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An energy of 24.6 eV is required to remove one of the electrons from a ground state neutral helium atom. What is the energy (in eV) required to remove both electrons from the ground state of a neutral helium atom?

A. 79.0
B. 51.8
C. 49.2
D. 38.2

Answer :

Final answer:

To fully ionize a ground state neutral helium atom by removing both electrons, the energy required is the sum of the first and second ionization energies, equating to 79.01 eV, hence correct answer is A) 79.0

Explanation:

This is the energy required to remove the first electron. The second ionization energy, which is needed to remove the second electron, is notably higher, measured at 54.42 eV. This significant increase is due to the fact that removing the second electron requires more energy as it is being removed from a positively charged ion (He⁺), which holds onto the remaining electron more tightly.

The combined total energy required to fully ionize a helium atom by removing both electrons (the sum of the first and second ionization energies) is thus 24.59 eV + 54.42 eV = 79.01 eV.

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