Question #3d690

1 Answer
Aug 28, 2016

#""_8^18O+""_1^1H->""_9^18F+""_0^1n#

Total charge (proton) in the LHS will be equal to total charge in RHS. Also the total mass no (proton + neutron) in LHS will be equal to total mass no in RHS

So the charge on the species to be placed in the blanc will be zero and its mass will be 1.

It will be neutron #""_0^1n#