Chemical Forums
Chemistry Forums for Students => Physical Chemistry Forum => Topic started by: Eddzzz_2012 on February 26, 2013, 07:46:31 AM
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2 isotopes 244Hb - 1st order decay, half life 10million years produce 240Hb
one atom of 244Hb takes 60sec in 1g of pure hibernium to decay to 240Hb
I need to calculate the natural abundance of 244Hb and also state any assumptions.
Any help on where to start would be appreciated! Thanks
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What is Hibernium? Never heard. Also the decay from 244 Hb to 240 Hb means it looses 4 neutrons, because the same element remaining. Strange decay.
You can calculate the 1g into moles and with the Avogadro constant to the amount of atoms. Every minute one atom converts. This you calculate to 10 miilion years, how many remaining.
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Every minute one atom converts. This you calculate to 10 miilion years, how many remaining.
That's a little bit risky idea - you are assuming activity won't change during this time. It doesn't have to be true in general.
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Right here's the actual question:
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So far I've worked out k, the mass, relative mass, volume for each and number of particles. I just have no idea how to get the natural abundance.
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Here's a new approach that I thought of. Can you check please?
The half-life of 244Hb is 10 million years.
The half-life of 244Hb is 5.2595×10^12 minutes.
The mean life of 244Hb is 7.58783677×10^12 minutes.
So if you had a sample of 7.58783677×10^12 atoms of 244Hb, you could expect about 1 decay a minute.
You have x grams of 244Hb and 1-x grams of 240Hb.
Thus you have about x/244 moles of 244Hb.
Thus you have about 2.468×10^21×x atoms of 244Hb.
Since this sample decays one atom per minute, we know 2.468×10^21×x = 7.58783677×10^12.
Or x = (1/ln 2)(10 million years / 1 minute) (244 grams/mole / ( N_A per mole * 1 gram ) = 3.07×10^-9
Natural abundance of 240Hb on a per atom basis is ( x / 244)/ ( ( x / 244) + ( (1-x) / 240) ) = 60 x/(61 - x). Why?
So the natural abundance of 240Hb is about 3.02×10^-9.
If there is no natural source of 244Hb, this implies that the sample of Hibernium is no more than about 283 million years old.
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The mean life of 244Hb is 7.58783677×10^12 minutes.
So if you had a sample of 7.58783677×10^12 atoms of 244Hb, you could expect about 1 decay a minute.
Are you sure about that? I must admit I have never used mean life, so I can't comment on its use, but my trials (based on two different methods) yield N0 of 5.25x1012 (which is in agreement with the approximate approach suggested by Hunter2).
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Here's a new approach that I thought of. Can you check please?
The half-life of 244Hb is 10 million years.
The half-life of 244Hb is 5.2595×10^12 minutes.
The mean life of 244Hb is 7.58783677×10^12 minutes.
So if you had a sample of 7.58783677×10^12 atoms of 244Hb, you could expect about 1 decay a minute.
You have x grams of 244Hb and 1-x grams of 240Hb.
Thus you have about x/244 moles of 244Hb.
Thus you have about 2.468×10^21×x atoms of 244Hb.
Since this sample decays one atom per minute, we know 2.468×10^21×x = 7.58783677×10^12.
Or x = (1/ln 2)(10 million years / 1 minute) (244 grams/mole / ( N_A per mole * 1 gram ) = 3.07×10^-9
Natural abundance of 240Hb on a per atom basis is ( x / 244)/ ( ( x / 244) + ( (1-x) / 240) ) = 60 x/(61 - x). Why?
So the natural abundance of 240Hb is about 3.02×10^-9.
If there is no natural source of 244Hb, this implies that the sample of Hibernium is no more than about 283 million years old.
Sadly, turns out it is not your original work, but something you have copied from other forum.
Topic locked.