Question

If the ionisation energy of hydrogen atom is $13.6eV$, what will be the ionisation energy of $H{e}^{+}$ and $L{i}^{2+}$ ions respectively?

• A. $27.2eV,40.8eV$
• B. $13.6eV,54.4eV$
• C. $54.4eV,54.4eV$
• D. $54.4eV,122.4eV$

Answer 1

The correct option is D $54.4eV,122.4eV$

Given:
$\text{Ionisation energy of Hydrogen atom = 13.6 eV}$
We know that,
$\text{Ionisation energy}=-\left(\text{Energy of the 1st orbit}\right)$
Energy of the $1^{st}$orbit of $H{e}^{+}=-13.6\times Z^2=-54.4eV$
(where, $Z=2$ for $H{e}^{+}$)
So,
Ionisation energy of $H{e}^{+}=-(-54.4eV)=54.4eV$
$\text{Energy of the}{1}^{st}\text{orbit of}L{i}^{2+},=-13.6\times 3^2=-122.4eV$
(where, $Z=3$ for $L{i}^{2+}$)
$\text{Ionisation energy of}L{i}^{2+}=-(-122.4eV)=122.4eV$