The ionization potential of the hydrogen atom is 13.6 eV. Its energy in n = 2 energy state is

-3.4 eV

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-6.8 eV

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-13.6 eV

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- 27 .2 eV

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Solution

The correct option is A -3.4 eV
Energy of the electron in the first orbit of the atom, $E_{1} = - 13.6 e V$
Its energy in the second orbit, $E_{2} = - \frac{13.6 e V}{(2)^{2}} = - 3.4 e V$
Hence, the correct choice is (a).

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13.6 e V

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The total energy of an electron in the first excited state of hydrogen is about -3.4 eV. Its kinetic energy in this state is

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