- #1
Eddzzz
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I thought this area was appropriate for this question from a previous exam paper which I just need checking. So:
Working out:
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.
Working out:
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.
