An energy of 24.6 eV is required to remove one of the electrons from a neutral helium atom. The energy (in eV) required to remove both the electrons f

On removing one of the electrons from a neutral helium atom, it is hydrogen like atom. 

The ionisation energy of the hydrogen like atom is given by E_ion = Z²Rhc

The ionisation energy of the helium atom is given by Z = 2Rhc = 13.6 eV

Required energy to emit second electron from helium is given by:

 E₂ = E_{ion =} 2² x 13.6 = 54eV

Therefore, the required energy is E₁ +  E₂ = 24.6 + 54.4 = 79.0Ev

Hence, option D is correct. ( 79.0 eV )