If the I.E. of He^{+} is 54.4 eV,then - | vedclass

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(A) The ionization energy $(I.E.)$ of a hydrogen-like species is given by the formula: $I.E. = 13.6 	imes Z^2 \ eV$,where $Z$ is the atomic number.Fo

Vedclass · Accepted Answer

$I.E.$ of $H$ is $13.6 \ eV$ and that of $Li^{2+}$ is $122.4 \ eV$

Vedclass · Answer

(A) The ionization energy $(I.E.)$ of a hydrogen-like species is given by the formula: $I.E. = 13.6 	imes Z^2 \ eV$,where $Z$ is the atomic number.For $He^{+}$ $(Z=2)$: $I.E. = 13.6 	imes (2)^2 = 13.6 	imes 4 = 54.4 \ eV$. This matches the given value.For $H$ $(Z=1)$: $I.E. = 13.6 	imes (1)^2 = 13.6 \ eV$.For $Li^{2+}$ $(Z=3)$: $I.E. = 13.6 	imes (3)^2 = 13.6 	imes 9 = 122.4 \ eV$.Therefore,option $A$ is correct.

If the $I.E.$ of $He^{+}$ is $54.4 \ eV$,then -

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