Bridget Jones Posted June 22, 2004 Posted June 22, 2004 Hi Please could anyone advise the best way to show 7 successions of alpha decay on: 234 U 92 I know the answer is 206 Hg 80 Should I show it as 7 different stages eg. stage 1 234 230 4 U ---- Th + He 92 90 2 Stage 2 230 226 4 Th ------ Ra + He 90 88 2 etc etc or can I simplify it as : 234 206 4 U ------- Hg + 7( He ) 92 80 2 Hope this makes sense, I would be most grateful for any advice. Thank you
Dave Posted June 22, 2004 Posted June 22, 2004 I think I'd show it in seperate stages rather than combine it all into one humongous equation.
Bridget Jones Posted June 22, 2004 Author Posted June 22, 2004 Thanks for that Dave. This is the first time I've used a forum so format of my post not that good, guess I have to work on that one!
Dave Posted June 22, 2004 Posted June 22, 2004 Easiest way is to use LaTeX: [math]\left._{92}^{234} \mbox{U} \right.[/math] Like I said in the other thread
Bridget Jones Posted June 22, 2004 Author Posted June 22, 2004 So let me get this right if I was for example expressing flourine, I should write: {9}^{19} F ?
swansont Posted June 22, 2004 Posted June 22, 2004 Hi Please could anyone advise the best way to show 7 successions of alpha decay on: 234 U 92 I know the answer is 206 Hg 80 Shouldn't 7 alpha decays be [math]^{206}_{78}\mbox{Pt}[/math] ? I realize this may just be an exercise in writing out the steps' date=' but there's no way that this result is physically real. [math']^{214}_{82}\mbox{Pb}[/math] only beta decays. You have to include that, along with some other (unlikely) branching, to arrive at [math]^{206}_{80}\mbox{Hg}[/math]
Bridget Jones Posted June 22, 2004 Author Posted June 22, 2004 Shouldn't 7 alpha decays be [math]^{206}_{78}\mbox{Pt}[/math] ? I realize this may just be an exercise in writing out the steps' date=' but there's no way that this result is physically real. [math']^{214}_{82}\mbox{Pb}[/math] only beta decays. You have to include that, along with some other (unlikely) branching, to arrive at [math]^{206}_{80}\mbox{Hg}[/math] Yes you're right, thanks. Glad it was hypothetical and not an actual answer to an assignment!
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