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:$E_{ion} = 13.6 \, Z^2 \, 	ext{eV}$where $Z$ is the atomic numb

Vedclass · Accepted Answer

$1 : 9$

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(C) The ionization energy $(E_{ion})$ of a hydrogen-like atom is given by the formula:$E_{ion} = 13.6 \, Z^2 \, 	ext{eV}$where $Z$ is the atomic number of the atom.For hydrogen $(H)$,the atomic number $Z_H = 1$.Thus,$(E_{ion})_H = 13.6 	imes (1)^2 = 13.6 \, 	ext{eV}$.For a hydrogen-like lithium ion $(Li^{2+})$,the atomic number $Z_{Li} = 3$.Thus,$(E_{ion})_{Li} = 13.6 	imes (3)^2 = 13.6 	imes 9 = 122.4 \, 	ext{eV}$.The ratio of the ionization energy of hydrogen to that of lithium 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's hydrogen atom to that of a Bohr's hydrogen-like lithium atom is:

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