Question

Iron-59 is a radioisotope that is used to evaluate bone marrow function. The half-life of iron-59 is 44.5 days. How much time is required for the activity of a sample of iron-59 to fall to 19.0 percent of its original value? days Submit Answer

Answer 1

Radioactive decay of the given nuclei follow the first order kinetics. Thus, first order rate law has been used in solving the question.
Solution :- wa In Ao =kt А efore solving the given problems, first we know the theory and formula used in this coactive decay follow the first order kinetes, which can be represented by general formula- where to is the initial amount A is the amount at teinet k is the reate constant for half life, the concentration of Ao falls to half ie A In to = K t =) en z = kth Ao/ ty= lira AK where ln2 = loge 2 Same formula is used for readis active decay, where No is the initial amount ln No = xt Nis the camount at teine t s ies the idecay iconstant just similar to k in above equation Thus, X/ In 2 where ln2 = loge 2

Solution of problem I :- It is given in question that - half life of Iron-59 = 44.5 days This means that it takes 44.5 days to decay the initial camount of Fe-59 (Iron-59) to half of its initial camount. We know the relation between the and & (decay Constant). ty = ln 2 = t = 0.693 3) 1 = 0.693 = 0.693 44.5 til = 0.0155 days decay constant = x= 0.018 Now, we write the general equation, In No = xt as the final amount decreases to 19%, lets take initial camount as low and corresponding final amount as 19. en 100 = 0.0155 xt = ln 5.26 = 0.0155 xt 19 = 1.66 = 0.0155xt

t = 1.66 0.0155 -= 107.09 days = 107 days teme required to fall its coriginal value to 19% is 107 days. Solution of Problem - 2 :- It is given that is half life of thallium - 201 = 72.9 hours Initial amount = 37.2 mg We have to calculate the mass of thalliem 201 rafter 189 hours. For this, we have to calculate the decay constant & x= 0.693 because In 2 = 0.693 th =) x = 0.693 72.9 = 0.0095 hours Again, we use the general formula, In No=xt = In 37.2 = 0.005 x 189 N In 37.2 = 1.7955 N

2hemum , as we know that log a = loga - logo = In 37.2 - lu N = 1.7955 anx = loge = loge 37.2 - lag N = 1.7955 3) 3.61 - lage N = 1.7955 3 logeN = 3.61-1-7955 = logeN = 1.8145 1.8145 N = e N = 6.138 = 6-14 mg Hence, mass of thallium-201 after 189 hours is 6. 14 mg.