The ionization energy of hydrogen atom in the ground state is

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13.6 MeV

13.6 eV

13.6 Joule

Zero

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Answer is Wrong!

This question was previously asked in

UPSC NDA-2 – 2017

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The ionization energy of a hydrogen atom is the minimum energy required to remove the electron from the ground state (n=1) and take it to the ionization limit (n=infinity).

The energy of an electron in the n-th energy level of a hydrogen atom is given by E_n = -13.6/n² eV. In the ground state (n=1), the energy is E₁ = -13.6/1² = -13.6 eV. At the ionization limit (n=infinity), the energy is E_infinity = -13.6/infinity² = 0 eV. The ionization energy is the difference between these energies: Ionization Energy = E_infinity – E₁ = 0 – (-13.6 eV) = +13.6 eV.

MeV (Mega electron Volt) and Joule are also units of energy, but 13.6 eV is the standard and correct value for the hydrogen atom’s ionization energy. 1 eV is approximately 1.602 x 10⁻¹⁹ J. 1 MeV is 10⁶ eV. 13.6 MeV is an extremely large amount of energy in this context. Zero ionization energy would mean no energy is required to remove the electron, which is incorrect for a bound electron.

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