Question

What is the number N0 of 99mTc atoms that must be present to have an activity of 15mCi?

Answer 1

Base on my research, within 2 hours you have a number of atoms which remain. 
N= N0*2^(-t/6.020 = N= N0*2^-0.33223= 07943 N0

So, the number of atoms that are being disintegrated is N0-N=N0*(1-0.79430)=0.2057 N0

It must be equal to 15 mCi = 15*3.7*10^7= 5.55*10^8 atoms 

N0= 5.55*10*8/0.2057  = 2.698*10^9 atoms

Therefore, 2.698*10^9 atoms is the number of N0