Ionisation energy of He+ is 19.6 ×10−18 J atom−1. The energy of the first stationary state (n = 1) of Li2+ is

A
$4 .41 \times 10^{- 16} J {atom}^{- 1}$
B
$- 4 .41 \times 10^{- 17}$
C
$- 2 .2 \times 10^{- 15} J {atom}^{- 1}$
D
$8 .82 \times 10^{- 17} J {atom}^{- 1}$

Solution:

$\begin{array}{l} IE = - E_{1} \\ E_{1} for {He}^{+} = - 19 .6 \times 10^{- 18} J {atom}^{- 1} \\ \frac{{(E_{1})}_{{He}^{+}}}{{(E_{1})}_{{Li}^{2 +}}} = \frac{{(Z_{{He}^{+}})}^{2}}{{(Z_{{Li}^{2 +}})}^{2}} \\ \frac{- 19 .6 \times 10^{- 18}}{{(E_{1})}_{{Li}^{2 +}}} = \frac{4}{9} \\ Or E_{1} ({Li}^{2 +}) = \frac{- 19 .6 \times 9 \times 10^{- 18}}{4} \\ = - 4 .41 \times 10^{- 17} J {atom}^{- 1} \end{array}$

Ionisation energy of ${He}^{+} is 19 .6 \times 10^{- 18} J {atom}^{- 1} .$ The energy of the first stationary state (n = 1) of ${Li}^{2 +} is$

Solution:

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