Question

If $1^{st}$ ionization energy of $Li\left(g\right)$ is 5.39 eV and
$L{i}_{\left(g\right)}\to L{i}_{\left(g\right)}^{3+}+3e^{-}\Delta H=+203.43eV$
Then the ionisation energy of $L{i}^{+}$ is

Answer 1

$L{i}_{\left(g\right)}\to L{i}_{\left(g\right)}^{+}+e^{-}\Delta H=+5.39eV$
$L{i}_{\left(g\right)}^{+}\to L{i}_{\left(g\right)}^{2+}+e^{-}\Delta H=xeV$
$L{i}_{\left(g\right)}^{2+}\to L{i}_{\left(g\right)}^{3+}+e^{-}\Delta H=13.6\times 9eV$ ( $L{i}^{2+}$ is H like species, So using formula $E_n=-13.6\frac{Z^2}{n^2}$)
Adding above equations, we get
$L{i}_{\left(g\right)}\to L{i}_{\left(g\right)}^{3+}+3e^{-}\Delta H=203.43eV$
Therefore
$(5.39+x+13.6\times 9)eV=203.43eV$
$\Rightarrow x=75.64eV$