Radioactive isotopes decay according to a first­order process. If Nt is the Numb

Question

Radioactive isotopes decay according to a first­order process. If Nt is the Number of atoms of a radioactive isotope at time t, then Nt=N0e ­kt .

a. Cu­64 (t1/2 = 12.8 hr) is used in brain scans for tumors and in studies of Wilson’s disease, a genetic inability to metabolize copper. How many days are required before the dose of Cu­64 administered to a patient drops to 0.10% of its initial activity? (Assume that Cu­64 isn’t leaving the body by any process other than its normal radioactive decay.)

b. U­238 decays to Pb­206 with a half­life of 4.51x10 9 years, A sample of ocean sediment contains 1.50 mg of U­238 and 0.460 mg of Pb­206. Estimate the age of the sediment. (Assume that Pb­206 is formed only as a result of the decay of U­238 you may also assume that Pb­206 is stable and does not itself decay.)

Explanation / Answer

a) we know that

decay constant (lamda ) = 0.693 / t1/2

so

lamda = 0.693 / 12.8

lamda = 0.05414

lnA = lnAo - (lamda x t)

lamda x t = lnAO - lnA

lamda x t = ln (AO/A)

given that

A= 0.001 Ao

so

Ao/A = 1000

so

we get

lamda x t = ln 1000

0.05414 x t = ln1000

we get

t = 127.59

so

the time taken is 127.59 hours

b)

now

U ---> Pb

given

amount of Pb formed = 0.46

amount of U remaining = 1.50

so

initial amount of U = 1.50 + 0.46

initial amount of U = 1.96

now

we know that

lnC = lnCo - (lamda x t)

now

lamda = 0.693/ t1/2

so

lamda = 0.693 / 4.51 x 10^9

lamda = 1.5366 x 10-10

now

ln 1.5 = ln 1.96 - ( 1.5366 x 10-10 x t )

t = 1.74 x 10^9

so

the age of the rock is 1.74 x 10^9 years

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