Posted by Matt on Tuesday, July 20, 2010 at 2:04pm.
A sample of sodium24 with an activity of 12 mCi is used to study the rate of blow flow in the circulatory system. If sodium 24 has a halflife of 15 hours, what is the activity of the sodium after 2.5 d ?

Chemistry  DrBob222, Tuesday, July 20, 2010 at 2:44pm
You can do this two ways.
No/N = 2^{n}
12/N = 2^{2.5*24/15}
12/N = 16
N = 12/16 = 0.75 mCi after 2.5 days.
Other way.
Calculate k from 0.69315/15 = 0.04621 and substitute into the following:
ln(No/N) = kt
No = 12
N = ??
k = from above.
t = 2.5 days in hours is 2.5 x 24 = 60
ln(12/N) = 0.69315/60
Solve for N = 0.75 mCi.

Chemistry  Anonymous, Saturday, January 28, 2012 at 2:44pm
1.0 mCi is the freakin answer.

Chemistry  Endium, Friday, August 7, 2015 at 11:17am
Ln(Ao/A)=kt/2.30
ln(12/A)=0.0462*2.5/2.30
A=21.47
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