The one electron species having ionisation energy of $54.4$ eV is:

H e^{+}

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H

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B e^{2 +}

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B e^{3 +}

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Solution

The correct option is A $H e^{+}$
Ionisation energy $= - 13.6 \frac{Z^{2}}{n^{2}} e V$
$= - 13.6 \times \frac{(2)^{2}}{(1)^{2}}$ (For $H e^{+}$ , $Z = 2$ )
$= - 54.4$ eV.

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