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# chemistry

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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

#2
+5

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
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#1
+17655
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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.

geno3141  Nov 22, 2015
#2
+5

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
+91045
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
edited by Melody  Nov 23, 2015
#4
+5

Not me Melody.

Neither of them, whatever the other one was.

Bertie

Guest Nov 23, 2015
#5
+91045
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

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