A sample of Ar41 t12 182 hours was produced from a 3 gram s

A sample of Ar-41 (t_1/2 = 1.82 hours) was produced from a 3 gram sample of pure argon-40 gas (sigma =0.65 barn) in a research reactor. If one in every 565 million of the original Ar-40 atoms in the sample absorbed a neutron to become Ar-41, compute the sample\'s activity when removed from the reactor. What will the activity be 6 hours later?

Solution

Here ,

total mass of sample , m = 3 gm

number of atoms in Ar-41 = number of moles * Avagodro number

number of atoms in Ar-41 = (3/41) * 6.022 *10^23

number of atoms in Ar-41 = 4.406 *10^22

Now , number of total radioactive atoms

No = 4.406 *10^22/(565 *10^6)

No = 7.799 *10^13

for the activity after 6 hours

activity after 6 hours = 7.799 *10^13 * (0.5)^(6/1.82) * (ln(2)/(1.82 * 3600))

activity after 6 hours = 8.396 *10^8 Bq

the activity after 6 hours is 8.396 *10^8 Bq