so if if i could perhaps just come round here and # so if you've got H-two-O okay you've got the you've got the gamma ray or or the X-ray coming in and that will give you H-two-O-plus and E-minus so you've generated the E-minus there the question is how do you convert this through to O-H and as # as you rightly said you can use H-two-O-two because as the electron enters H-two-O-two it causes it to fall apart into two bits and there's the O-H radical that's the thing we're after the other there is another species

that's very very good at <trunc>c</trunc> at converting one into the other any offers on that well known in mass spectrometry or it's it's actually N-two-O nitrous oxide so laughing gas in fact what what happens there is that you get this <trunc>t</trunc> taking place and that simply picks up the proton from the water to give you O-H so in a sense # you were certainly right to pick on on H-two-0-two so that <trunc>th</trunc> that that was # <trunc>th</trunc> that was you know fine but the <trunc>f</trunc> the fullest picture is to <trunc>ge</trunc> is to do that you get the E-minus you get the E-minus either with with H-two-O-two as you said or with N-two-O and both of those routes take you through to to O-H-minus anyone got any problems with that okay well that's okay so so you you you you you were certainly <trunc>i</trunc> with H-two-O-two you were in the right kind of ballpark but # okay # oh i've managed to lose my sheet already so it's that's yours okay next # oh the <trunc>oth</trunc> the next one is going the other way how do you <trunc>co</trunc> how how can we possibly convert O-H back to E-minus any any offers on that did you have a thought on that one </u><u who="sf5045"> no i didn't </u><u who="nm5033">

you you # yeah at least yeah that was quite a tricky one # okay well i'll i'll give you i'll give you i'll give you the kind of the straight answer to that because it # sorry it's # getting in the way to get O-H through to E-minus that's the what's the what's the way if you're doing that well oddly enough and you you may have to think about this quite hard if you take <trunc>mole</trunc> if you take molecular hydrogen if you actually put hydrogen into water # <gap reason="inaudible" extent="1 sec"/> to atmospheric pressure or under under higher pressure what do people think O-H radical would do to to molecular hydrogen what does O-H radical love doing what's its favourite # pastime </u><u who="sm5044"> oxidizing </u><u who="sf5046"> <gap reason="inaudible" extent="1 sec"/> </u><u who="nm5033"> sorry </u><u who="sm5044"> oxidizing </u><u who="nm5033"> it <trunc>o</trunc> loves oxidizing so if it was going to oxidize H-two how would it oxidize it </u><u who="sf5047"> <gap reason="inaudible" extent="4 secs"/> </u><u who="nm5033"> it would attract yeah absolutely right it would <trunc>s</trunc> it it would abstract hydrogen so if you write <trunc>hy</trunc> if you write hydrogen like that what will happen is you'll get H-two-O plus a hydrogen atom okay that's the hydrogen atom and the hydrogen atom is linked to the electron in what way what what what <trunc>w</trunc> how do we link hydrogen with with # hydrogen atom with electrons </u><u who="sf5048"> it's a proton so </u><u who="nm5033"> yeah it's a <trunc>pro</trunc> <trunc>i</trunc> it that's right basically # if you actually have this in base O-H-minus you will get E-minus out of that in other words E-minus is related to if i write it more fully E-minus plus a proton gives you a hydrogen atom so this is the most powerful reducing agent there is that's a hydrogen atom which is the

simplest chemical # simplest atom that there is altogether and that's the simplest ion that there is and all three are interrelated by an acid-base equilibrium so there is in fact a P-K for the hydrogen atom actually measured the # change in the spectrum in pulse radiolysis by changing the P-H of the solution and you get E-minus at P-H seven and six and as you go down to at sort of P-H-two it's a complete loss including to the hydrogen atoms okay so that that that was that one there so that that was the the answer was just if you deluge the solution of hydrogen gas you will convert round </u><u who="sf5049"> could you put that one back on top please </u><u who="nm5033"> okay you want it back on right there there we are </u><u who="sf5049"> cheers </u><u who="nm5033">

yeah right while we're while we while you're furiously making notes we'll get another transparency out # oh the <trunc>ne</trunc> the next one you'll <gap reason="inaudible" extent="1 sec"/> </u><u who="sf5045"> i don't know </u><u who="nm5033"> <gap reason="inaudible" extent="1 sec"/> </u><u who="sf5045"> what do you think <gap reason="inaudible" extent="3 secs"/> </u><u who="nm5033"> <trunc>ba</trunc> <trunc>ba</trunc> <trunc>ba</trunc> <trunc>ba</trunc> basically you're being you're being asked to convert one thing into another and in fact well the third one is i've actually done it there because # if you look this the third problem the third problem is actually how do you convert electrons to hydrogen atoms and oddly enough that is the answer you simply # instead of pulse radiolyse water you pulse radiolyse dilute acid dilute sulphuric acid dilute H-C-L dilute nitric acid dilute anything really providing there's water there the water generates the minus and the E-minus will then get protonated to give the H atom so the third problem we've actually solved as it were en passant okay do you have any idea about the <trunc>ins</trunc> methyl chloride </u><u who="sf5045"> no </u><u who="nm5033"> no okay right <trunc>an</trunc> any offers on <trunc>w</trunc> on the methyl chloride one </u><u who="sf5050">

you just add radiation that would split from this <gap reason="inaudible" extent="1 sec"/> wouldn't it </u><u who="nm5033"> sorry the </u><u who="sf5050"> just that radiation would split <gap reason="inaudible" extent="1 sec"/> radical </u><u who="nm5033"> yeah <trunc>a</trunc> if you can have solid methyl chloride you probably would get # some some some # disruption like that so if i take that off there i'd be what what what # <gap reason="name" extent="1 word"/> says if <trunc>w</trunc> if we have a C-H-three-C-L and if it's that was a solid with a gamma ray we would get we would probably get something like this happening but in fact there is a a rather cleaner way of doing this # of splitting that up to give you C-H-three-dot <trunc>ins</trunc> in fact the other the <trunc>c</trunc> the thing this time is C-L-minus what what do we need to drive methyl chloride through to that into that process taking methyl chloride through to C-H-three C-L-minus what do you need to add to methyl chloride to push it over to the right what's the what's the missing missing bit </u><u who="sf5051"> an electron </u><u who="nm5033"> sorry </u><u who="sf5051"> an electron </u><u who="nm5033"> an electron yes yes an electron if you add if you put an electron if you add an electron to this as the electron enters

the methyl chloride system you're getting basically a very transient sort of species like this and it essentially it splits off there to give you C-H-three radical and C-L-minus so the conversion there was to to add add one electron to the system and whenever you've got # any kind of chloro organic anything with a C-C-L bond <gap reason="inaudible" extent="1 sec"/> chloride or chlorobenzene # adding electrons always gives the C-C-L bond <trunc>tha</trunc> that that is a a stronger <trunc>exo</trunc> # odonic or exothermic reaction # and you end up by sticking up this very very stable species and you're left behind with with with with with a C-H-three okay # so that that was # was methyl chloride # </u><u who="sf5050"> do you think </u><u who="nm5033"> any any thoughts on the next one </u><u who="sf5045"> # photo induced electron transfer </u><u who="nm5033"> yeah okay electron transfer do you do you want to # you want do you want to write it out then # you can see so we # did # did group A have a chance to talk together before they came not really </u><u who="sf5051"> no </u><u who="nm5033"> no </u><u who="sf5052"> we only <gap reason="inaudible" extent="1 sec"/> </u><u who="nm5033"> okay

so what we've got a conversion now which is platinum platinum-four going down to platinum-three right so but the question is what is the reagent that pushes it over # to to do that we're </u><u who="sf5045"> we need one electron </u><u who="nm5033"> you okay you want one electron that's basically it really if you pulse radiolyse you know you're right there if you pulse radiolyse aqueous platinum-four # you generate electrons and they will reduce platinum-four to platinum-three so you get this <trunc>un</trunc> you get this very unusual oxidation state of platinum and electrons # can produce all sorts of # strange species any offers on other strange species that electrons can produce that # what what unusual oxidation states can you get with electrons for example any <trunc>on</trunc> any offers from anyone </u><u who="sf5050"> do you get a peroxide <gap reason="inaudible" extent="1 sec"/> </u><u who="nm5033"> sorry </u><u who="sf5050"> # peroxides in <gap reason="inaudible" extent="1 sec"/> you've got <gap reason="inaudible" extent="1 sec"/> </u><u who="nm5033"> # i'm <trunc>re</trunc> well i'm thinking about metals <trunc>me</trunc> <trunc>me</trunc> metal metal oxidation states any any # </u><u who="sf5050">

iron-one </u><u who="nm5033"> iron yeah iron-one yes that that that that's that's a good answer iron-one yes because if you take iron-two and react it with electrons you <trunc>c</trunc> you drive it down to iron-one and you can also drive manganese down two down to manganese-one # and zinc-two down to zinc-one <trunc>c</trunc> cobalt-two down to cobalt-one and silver-one you can drive down to silver-nought so the electron being the most powerful reducing agent in the whole of chemistry it has no problem in entering the metal orbitals and giving us very unusual oxidation states so certainly that was right that you add # you add a <trunc>wo</trunc> one electron change there # now the next one the next # problem is have we got some space on there a little bit yeah # did anyone have any thoughts about these conversions with with methanol we've got two methanols # # derived # reactions there how how well how do we convert methanol to methoxy any

<trunc>off</trunc> any offers from anyone there </u><u who="sf5050"> <gap reason="inaudible" extent="1 sec"/> a hydroxyl radical </u><u who="nm5033"> well <trunc>ac</trunc> well actually it's it's not a it's not a bad answer a <trunc>hy</trunc> <trunc>hydroc</trunc> hydroxyl radical is a very powerful species a very powerful oxyl species it will attract methanol but in fact O-H will pull off the hydrogen from the carbon you get C-H-two-O-H so to get methoxy # which is a itself a little bit like hydroxy it's a very very high energy species very very reactive so if we had some <trunc>meth</trunc> if we had some methonal in # a container how would we go about converting it to # # methoxy any any offers from anyone what what sort of thing do you need don't forget so far all the questions have been about reagents and <gap reason="inaudible due to overlap" extent="1 sec"/> </u><u who="sm5038"> <gap reason="inaudible due to overlap" extent="1 sec"/> </u><u who="nm5033"> some yeah almost yeah but what not quite you need a bit more energy than that so what's the next highest energy along </u><u who="sm5038">

pulse radiolysis </u><u who="nm5033"> sorry </u><u who="sm5038"> radiolysis </u><u who="nm5033"> radiolysis yeah if you <trunc>re</trunc> if you radiolyse so if we write down # methanol as a radiolytic symbol we can either use an X-ray or or a gamma ray # it appears the <trunc>f</trunc> the first process we get is C-H-three-O-H-plus # plus an electron and that then undergoes a very fast acid-base reaction to give us C-H-three-O-H-two-plus that's just ordinary protonated methanol and you get that thing formed there so this this the radiolysis takes us through to # to methoxy and # that was thought thought to be </u><u who="sm5053"> Professor could you just stand to the side slightly <gap reason="inaudible" extent="1 sec"/></u><u who="nm5033"> sorry i was just standing in the standing in the way was i okay yeah so <trunc>th</trunc> there's the there's there's methanol going through to to methoxy radical # and you might have thought well that that's the end of the story but what do you think the methoxy might do what what what why do you think the methoxy might # do something i said it was

extremely reactive so what what what what what are its possibilities </u><u who="sf5054"> recombine with H at source </u><u who="nm5033"> well i've said it was very much like hydroxy and hydroxy we heard earlier on is is extremely good at # </u><u who="sm5038"> <gap reason="inaudible" extent="1 sec"/> </u><u who="nm5033"> taking hydrogen atoms off things so so what will what will the methoxy do </u><u who="sm5038"> <gap reason="inaudible" extent="1 sec"/> take the proton on one of the methanols </u><u who="sm5038"> yeah it will take it will take a hydrogen atom <trunc>ou</trunc> out of one of the methanols so the <trunc>ne</trunc> the next stage is that this thing will come along it will attack another methanol and it will pull one of these off so you end up by getting C-H-three-O-H and this thing will then give you C-H-two-O-H like that so the primary <trunc>pro</trunc> products are E-minus and C-H-three-O and C-H-three-O turns around attacks another nearby methanol a neighbouring one to give that and that even occurs at temperatures of fifty kelvin so that even if you have # extremely low temperatures you will get that process occurring and you can monitor

it at fifty kelvin by E-S-R spectroscopy </u><u who="sf5055"> why is it taking it off the carbon more than off the <trunc>o</trunc> oxygen </u><u who="nm5033"> what why is it taken off the carbon not <trunc>o</trunc> <trunc>o</trunc> not off off off the oxygen whoops let's push that up a bit well the <trunc>s</trunc> in fact the the C-H bond # is very slightly weaker than the than the O-H bond # so serodynamically # there's a lot going for this to go through this conversion here you get you actually get release of heat it's simply bond energies O-H is stronger than C-H # you might wonder well why do you get C-H-three-O in the first place shall i put a dot there to just to emphasize the <gap reason="inaudible" extent="1 sec"/> cation the the main reason there is that # # in a gamma ray or an X-ray you've got all the energy in the world that you want you've released an electron from here you've left that thing behind or you've you've forced out an electron because of the the sheer amount of energy and # the deprotonation of this thing # the proton # comes off the most acidic point in molecule so

you we'll get C-H-three-O and not C-H-two-O-H that's simply the acidity of hydrogen bonded to oxygen so here it's driven by acidity you're getting <trunc>g</trunc> in getting from here to here but you're getting from there down to there 'cause it's driven by thermodynamics because the C-H bond energy is # lower than the <trunc>O</trunc> the O-H bond energy so that radical reaction # releases a lot of heat therefore there's a driving force for it to go okay yeah # and then the the next question # which in fact we've answered already that's one two three four five <trunc>nu</trunc> number one two three four five six the seventh one we've just answered to convert this through to through to that you just let it react with methanol so it's a secondary reaction so we've really ready to finish group A and the last one is # actually it was a misprint there was <trunc>a</trunc> there was R so if you didn't a little R # <trunc>chr</trunc> chromium and the thing on the right that little L should be a big L but otherwise # it's fine # so any any thoughts on that one did you how do you get from quartet to doublet </u><u who="sf5055">

# well it's intersystem converting <gap reason="inaudible" extent="1 sec"/></u><u who="nm5033"> it's intersystem crossing that would be <trunc>g</trunc> it's so # that that is the actual conversion so # in fact in a sense you could say you don't actually need to do anything at all it just happens spontaneously but # i suppose a more sophisticated answer would be to say that if you <trunc>st</trunc> if you have the quartets a very complex <trunc>p</trunc> ground state you need to excite it with # could be with visible light actually 'cause chromium being a green compound all chrome compounds are green # it will absorb in the visible # and be promoted into a higher quartet state which then crosses spontaneously to <trunc>d</trunc> doublet state so i think the the perfect answer would be # visible light stimulation to generate the initial # chromium complex and <trunc>th</trunc> and then a crossover so # i need i need another <gap reason="inaudible" extent="1 sec"/> transparency okay so # <gap reason="inaudible" extent="1 sec"/> # you can you go and sit down now 'cause you've thanks very much indeed # very briefly saying that # the the ground state of the chromium complex is a quartet so if i put quartet

there chromium there's the excited state which is a quartet as well and the doublet state is over here and essentially you you you put you use visible light to get up to there and then essentially you get intersystem crossing to take you over to the doublet state which is then the one which the which actually phosphoresces okay so that was the that was the picture that # would would have been as it were a a very complete answer with all of these questions of course # i you want to look at the time you've got you want to say well if there are <trunc>s</trunc> if i've got forty minutes and there's seven parts you know seven sixes are forty-two <trunc>s</trunc> so i've got just under six minutes five minutes to answer each part and you want to make sure you don't spend fifteen minutes answering one little bit because you know it will simply # you can get you might get sort of five out of six but you're not going to get eight out of six # so you know you want to kind of ration your time so that you don't use it # # unsuccessfully okay well that that well that was # question one that was done

by group A what what other groups are well represented here # apart from what what what what have we got # <gap reason="inaudible" extent="1 sec"/> this group over here </u><u who="sm5056"> we're group B </u><u who="nm5033"> yes and and you've got some worked answers </u><u who="sm5056"> # </u><u who="nm5033"> yes </u><u who="sm5056"> yes some </u><u who="nm5033"> yes okay do you want to come up they're not on overheads are they just on paper </u><u who="sm5056"> all right just ordinary paper </u><u who="nm5033"> <gap reason="inaudible" extent="1 sec"/> up </u><u who="sm5056"> <gap reason="inaudible" extent="1 sec"/> </u><u who="nm5033"> okay right do you want to come up oh you've got them on a okay </u><u who="sm5056"> i've written it all down i just hope you can read it </u><u who="nm5033"> right that's excellent </u><u who="sm5056"> <gap reason="inaudible due to overlap" extent="1 sec"/> </u><u who="nm5033"> <gap reason="name" extent="1 word"/> do you want to come up and show them to us all </u><u who="sm5056"> <gap reason="inaudible" extent="1 sec"/> </u><u who="nm5033"> oh oh i think </u><u who="sm5056"> <gap reason="inaudible" extent="2 secs"/> </u><u who="nm5033"> i think # i i can't see you as as a # a shrinking violet so okay <trunc>whi</trunc> which which which question was this </u><u who="sm5056"> <gap reason="inaudible" extent="1 sec"/> D </u><u who="nm5033"> <gap reason="inaudible" extent="1 sec"/> D </u><u who="sm5056"> yeah </u><u who="nm5033"> okay </u><u who="sm5056"> <gap reason="inaudible" extent="1 sec"/> </u><u who="nm5033"> so we've got some we got some answers here and do you do you want to take us through them okay do you want to do you want to speak to </u><u who="sm5056"> <gap reason="inaudible" extent="1 sec"/> </u><u who="nm5033"> there's a question there then we'll go on to then talk to your overhead </u><u who="sm5056"> # well it's # it's self-explanatory of course <gap reason="inaudible" extent="1 sec"/> <gap reason="inaudible" extent="1 sec"/>

</u><u who="nm5033"> well the the first question was explain why ruthenium <trunc>c</trunc> # the ruthenium complex with three bipyridines two <gap reason="inaudible" extent="1 sec"/> both are a better oxidant and a better reductant when you excite it why why is it that you've got better oxidizing properties and better reducing properties at the same time it seems almost illogical that that could be the case and so we were looking for a # a kind of explanation for this and we've got an answer # which says ruthenium excitation moves moves the well we've got various <gap reason="inaudible" extent="1 sec"/> conduction band i mean the # the <trunc>f</trunc> the first thing to say is <gap reason="inaudible" extent="1 sec"/> is that i there is some confusion here # this is this is a complex which is a kind of discrete molecule yeah it's it's not a semiconductor semiconductors tend to be # things like metal oxides metal sulphides # binary compounds like that which are crystals only a crystal can be a semiconductor you can't have the single molecule <trunc>be</trunc> <trunc>be</trunc> </u><u who="sm5057"> we are talking about the T-I-O-2 i mean talk about the <trunc>semi</trunc> semiconductor promotes the electron </u><u who="nm5033">

oh </u><u who="sm5057"> it's the electron forms the semiconductor isn't it that's promoted <gap reason="inaudible due to overlap" extent="1 sec"/></u><u who="nm5033"> well the the well the question A was why why is the rithian complex a better oxidant and a better reductant and it's no mention of semiconductors okay </u><u who="sm5057"> oh right yeah </u><u who="nm5033"> yeah # so but i <trunc>n</trunc> i think there is some sort of # # okay you've got this strong oxidizing agent <gap reason="inaudible" extent="1 sec"/> when it gets to ruthenium-three okay you so you're talking about the excitation here </u><u who="sm5057"> yeah </u><u who="nm5033"> # if you excite this complex you basically push an electron from the ruthenium on to the ligand and so # in the excited state what you've got is essentially if i i i i'd have to i'd remove that for the moment in the excited state what you've got you've got R-U-bipy where the electron has gone from here onto one of these ligands okay so if i write bipy-two you have taken an electron out of the complex so you've basically got a ruthenium-three

there now and you've got an odd electron on one of the ligands okay so if you look at the excited state configuration like that clearly the ruthenium core has become a powerful oxidant and you can do exactly what's been written there but conversely the ligand has become semi reduced and this is a powerful reductant so you've taken a molecule where there's been a roughly equal charge distribution all over the place on excitation you've taken one electron from the centre and put it on the outside so that you have got a oxidizing core in the ruthenium ruthenium-three and you've got a very powerful reducing ligand bipy so that that would be # a kind of # microscopic explanation it looks at the in detail at the metal and it looks in detail at the ligand and once you've done that of course this can oxidize O-H-minus to oxygen this can <trunc>re</trunc> this bit can reduce protons down to hydrogen so that that that is one explanation but there is another completely different way of looking at it and # i don't know whether anyone's got any thoughts on a different a completely different way of looking at it

any any <trunc>an</trunc> # a much more general way i mean i gave that as a particular complex but <trunc>i</trunc> this would do for any <trunc>com</trunc> <trunc>i</trunc> in fact if almost any molecule so perhaps you've got a molecule and you've got a lowest a highest occupied molecular orbital and a lowest unoccupied molecular orbital okay now if you excite it with # a light quantum you push one electron up into here and you're left with one behind down there and so what you've got also down here is an empty hole so i could write that as H-plus and i could write that as E-minus so it's very much like the semiconductor situation but it is a single molecule okay we're talking about one we're talking about one one molecule so <trunc>i</trunc> if that's if you could represent the excited state like this # would anyone like to tell me why is it awhy is it a better reductant if you promote an electron into a into this upper orbital any offers why is it a better reductor </u><u who="sm5057"> 'cause it's got an electron and higher energy so it's <gap reason="inaudible" extent="1 sec"/> </u><u who="nm5033">

yeah yeah the electron that's absolutely right the electron is # in a <trunc>th</trunc> it's a it's a higher energy level it's very near the ionization level i mean you could imagine that somewhere at the top here off the <trunc>t</trunc> off the top is so called ionization level okay and if the electron had got as far as there it would have escaped altogether and you'd have had a free electron so it would be extremely powerful reducing now to get the electron to leave the molecule from here you've got to push it all the way from there up to here but to get it to go from here to there it only goes up that small distance and so you have got as as you say a much more pwerfully reducing species once it's excited 'cause the electron is much nearer the point at which you can leave the molecule so it's a much more powerful reducing agent okay now what about oxidizing properties why why is it a better oxidant any any offers from this side why is it a better oxidant now than it was when it started out any offers from group D what

what <trunc>a</trunc> what about # someone else why why you've explained why it's a better reductant why can you explain why it's become a better oxidant </u><u who="sm5058"> 'cause it's got a vacant hole </u><u who="nm5033"> it's got <trunc>ye</trunc> it's got a vacant hole in other words if if if you were actually # wanting this molecule to oxidize something suppose this is this is the ground state here so if i just sort of just remind you that's that's the ground state there this is the excited state here now for this for this ground state to oxidize something it's got to pull an electron out of it and the electron will go where if this puts up an electron where's it going to go what orbital yes it will go in here so for this to oxidize something it must pull an electron out of a substrate and it will tuck into into that orbital there okay right but if you if this excited state was going to oxidize something where would the electron go now </u><u who="ss"> <gap reason="inaudible, multiple speakers" extent="1 sec"/> </u><u who="nm5033"> it goes in the hole so you see this <trunc>na</trunc> this excited state is now a powerful oxidant because it can

put an electron into this hole with a much greater # energy advantage that put the electron up here if you can stuff it in there you're you're you're in you're in a very good # position so this charged separation you see it gives you added reducing power and added oxidizing power so that that is the kind of general answer # and for the ruthenium complex that's just one example out of a thousand that one could have one could have taken okay that's fine now # in <trunc>s</trunc> the second part of that question # what are the physical properties of # where's where's the # i've i've # put your answer down somewhere and temporarily lost it okay # okay ah yeah this is all about semiconductors isn't it in a way </u><u who="sm5056"> <gap reason="inaudible" extent="1 sec"/></u><u who="nm5033"> yeah or is it is it the ruthenium complex # maybe it's the ruthenium complex as well it probably applies to both # why <trunc>wha</trunc> the question was why what other physical properties has the ruthenium complex got that make it a target for so much research and the <trunc>res</trunc> the research is what what sort of research is it is it being used for

</u><u who="sf5059"> renewable energy source </u><u who="nm5033"> sorry </u><u who="sf5059"> renewable energy source <gap reason="inaudible" extent="1 sec"/></u><u who="nm5033"> renewable energy source yes it's used as a renewable energy source # it's # been # put in a number of target systems # one is photoelectrolytic cells that are driven by sunlight so you get perpetual electricity from a cell based on the ruthenium complex and the other one is splitting water into hydrogen and oxygen using the hydrogen as fuel so the big interest in this complex well we got # got some answers from # group D and # it's it's got a high quantum yield certainly the i mean the phosphorescence quantum yield which is the measurement of the number of excited states you've got is quite good # it's got a high turnover number # it's just likely to turn to dioxide in that way it's a ruthenium complex it # it can be used fairly exhaustively i mean you can use many thousands of times but ultimately it will degrade you will acually degrade the ligands # so it's not perfection but it's about as near as we've got to

perfection # and then finally it has really got a very good absorption spectrum it's an orange compound and it matches the solar spectrum fairly well so if you can imagine the solar spectrum at earth level it's looking a little bit like this where this is sort of # maybe about # a thousand nanometres at P infrared this is a visible that the <trunc>spe</trunc> the # spectrum of the ruthenium complex is actually going to be very much in the visible because it's orange and it will be looking a little bit like in fact it looks a bit like that so it's <trunc>n</trunc> it's not a bad match that's the overlap region that's the bit we're interested in where the sunlight spectrum at earth level matches the absorption spectrum <trunc>o</trunc> of the ruthenium complex now the <trunc>l</trunc> the last question was a bit nasty # how can ruthenium complexes be # tethered to # to <trunc>semicondu</trunc> # semiconductor services # <trunc>ano</trunc> another typo leaped out at me # how can you tether them to surfaces and then it says why why should you want to but let's let's hear it how you can tether them to start with if you want

to tether a complex to a polar surface what's the best thing to do to it any <trunc>t</trunc> any suggestions after i've got say suppose you've got a a <trunc>g</trunc> a silica surface or an alumina surface or a titania surface and you want to get a metal complex to lock on to that surface and stick # really hard ruthenia bypyridium won't be awfully good at doing this it will just wash off but so how what what do you want to do to the ligand to make it bind to a to to a surface </u><u who="sf5050"> <gap reason="inaudible" extent="1 sec"/> </u><u who="nm5033"> yeah # yeah put a polar group on it # is a suggestion that is that is a good suggestion actually # what we would want to do i'll pick yet another one of oh that's <gap reason="inaudible" extent="1 sec"/> i'll do it here if your surface is down here so basically it's a metal oxide surface <trunc>M</trunc> i'll call it M-O you've got on the surface you've got O-two-minus ions all the way down and you've got M-two-plus ions say it's suppose it's # a binary # # i'll i'll put it as M-two-plus if you've got a ligand say by <trunc>p</trunc> the ruthenium complex with a ligand and if you've with if you put on the ligand a carboxy group or a sulphinate group or a phosphate group so up here put two-minus if you have

any of these groups on the ligand periphery they will ligate strongly to the to the M-two-plus species that <trunc>i</trunc> you that <trunc>i</trunc> you you'll get a a strong # <gap reason="inaudible" extent="1 sec"/> interaction between the # this # group on the surface and the metal okay like so you will be getting that sort of interaction so the answer is you modify the ligand to put on a polar group you take a hydrogen atom and bang in a carboxy or a sulphinate or something like that okay so that was the right answer there that's how you can do it now why would you want this ruthenium complex or something like it why would you want it on # on a semiconductor surface anyway what would you want to do that any offers why should why should you want to do it </u><u who="sf5060"> if you can do it on a semiconductive surface you can get electrons into it easier with the semiconductor sites than using electrons from </u><u who="nm5033"> yeah </u><u who="sf5060"> one or the other to <gap reason="inaudible" extent="1 sec"/></u><u who="nm5033"> yeah that that that's absolutely right </u><u who="sf5060"> mm </u><u who="nm5033"> # one of the problems with semiconductors is that they

often don't have a very good match with the <trunc>so</trunc> with the solar spectrum they have a U-V absorption many of them the <gap reason="inaudible" extent="1 sec"/> O-two in particular and there's not a lot there's really not a lot of U-V in sunlight i mean it's enough to make you get sunburn but there's really in energy terms only a <trunc>s</trunc> only a few per cent of the total solar output <trunc>i</trunc> reaching earth that's actually U-V so what would be nice to try and get better devices would be to get a better matching of the # action spectrum of your semiconductor to sunlight and what you can do is to coat the semiconductor with a monolayer of one of these complexes because you will then get the matching of the spectrum of the tethered complex to sunlight you get a very good match because you can use one of these ruthenium complexes or you can use a <gap reason="inaudible" extent="1 sec"/> complex or various other ones that have been tried out and so <trunc>y</trunc> the reason why you want to do it is to improve the action spectrum and the efficiency of the semiconductor towards sunlight # and of course you need a monolayer because as the sun as the sunlight falls onto the ruthenium

complex what you're hoping will happen is that it will then inject an electron into the conduction band of the semiconductor so you activate the semiconductor using light the semiconductor would normally completely disregard it wouldn't <trunc>abs</trunc> you know the band gap would be too big but you're able to overcome that by having this complex on there <trunc>o</trunc> okay <trunc>a</trunc> any have you all got the that idea yeah okay # right so that was a question for group D and then we had # <gap reason="inaudible" extent="1 sec"/> we won't have the time <gap reason="inaudible" extent="1 sec"/> we've still got quite a bit of time left # group B is group B about is group have you got any sort of consolidated answers do you want to come out and have a shot okay come come and just come and tell us what you've got it's # what # what are the products on in irradiating these compounds okay right so as you can # speak to it <gap reason="inaudible" extent="1 sec"/> quite well prepared okay right what are the products you'd expect to find on radiolysis of and there it there are three compounds okay so if you # want to # take us through it </u><u who="sm5061"> yeah # take it well yeah # i just took it to be products and then worked out a mechanism and then #

how do you prove the mechanism is looking at the intermediates it's E-S-R # <gap reason="inaudible" extent="1 sec"/> straight from the notes </u><u who="nm5033"> yes okay it it was <trunc>ba</trunc> basically very much a textbook answer and you've got a set of products and then you've you've got a mechanism there and you've relied basically on on E-S-R # what what <trunc>o</trunc> what <trunc>o</trunc> i mean what other methods were i mean E-S-R has told us a bit about the mechanism because it has identified the intermediate radical but # what what other physical methods would you <trunc>l</trunc> use for studying glycine suppose you had some crystalline glycine and you went away and give it # maybe on its own or indeed # in <trunc>a</trunc> aqueous solution # it says aqueous glycine so suppose you've got some glycine in water and you gamma irradiate it or X-radiate it how do you follow the reaction what what what what physical techniques would you use to follow that how would we know we've got those things that are on the board what would be the favoured method

</u><u who="sf5062"> <gap reason="inaudible" extent="1 sec"/></u><u who="nm5033"> well no i i'm really thinking of actual products we've got H-two N-H-three C-H C-O-two and formaldehyde <trunc>e</trunc> N-M-R yeah you you would get in in the aqueous solution you would actually get the formaldehyde protons coming up all you'd get the formaldehyde # C-thirteen # what about the gases how would you know they were there what what what what's the favoured method for going after gases gas gaseous products would you say </u><u who="ss"> <gap reason="inaudible, multiple speakers" extent="1 sec"/></u><u who="nm5033"> sorry </u><u who="ss"> <gap reason="inaudible, multiple speakers" extent="1 sec"/></u><u who="nm5033"> you <trunc>g</trunc> well you can use G # G-O-C gas <gap reason="inaudible" extent="1 sec"/> # that would be quite good or you can actually combine it with mass spectrometry G-C-M-S # an awful lot of systems which have been subject to either photolysis or radiolysis or indeed # microbial action if you want to look at the

gases coming off G-C-M-S is is much the best way because these are very # microelectronic species hydrogen ammonia C-O-two and they they show up extremely well # on with with with G-C-M-S you get really quite good separation okay so we've got E-S-R looking at the intermediates and we reduce G-C-M-S for looking at the products okay so that that's quite good how about frozen ethanol did you <gap reason="inaudible" extent="1 sec"/> frozen ethanol okay do you want to do frozen ethanol </u><u who="sm5063"> it must it's we assumed it would be the same as methanol <gap reason="inaudible" extent="1 sec"/> </u><u who="nm5033"> yeah not a bad not a bad assumption </u><u who="sm5063"> # yeah got frozen methanol ethanol # # you it's it's <trunc>ra</trunc> radiolysed to get to this stuff and then there's various routes it takes ending up with hydrogen # this whatever that is <gap reason="inaudible" extent="1 sec"/> thing and that which it is ethanol # to prove this you can find the intermediates by E-<trunc>S</trunc> E-S-R again that's how it would show up but with the other products i said could be done with normal means like <gap reason="inaudible" extent="1 sec"/> or N-M-R </u><u who="nm5033"> okay any any observations from anyone on that we could start just

start with <gap reason="inaudible" extent="1 sec"/> you've got # <gap reason="inaudible" extent="1 sec"/> you've got the intermediates # forms you've got # okay # yeah you've got C-two-H-five-O here which # is # an alkoxy radical # and then you've got this reacting ultimate going down here we have to give # yeah okay you've got down here we have to give # # yes # yeah okay # # you've got this conversion here you've got an alcoxy through to a # this radical here # that's it's quite a good quite a good answer but it's not exactly right # i'll explain why # suppose you've got ethanol and you're going to pull a hydrogen atom off one of the carbons there # the way that our speaker had it was he pulled off this one and got C-H-two-C-H-two-O-H # in fact # the C-H bond strength of that C-H bond is very slightly less than the C-H bond strength of the methyl and so <trunc>a</trunc> actually what you do get out of this is # when you get the the alcoxy coming down there's the there's the alcoxy when this attacks that it gives you C-H-three-C-H-dot-O-H and this will then of course what will

that do once it's been formed well you've got you've got it's got dimerization here which is # quite a good answer that would dimerize and so basically that would go through to C-H-O-three-C-H-O-H C-H-three-C-H-O-H but you probably got two-thirds of the marks for that i think that's really you know quite a i mean you've done methanol you've got to be able to jump from one a molecule that you know about to one that you've not seen before but there is enough similarity to carry the ideas over you've got to be able to extemporize a bit in in doing some of these answers # yeah so in fact if you looked at the E-S-R of that and you do get it # you've got four protons actually which are very very close together so you you actually end up by getting a quintet it's like four protons in N-M-R four equivalent protons in N-M-R gives you a one-four-six-four-one quintet and you get a one-four-six-four-one quintet in the E-S-R spectrum just just as you would in the N-M-R spectrum for # a molecule with with four equivalent protons but # so going talking about

E-S-R was the right thing # not <trunc>qu</trunc> not quite the white right not quite the right radicals but then the <trunc>ana</trunc> <trunc>anal</trunc> analytical methods that you mentioned are really absolutely the right ones okay well thank you for that that was # really quite reasonable and then finally crystalline D-N-A did someone have crystalline D-N-A do you want to come and talk about crystalline D-N-A briefly tell us about # what happens when you take <trunc>fi</trunc> fibrous crystalline D-N-A and subject it to to gamma gamma radiolysis what what are you going to get ah </u><u who="sm5061"> # i wasn't quite sure what you were asking because you've got loads of # information from the notes </u><u who="nm5033"> yes </u><u who="sm5061"> so just </u><u who="nm5033"> well it's not an aqueous solution it's <trunc>ac</trunc> just pure crystalline D-N-A so it's # if it's not an aqueous solution you don't get the indirect effect you must that only happens if you've got aqueous D-N-A aqueous D-N-A is important in <trunc>st</trunc> looking at at cells responsive cells because

our bodies are made up of cells that contain eighty ninety per cent water and so the indirect effect is very important with with cells in our body if you took fibrous D-N-A suppose you had a cell a <trunc>c</trunc> # you had some D-N-A in an environment without water within a cell and you took this crystalline D-N-A and you you damaged it what sort of things do you get well what have you got you've got functional groups on purines may be irreversibly altered well that's certainly true # extensive that they may be lost in the D-N-A molecule yeah you can actually split them out and lose them you get a radical a base salt that that <trunc>als</trunc> also encourages they break they break the sugar-phosphate backbone you get single strand breaks you get double strand breaks that's that's that's quite a good summary # i'd like to ask one question of of of the whole class actually # that that that's quite reasonable if you've got D-N-A and you're subjecting it to gamma radiolysis # so if i try and write the D-N-A in shorthand guanine thymine adenine thymine guanine adenine cytosine cytosine thymine adenine

guanine cytosine thymine adenine right that's so that's just one little section of a D-N-A molecule # and i'd like to ask the question where are where are the damaged sites if you've got that long string of # nuclear <gap reason="inaudible" extent="1 sec"/> put together and you're <trunc>s</trunc> submitting it to ionizing radiation you get charge you get ionization occuring okay you get charged separation and you get holes and electrons don't you holes and electrons so where where do the where do the # where do the electrons tend to go out of those four bases where do the electrons love to go more than anywhere else </u><u who="ss"> T </u><u who="nm5033"> they go to T yeah the <trunc>elec</trunc> the electrons will go onto them and the electrons will end up on T residue so you get T-dot-minus-A T-dot-minus-G-A C-C-T-dot-minus okay blah blah blah blah blah and the T-dot-minuses do you know what happens to them afterwards after after you get the electron on to the thymine what happens next to it any offers </u><u who="sf5064"> it goes on to the phosphate </u><u who="nm5033">

sorry it # picks up a proton picks up a proton and you end up now by getting essentially G-T-H with a dot A-T-H with a dot G-A-C-C-T-H with a dot blah blah blah blah blah okay well well that's the that's the electrons where do the holes go guanine guanine the holes go in the guanine so we end up by getting the holes so if i write H-plus here we'll end up by getting G-dot-plus T-A-T-G-dot-plus A-C-C et cetera et cetera et cetera yeah so that that's the that's the story there basically the thymine has the highest electron affinity # and the guanine has got the lowest ionization energy so you end up by getting that okay now # are we about to i've lost my question sheet A B <trunc>s</trunc> is group C </u><u who="sm5063"> how do you how do you prove that </u><u who="nm5033"> by E-S-R </u><u who="sm5063"> E-S-R </u><u who="nm5033"> yeah E-S-R will show up the timing radical which has got an optic spectrum don't ask me to explain why i <trunc>ca</trunc> i could explain why but it will take me about twenty minutes # you get an optic spectrum on with T and you the the guanine also has got catrius E-S-R in fact what you can do you can take the pure nucleotides and gamma irradiate and get and get an optic spectra for all

possible anions and cations and then match them to <trunc>d</trunc> D-N-A spectrum and you can and you can synthesize it from the components # by computer group C </u><u who="sf5060"> # we don't have anything prepared but </u><u who="nm5033"> okay do you want a quick # resume have you got a a i've got a transparency or did you want to speak to </u><u who="sf5060"> no no i haven't got anything </u><u who="nm5033"> you haven't got anything </u><u who="sf5060"> i literally i found this this morning <gap reason="inaudible" extent="1 sec"/></u><u who="nm5033"> you found it this morning have you got any thoughts about </u><u who="sf5060"> yeah phosphorescence is rare 'cause you get # they recombine </u><u who="nm5033"> you get two triplets recombining yeah </u><u who="sf5060"> <gap reason="inaudible due to overlap" extent="1 sec"/> it reacts with oxygen </u><u who="nm5033"> reacts with oxygen oxygen perturbs the excited state level yes </u><u who="sf5060"> and there's a third one </u><u who="sm5065"> <gap reason="inaudible" extent="1 sec"/></u><u who="sf5060"> <gap reason="inaudible" extent="1 sec"/> thank you <gap reason="inaudible" extent="1 sec"/> done that </u><u who="nm5033">

you get <gap reason="inaudible" extent="1 sec"/> and what about the solvents anything to do with solvent </u><u who="sf5060"> hydrogen extraction </u><u who="nm5033"> hydrogen extraction yeah that's very good so # you should be that you're you're obviously quite well aware of that # how does a small amount of naphtholene protect solvents from attack </u><u who="sf5060"> it takes the triplet state and absorbs the energy and moves it on delocalizes it round the <gap reason="inaudible" extent="1 sec"/></u><u who="nm5033"> okay that's that seems all right so what what what what we're told is that if you take a triplet and if i call it R-one-R-two-C-double-bond then there's our triplet ketone this will normally attack a solvent R-S-A to R-one-R-two-C-O-H-dot-S-S-H so you degrade the solvent if you put in naphtholene naphtholene C-ten-H-eight what happens is the triplet he migrates from there to there and we will now get ground

state ketone which is completely innocuous triplet naphthalene which does not attack solvents at all it hasn't got an odd electron on an oxygen so it's is perfectly harmless it just dissipates it as heat okay </u><u who="sm5066"> can you move it up a bit </u><u who="nm5033"> sorry </u><u who="sm5066"> can you move it up up a bit </u><u who="nm5033"> oh move it up a bit sorry yeah there we are okay well we've # used up all our time # i think i'm due am i due to give you <trunc>t</trunc> yet another revision class </u><u who="ss"> yes </u><u who="nm5033"> yes okay well i will i will contrive to produce another sheet of questions and we'll we'll <trunc>d</trunc> we'll do this again but i'll i'll next time i'll give you a bit more notice it's mainly because i suppose <gap reason="name" extent="1 word"/>'s been off and i've really been doing half of his job as well as my own but he's back now <gap reason="name" extent="2 words"/> is back he's fully fully recovered okay thanks very much then