Answer:
\[\Delta E={{E}_{\infty }}-{{E}_{5}}\]
=\[0-\left[
\frac{-{{Z}^{2}}}{{{n}^{2}}}\times 2.18\times {{10}^{-18}}\text{J} \right]\]
\[=-\left[
\frac{-{{1}^{2}}}{{{5}^{2}}}\times 2.18\times {{10}^{-18}}\text{J} \right]\]
\[=\frac{1\times 2.18\times
{{10}^{-18}}}{25}=8.72\times {{10}^{-20}}J\]
Ionisation enthalpy of hydrogen atom
\[=2.18\times {{10}^{-18}}\text{J}\]
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