A medical technician is working with the four samples of rad

A medical technician is working with the four samples of radionuclides listed in the table below. Initially, each sample contains 3.00 .nol of the radionuclide. First, order the samples by decreasing initial radioactivity. Then calculate how long it will take for the amount of radionuclide in each sample to decrease to 1/32 of the initial amount. radionuclide time for amount of initial radioactivity radionuclide to decrease to sample half-life 14 days (choose one) 60.0 days(choose one). symbol 1 /32 of initial amount 32 15 days days minutes hours 125 75, 98.0 minutes (choose one) 62 | 9.0 hours (choose one).

Solution

We know that the radioactivity is 1st order process:

So, for 1st order process

lnC_Ao/C_A = kt

t_1/2 is half life,when C_A = C_Ao/2

for half life lnC_Ao/C_A = ln (C_Ao/C_Ao/2) = ln2

k = (lnC_Ao/C_A) / t

A .

k = (ln2)/t_1/2 = 0.0495 day^-1

ln2 = 0.693

Initial rate = kC_A = 0.0495 day^-1 x 3.00 umol = 0.1485 umol/day

Similarly for B,C,D

for 1/2 times the initial conc. ,time taken = 14 d
for 1/2 of 1/2 = 1/4 time taken will be 14 x 2 = 28 d

1/2x1/2x1/2x1/2x1/2 = 1/32 times

14 x 5 = 70 days

B.

k = 0.693 / 60 d = 0.01155 d^-1

Initial rate = kCA = 0.0115 d-1 x 3 umol = 0.03465 umol/d

1/32 = 60 x 5 = 300 days

C.

k = 0.693 / 98 min = 0.00707 min^-1

initial rate = kCA = 0.00707 min^-1 x 3 umol = 0.0212 mol/min

1/32 in 5 x 98 min = 490 min

D.

k = 0.693 / 9 h = 0.077 h^-1

Initial rate = 0.077 h^-1 x 3umol = 0.231 umol/h

1/32 CAo will take 9h x 5 = 45h