chemistry

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!.

A formula for exponential growth and decay is:  A  =  A₀e^kt

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 = A₀e^kt

Dividing both sides by A₀:              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 = 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.

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

Not me Melody.

Neither of them, whatever the other one was.

Bertie

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. :))