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 ?

Answer 1

You can do this two ways.
No/N = 2ⁿ
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.

Answer 2

1.0 mCi is the freakin answer.

Answer 3

Ln(Ao/A)=kt/2.30
ln(12/A)=0.0462*2.5/2.30
A=21.47