Given, Half-life of Au198,T1/2=2.7 days Amount of Au198,m=1.5mg=1.5×10−3g Atomic weight of Au198,A=198gmol−1 We know that activity, A0=λN0 where, λ= decay constant =
ln2
T1/2
N0= initial number of atoms =
m
A
NA ⇒=A0λ⋅
m
A
NA=
ln2
T1∕2
×
1.5×10−3
198
×6.023×1023 ⇒A0=
ln2×1.5×10−3×6×1023
2.7×3600×24×198×3.7×1010
Ci [∵1 curie (Ci)=3.7×1010Bq] ⇒A0=3.65×102Ci=365Ci The answer is close to 357Ci.