The binding energy of an electron in the gorund state of He atom is equal to `E_0 = 24.6 eV.` Find the energy required to remove both electrons form the atom.

The binding energy of an electron in the gorund state of He atom is eq

1.	The binding energy of an electron in the gorund state of He atom is equal to `E_0 = 24.6 eV.` Find the energy required to remove both electrons form the atom.
Answer» Energy required to remove first electron is 24.6 eV. After removing first electron form this atom, it will become `He^+` `E_1 =- (13.6)(2)^2 (as E prop Z^2 and Z=2)` = - 54.4 eV `:. Energy required to remove this second electron will be 54.4 eV.` `:. Total energy required to remove both electrons` =24.6 +54.4 =79 eV.

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The binding energy of an electron in the gorund state of He atom is equal to `E_0 = 24.6 eV.` Find the energy required to remove both electrons form the atom.

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