radium has a half life of 1600 years. what percentage of a sample would be left after 2.07*10^3 years ?

Guest Nov 22, 2015

#1**+5 **

A formula for exponential growth and decay is: A = A_{0}e^{kt}

where A is the final amount, A_{0} is the initial amount, k is the proportionality constant, and t is the time.

For half life, A = 1/2A_{0}.

So, substituting into the original formula: 1/2A_{0} = A_{0}e^{kt}

Dividing both sides by A_{0}: 1/2 = e^{kt}

Since the time is 1600 years: 1/2 = e^{1600k}

Take the ln of both sides: ln(1/2) = ln(e^{1600k})

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 = A_{0}e^{-0.000433t}

Let's assume we start with an initial amount of 1 (100%) ---> A = 1e^{-0.000433t}

After 2.07 x 10^{3} years ---> A = e^{-0.00043(2070)}

So, the amount left is .408, or about 41% of the original amount.

geno3141 Nov 22, 2015

#2**+5 **

Best Answer

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!.

Guest Nov 22, 2015

#3**0 **

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

Melody Nov 23, 2015

#5**0 **

**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. :))

Melody Nov 23, 2015