Question

Calculate the total number of atoms and total mass of 99mTc present in 15 mCi (555 MBq) 99mTc activity (t1/2 = 6 h).

Answer 1

$A=\frac {N\ln2}{T_{1/2}},\implies$

$N=\frac{AT_{1/2}}{\ln 2}=17.3\cdot 10^{12};$

$N=\frac mMN_A,\implies$

$m=\frac{NM}{N_A}=2.85\cdot 10^{-12}~kg.$