29sir .. i solved the problem like this
1g Mg contains 1/24 moles of Mg
so i converted the ionisation energy in KJ/g so IE1= 31KJ/g and IE2= 60 KJ/g
since we are supplying 50KJ energy so firstly all Mg atoms will be converted to Mg+..this process requires 31KJ energy
now the remaining 19KJ energy will be absorbed by some atoms to convert into Mg2+
fraction of atoms converting from Mg+ to Mg2+ = 19/60 = 0.32 approx
so % of Mg2+ = 32% and Mg+ = 68%
