The binding energy of an electron in the ground state of an atom is equal to $24.6 e V$ . Find the energy required to remove both the electrons from the atom.

79 e V

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89 e V

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99 e V

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69 e V

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Solution

The correct option is A $79 e V$
The binding energy of Helium atom $= 24.6$ $e V$

When one electron is removed from it, helium atom, it becomes a single electron species.

Ionization energy of $H e^{+}$ atom $= - 13.6 \times 4 = - 54.4$ $e V$ .

$∴$ Energy required to remove both the electrons $= 54.4 + 24.6 = 79$ $e V$ .

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The binding energy of an electron in the ground state of an atom is equal to 24.6 eV. Find the energy required to remove both the electrons from the atom.

The correct option is A 79 eVThe binding energy of Helium atom=24.6 eVWhen one electron is removed from it, helium atom, it becomes a single electron species.Ionization energy of He+ atom =−13.6×4=−54.4 eV.∴ Energy required to remove both the electrons =54.4+24.6=79 eV.

The binding energy of an electron in the ground state of an atom is equal to $24.6 e V$ . Find the energy required to remove both the electrons from the atom.