The radioactive isotope 198Au has a half-life of 64.8 hr. A sample has an initial activity of 41.5 µCi?

Question

The radioactive isotope 198Au has a half-life of 64.8 hr. A sample containing this isotope has an initial activity (t = 0) of 41.5 µCi. Calculate the number of nuclei that decay in the time interval between t1 = 13.9 hr and t2 = 14.9 hr.

Shouldn't I just be able to find the activity after between T1 and t2, convert it to nuclei using :

N= R/r

where, R is the activity, and r is the decay rate.

r= ln 2/ half life.

But i am not getting the correct answer, so I am doing it incorrectly.

Answer 1

You don't say what the "correct" answer should be.  To three significant digits, I get 1.29 x 10^5 particles decaying in that one-hour interval.

Compute the activity at 13.9 hrs (50,040 s), and the activity at 14.9 hrs (53,640 s).  Use your N = A / k where N is the total number of nuclei, A is the activity at time t, and k is the decay constant.  Then subtract the two values of N

k = 0.693 / t(1/2) = 0.0107 1/hr

0.0107 1/hr x (1 hr / 3600 sec) = 2.972 x 10^-6 1/sec

A = Ao e^-kt

A1 = 41.5 uCi x e^(-0.01071/hr x 13.9 hr) = 35.76 uCi

A2 = 35.38 uCi

N1 = A1 / k = 35.76 / 2.972 x 10^-6 = 1.2032 x10^7 nuclei

N2 = A2 / k = 36.38 / 2.972 x 10^-6 = 1.1904 x 10^7 nuclei

N1-N2 = 1.286 x 10^5 disintegrations during the one hour interval from 13.9 to 14.9 hrs.

Or you can get the difference in the two activities at 13.9 and 14.9 hrs, 0.38 uCi, and divide by the decay constant.

N = 0.38 uCi / 2.972 x 10^-6 1/sec = 1.3 x 10^5 disintegrations.

This last figure has lost some precision in the subtraction.

Answer 2

the mass is 238.029 from the starting piont but the radioactive dateing the last part sold be 0.00001