After the scan. Is there any mass left after interactions?

No - positron and electron annihilate each other, generating two gamma photons.

No - positron and electron annihilate each other, generating two gamma photons.
If annihilation produces E=mc^2 why does the body not explode?

Do the math:

E=mc^2, where m = mass of positron + mass of electron

E = 2(9.1 \times x 10^{-31 }\ Kg) \times (3.0 \times 10^8 \ \frac m s )^2 = 1.6 \times 10 ^{-13} J

So the energy of a positron-electron annihiliation is minuscule.

Do the math:

E=mc^2, where m = mass of positron + mass of electron

E = 2(9.1 \times x 10^{-31 }\ Kg) \times (3.0 \times 10^8 \ \frac m s )^2 = 1.6 \times 10 ^{-13} J

So the energy of a positron-electron annihiliation is miniscule.
Since there are millios of annihilations why do we no see the sum(E)?
Sum(E) may not be explosive; but what about collateral thermal damage?

Even if there were 100 million annihilations the total energy released would only be 1.6x10^-5 J or 16 microjoules. This is still a minuscule amount. Given that the specific heat of water is 4.2 joules per gram per degree C and assuming that all the energy was concentrated in 1 milligram of water ( in reality it's all over the body) it would only raise the temperature of that milligram of water by about 4thousandths of a degree C. Definitely no thermal damage occurring

Even if there were 100 million annihilations the total energy released would only be 1.6x10^-5 J or 16 microjoules. This is still a minuscule amount. Given that the specific heat of water is 4.2 joules per gram per degree C and assuming that all the energy was concentrated in 1 milligram of water ( in reality it's all over the body) it would only raise the temperature of that milligram of water by about 4thousandths of a degree C. Definitely no thermal damage occurring
Thanks