Q. Ionization energy of $He^{+}$ ion at minimum position is
A) 13.6 $eV$
B) 27.2 $eV$
C) 54.4 $eV$
D) 68.0$\ eV$
Explanation
(c) Energy of helium ions.
$E_{n}\ = \ -$ $\frac{13.6\ Z^{2}}{n^{2}}\ eV$
In minimum position, $n$=1
For $\ He^{+},\ Z = 2$
$E = \frac{- 13.6\ {\times \ (2)}^{2}}{1}\ eV$
$E = 54.4\ eV$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.