A single electron system has ionization energy $11180$ $k J$ ${m o l}^{- 1}$ . Find the number of protons in the nucleus of the system.

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Solution

$Z = 3$ .
$I E = \frac{Z^{2}}{n^{2}} \times 21.69 \times 10^{- 19} J$
$\frac{11180 \times 10^{3}}{6.023 \times 10^{23}} = \frac{Z^{2}}{1^{2}} \times 21.69 \times 10^{- 19}$
$Z = 3$ .

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