Good afternoo, I have been struggling with this problem for a few hours now and

Question

Good afternoo, I have been struggling with this problem for a few hours now and would really appreciate if someone could work it out clearly step by step and help me understand that would be very much appreciated

3. G+ 0/1 points I Previous Answers BUElemPhys1 29, P026 My Notes A supply of fluorodeoxyglucose (FDG) arrives at a positron emission tomography clinic at 8 am. At 4 pm, when you arrive at the clinic to get a PET scan, the activity level of the FDG has dropped significantly because of the 110 minute half-life of the radioactive fluorine. Considering equal masses of FDG at 8 am and 4 pm, by what factor has the activity level been reduced by 4 pm? Note that another way to ask the same question is to say that we're looking for the ratio of the initial activity to the activity at 4 pm. The level has been reduced by a factor of 7e-14 Additional Materials eBook

Explanation / Answer

Let the initial activity ( 8 AM) of FDG be Ao

Let the final activity (4 PM) of FDG be A

A = Aoe-t

A/Ao = e-t

t1/2 of FDG = 110 min

t = 4 pm - 8am = 8 hours = 8x60 = 480 min

= 0.693 / t1/2 = 0.693 / 110 = 0.0063 min-1

A/Ao = e-(0.00693 x 480)

A/Ao = e-(3.024)

A/Ao = 4.86x10-2

The level of the activity is reduced by a factor of 4.86x10-2.

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