Question

The iodine isotope 13153I is used in hospitals for diagnosis of thyroid function. If 772 μg are ingested by a patient.

I only need help for part c but I would appreciate it if you could show how you did parts a and b. :)

A. Determine the activity immediately. Use Appendix F in the textbook.

∣dNdt∣∣0 = 3.55×10¹² decays/s  

B. Determine the activity 1.00 h later when the thyroid is being tested.

∣dNdt∣∣ =  3.54×10¹² decays/s  

C. Determine the activity 4.4 months later.

Answer 1

given,

M = 772 micro g

= 772*10^-6 g

= 772*10^-9 kg

we know, half life time of Iodine, T1/2 = 8.02 days

= 8.02*24*60*60 s

= 6.92928*10^5 s

mass of Iondine atom, m = 2.107*10^-25 kg

no of atoms present in the sample,

No = M/m

= 772*10^-9/(2.107*10^-25)

= 3.66*10^18

decay constant of Iodine, lamda = 0.693/(T1/2)

= 0.693/(6.92928*10^5)

= 1*10^-6 s^-1

A) initial activity, Ao = lamda*No

= 1*10^-6*3.66*10^18

= 3.66*10^12 decay/s <<<<<<<<<<<<<<-----------------Answer

B) after t = 1 hour = 60*60 = 3600 s

A = Ao*e^(-lamda*t)

= 3.66*10^12*e^(-1*10^-6*3600)

= 3.64*10^12 decay/s <<<<<<<<<<<<<<-----------------Answer

C) after t = 4.4 months = 4.4*30*24*60*60 s = 1.14*10^7 s

A = Ao*e^(-lamda*t)

= 3.66*10^12*e^(-1*10^-6*1.14*10^7)

= 4.10*10^7 decay/s <<<<<<<<<<<<<<-----------------Answer