From the formula for the activity of a substance
"C=\\frac{0.693 \\cdot N_A \\cdot m}{M \\cdot T_{1\/2}}"
Where
"N_A=6.022 \\cdot 10^{26}" - Avogadro number
"M" - atomic mass of the radioactive element
"T_{1\/2}=4.77 \\cdot 60=286.2 s" - half-life of a radioactive element
"m" -mass of a radioactive element
Note
"m=\\frac{C \\cdot M \\cdot T_{1\/2}}{0.693 \\cdot N_A}"
Given that
"1Ci=3.7 \\cdot 10^{10}{decays}\/s"
"m=\\frac{1500 \\cdot 206.977 \\cdot( 4.77\\cdot 60 )\\cdot 3.7 \\cdot 10^{10} \\cdot 10^{6}}{0.693 \\cdot 6.022 \\cdot 10^{26}}=7.878 \\cdot 10^{-3}\\mu g"
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