Question

The first excitation potential of $H{e}^{+}$ ion is $n$, and the ionization potential of $L{i}^{++}$ ion is $m$ then find out value of $\frac{m}{n}$

Answer 1

First excitation potential of

$H{e}^{+}=10.2\times z^2$

$z=2forH{e}^{+}$

$n=40.8V$

Ionization potential for

$L{i}^{++}=13.6\times z^2$

$m=122.4$

$\frac{m}{n}=\frac{122.4}{40.8}=3$