Question

Given data: Ln2=0.693, 1Ci=3,7.10¹⁰ des/s

¹⁴⁰Ba radionuclide with half-life T_p=306 hours decays to its daughter ¹⁴⁰La with a half-life of T_d=40.3 hours and to stable granddaughter ¹⁴⁰Ce. The scheme of the decay is illustrated as follows:

¹⁴⁰Ba → ¹⁴⁰La → ¹⁴⁰Ce

The parent, ¹⁴⁰Ba, has an initial activity of A₀=5mCi. The initial number of nuclei (at t=0) of the daughter ¹⁴⁰La is N₀ = 0, and t_m is the time when the daughter activity reaches its maximum value.  The equations of the activity of the parent and daughter at any time are given as follow:

Do your calculation and circle the correct answer for the following:

• At what time, t, the ¹⁴⁰Ba activity will be 25% of the initial activity (give time in hours)?
• At what time, t_m, the daughter activity A_d reaches its maximum (give time in hours)?
• When t=t_m, we separate the parent and daughter, what would be the maximum activity, A_m, in mCi of ¹⁴⁰La at t=t_m?

1. t= 0.612 hr _,t_m= 0.1355 hr _,A_m=0.36 mCi
2. t= 612 hr _,t_m= 13.5 hr _,A_m=5 mCi
3. t= 306 hr _,t_m= 0.1355 hr _,A_m=0.36 mCi
4. None of the above

Answer 1

Another feature of each radionuclide is its half-life. Half-life is the length of time it takes for half of the radioactive atoms of a specific radionuclide to decay. A good rule of thumb is that, after seven half-lives, you will have less than one percent of the original amount of radiation.