Question

The one electron species having ionization energy of 54.4 eV is

• A. $H$
• B. $H{e}^{+}$
• C. $B^{4+}$
• D. $L{i}^{2+}$
• E. $B{e}^{2+}$

Answer 1

Answer: (A)

$H$

$I.E=\frac{13.6Z^2}{n^2}eV=13.6Z^2$ for one electron species.

$13.6Z^2=54.4$ or $Z^2=4$ or $Z=2$ i.e., $H{e}^{+}$.