The ratio of the ionization energy of a Bohr | vedclass

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(C) The ionization energy $(E_{ion})$ of a hydrogen-like atom is given by the formula:$E_{ion} = 13.6 	imes Z^2 	ext{ eV}$where $Z$ is the atomic nu

Vedclass · Accepted Answer

$1:9$

Vedclass · Answer

(C) The ionization energy $(E_{ion})$ of a hydrogen-like atom is given by the formula:$E_{ion} = 13.6 	imes Z^2 	ext{ eV}$where $Z$ is the atomic number of the atom.For a hydrogen atom,$Z_H = 1$. Thus,$(E_{ion})_H = 13.6 	imes (1)^2 = 13.6 	ext{ eV}$.For a lithium atom (hydrogen-like,$Li^{2+}$),$Z_{Li} = 3$. Thus,$(E_{ion})_{Li} = 13.6 	imes (3)^2 = 13.6 	imes 9 	ext{ eV}$.The ratio of the ionization energy of the hydrogen atom to the lithium atom is:$\frac{(E_{ion})_H}{(E_{ion})_{Li}} = \frac{13.6 	imes (1)^2}{13.6 	imes (3)^2} = \frac{1}{9}$.Therefore,the ratio is $1:9$.

The ratio of the ionization energy of a Bohr hydrogen atom to that of a Bohr hydrogen-like lithium atom is:

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