The binding energy of the electron in the lowest orbit of the hydrogen atom is 13.6 eV. The energies required in eV to remove an electron from the three lowest orbits of the hydrogen atom are

Question

Dr. Abubacker · Accepted Answer

(d) Lowest orbit is $n = 1$. Three lower orbits correspond to $n = 1.\ 2.\ 3$
$	herefore E_{1} = \frac{13.6}{1^{2}} = 13.6\ eV\ ,$
$E_{2} = \frac{13.6}{2^{2}} = 3.4\ eV,\ E_{3} = \frac{13.6}{3^{2}} = 1.5\ eV$

Q. The binding energy of the electron in the lowest orbit of the hydrogen atom is 13.6 eV. The energies required in eV to remove an electron from the three lowest orbits of the hydrogen atom are

Explanation

Reviewed by Dr. Abubacker

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