Question

if a vial of generator eluate contains 247 of Tc-99m and 27 uCi of Mo-99, what is the moly concentration?

Answer 1

The activity of Tc-99 is 247 Ci = 247*10^-6 * 3.7*10¹⁰ =9139000 Bq

and the activity of Mo-99 is 27Ci .= 999000 Bq

The activity is given by

Where is the rate of radioactive decay given by

(half life)

For Tc-99, half life is 6 hrs = 21600 s

For Mo- 99 , half life is 66 hrs = 237600 s

N = A*t_1/2/0.693

N_Tc = 9139000*21600/0.693 = 2.8485*10¹¹ atoms

N_Mo = 999000*237600/0.693 = 3.425*10¹¹ atoms

So, mole fraction = moles of Mo/Total moles = N_Mo /(N_Mo +N_Tc ) = 3.425/(3.425+2.8485) = 0.5459