From reactivity law R= N/(T_1/2)
N( constration)=(1gram/Atomic mass)X Ave.number ... atm/g
Return to activity it will be in (decay/sec)
Decay/sec to be changed to Bq
Bq×(1/10^6) = Bq/ppm
Dear Mohamed
Thanks for your reply, I am familiar with the above mentioned equation. But I thought there is another methodology or empirical formula because the above equation gives extremely low values and it depends upon the fact of natural abundance. So it gives you good results in naturally radioactive materials not artificially produced like Cs, Co, and Eu etc....
Muhammad S. Mansy You expect extremely low numbers. Compare the specific activity of natural emitters to artificially produced.
@Francios Tondeur
Thanks for your interest, yes I mean aqueous solution
With C = activity concentration in Bq/l, volume V[l]= Mw mass of water in kg
and MCs= mass of Cs137, Avo= Avogadro number 6.02*10^23 atoms/mol
Atm= atomic mass 137 g/mol , T1/2 in sec
activity = C * Mw[kg] = ln2 *MCs[mg] * Avo / (T1/2 * 10^3 * Atm)
ppm = MCs[mg]/Mw[kg] = C[Bq/l] * T1/2 * 10^3 * Atm / (ln2 * Avo)
A very low concentration indeed, even for a high activity.
Specific radioactivity of Cs-137=3.2*10^12 Bq/g. This means 1 Bq~ 10^-11 g
and 10^-3 g/L= 1ppm so 1 Bq/L of Cs-137 becomes 10^-8 ppm roughly.
similar calculation can be done for Eu-152 using its specific activity
Thanks
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