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

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