Question

The iodine isotope 131 53 I (half-life = 8 days, atomic mass = 131 u) is used in hospitals for diagnosis of thyroid function. (1 u = 1.66 x 10-27 kg). If 682 μg (1 μg = 10-6 g) are ingested by a patient, determine the activity

(a) Immediately

(b) One hour later, when the thyroid is being tested

(c) Six months later (assume the iodine is still in the patient’s body)

Answer 1

number of Iodine nuclie present initially,

No = mass of the sample/mass of one nuclie

= 682*10^-9/(131*1.66*10^-27)

= 3.136*10^18

given
half life time, T1/2 = 8 days

= 8*24*60*60

= 691200 s

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

= 0.693/691200

= 1.0016*10^-6 s^-1

a) Immediately,

initial activity, Ao = No*lamda

= 3.136*10^18*1.0016*10^-6

= 3.14*10^12 decay/s <<<<<<<---------Answer

b) one hour later,

t = 60*60 s

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

= 3.12*10^12*e^(-1.0016*10^-6*60*60)

= 3.11*10^12 decay/s <<<<<<-----------Answer

c) six months later,

t = 6*30*24*60*60 s

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

= 3.12*10^12*e^(-1.0016*10^-6*6*30*24*60*60)

= 5.36*10^5 decay/s <<<<<<-----------Answer