Question

Calculate the ionisation energy of (a) one ${Li}^{2+}$ ion and (b) one mole of ${Li}^{2+}$ ion.

(Given: $R=1.0974\times {10}^{-7}{m}^{-1}$).

Answer 1

$a.$ Energy of $n^{th}$ orbit $=\frac{-21.76\times {10}^{-11}}{}{n}^2\times z^{2}J$

In $L{i}^{2+}$ , $n=1$ as it id a one-electron species.

$\therefore E_1=\frac{-21.76\times {10}^{-19}}{1}\times 9=-195.84\times {10}^{-19}J$

Energy at infinity $=E_{\infty }=0$

$\therefore$ Ionization energy$=E_{\infty }-E_1=195.84\times {10}^{-19}J$

$b.$ Ionization energy of one $L{i}^{2+}$ atom $=1178.9\times {10}^{4}J=11789KJ$