1. half-life

I need help setting up this problem

The radioactive decay of Rubidium-87 into Strontium-87 has a half-life of 48.8 billion years, making it one of the most useful decay modes for obtaining the ages of very old igneous rock. A sample of some of the oldest igneous rock found on Earth was chemically analyzed and found to contain 53 atoms of Sr 87 for ever 947 atoms of Rb 87. Assuming that no Sr 87 was present when this sample solidified, how old is this rock?

2. Originally Posted by viet
I need help setting up this problem
Since one atom of Rb turns into one atom of Sr, there were initially 87+ 947= 1034 atoms of Rb. Now there are 947 atoms left.

Saying that it has a half life of 48.8 billion years means the amount is multiplied by 1/2 every 48.8 billion years. If T is the number of billions of years that have passed, then T/48.8 is the number of "half lives" and so that amount left after T billions of years is $A(1/2)^{T/48.8}$ where "A" is the initial amount. Your equation is $1034 (1/2)^{T/48.8}= 947$

Solve that for T.

3. please check me calculation.

$1034 (\frac12)^{T/48.8}= 947$
$(\frac12)^{T/48.8}=.91586$

$\frac {T}{48.8} = \frac {log(0.91586)}{log(\frac{1}{2})}$

$\frac {T}{48.8} = 0.12680$

$T = 6.1878$

4. The method is correct; however, because there were 53 atoms of Sr for every 947 atoms of Rb, that means the ratio of the current amount of Rb to the original amount of Rb is 947/1000, not 947/1034. This will provide you with a slightly different answer.

5. Originally Posted by icemanfan
The method is correct; however, because there were 53 atoms of Sr for every 947 atoms of Rb, that means the ratio of the current amount of Rb to the original amount of Rb is 947/1000, not 947/1034. This will provide you with a slightly different answer.
so the answer would be T = 3.83 billion years?

6. Originally Posted by viet
so the answer would be T = 3.83 billion years?
Yes, that is correct.