QJEEM18Ph04e9: JEE Main 2018 Physics - Class 12 JEEM18Ph04e9 Solution: The energy required to remove the electron from a singly…

Energy required to remove e ^$-$ from singly ionized Helium atom

E₁ = ${{13.6{z^2}} \over {{n^2}}}$ = ${{13.6 \times {2^2}} \over {{1^2}}}$ = 54.4 eV.

Let, E₂ = energy required to remove e ^$-$ from He $-$ atom

$\therefore\,\,\,\,$ According to q vertion.

E₁ = 2.2 E₂

$\therefore\,\,\,\,$ E₂ = ${{54.4} \over {2.2}}$ = 24.72 eV

$\therefore\,\,\,\,$ Total energy required to ionize Helium atom completely

= (54.4 + 24.72) eV

= 79.12 eV