Posted by **Matt** on Tuesday, July 20, 2010 at 2:04pm.

A sample of sodium-24 with an activity of 12 mCi is used to study the rate of blow flow in the circulatory system. If sodium -24 has a half-life 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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