Answer:
\[IE={{E}_{\infty
}}-{{E}_{n}}\]
\[=0-(-2.18\times
{{10}^{-18}})=2.18\times {{10}^{-18}}J\]
Ionisation
enthalpy per mole
\[=2.18\times
{{10}^{-18}}\times 6.023\times {{10}^{23}}\]
\[=1.313\times
{{10}^{6}}J\]
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