Question

If the $1^{st}$ ionization energy of fH atom is 13.6 eV, then the $2^{nd}$ ionzation energy of He atom is

• A. 27.2 eV
• B. 40.8 eV
• C. 54.4 ev
• D. 108.8 ev.

Answer 1

The correct option is C 54.4 ev

No. of electron in H =1

no. of neutron in H =0

ionization energy for H atom =13.6eV

as no. of electron in helium =2

and no. of neutron in helium =2

ionization energy for Helium atom =54.4eV.