radium has a half life of 1600 years. what percentage of a sample would be left after 2.07*10^3 years ?
A formula for exponential growth and decay is: A = A0ekt
where A is the final amount, A0 is the initial amount, k is the proportionality constant, and t is the time.
For half life, A = 1/2A0.
So, substituting into the original formula: 1/2A0 = A0ekt
Dividing both sides by A0: 1/2 = ekt
Since the time is 1600 years: 1/2 = e1600k
Take the ln of both sides: ln(1/2) = ln(e1600k)
Inside an ln expression, the exponent comes out as a multiplier: ln(1/2) = 1600k ln(e)
ln(e) = 1 ln(1/2) = 1600k
Divide both sides by 1600 and simplify: k = -0.000433
The formula becomes: A = A0e-0.000433t
Let's assume we start with an initial amount of 1 (100%) ---> A = 1e-0.000433t
After 2.07 x 103 years ---> A = e-0.00043(2070)
So, the amount left is .408, or about 41% of the original amount.
radium has a half life of 1600 years. what percentage of a sample would be left after 2.07*10^3 years ?
2070/1600=1.29375 half-lives, therefore,
2^-1.29375=0.40789.....X 100=40.789% of the original amount left.
Geno3141: Please, always bear in mind Occam's razor!.
Thank you for those 2 excellent answers.
I would not have thought to do it our guests method but was much quicker that way.
I am with Geno, I usually take the scenic tour rather than the expressway :)
https://simple.wikipedia.org/wiki/Occam%27s_razor
Please guest, won't you identify yourself ? :)
This is the second occurance of a mysterious guest today.
The first time I thought it was Bertie.
This time I suspect it is Nauseated, but I am not sure either time.
Nauseated would usually identify himself so maybe it is Bertie both times ://
Or maybe it is someone else altogether :D
Hi Bertie,
Andre Massow was trying to find your access problem, there is a couple in you position but not many.
As far as I know he has not been able to reproduce the problem so I guess he hasn't solved it either. ://
It is nice to say hello to you though :D
The other one was an answer to a diophantine equation. :)
I suggested it just as a play question. :))