ENERGY of an ELECTRON \[E=\frac{-{{E}_{0}}}{{{N}^{2}}}\] For energy level \[(n=2)\]      \[E=-\frac{13.6}{{{(2)}^{2}}}=\frac{-13.6}{4}=-3.4eV\]. "> ENERGY of an ELECTRON \[E=\frac{-{{E}_{0}}}{{{N}^{2}}}\] For energy level \[(n=2)\]      \[E=-\frac{13.6}{{{(2)}^{2}}}=\frac{-13.6}{4}=-3.4eV\]. ">
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The ionisation potential of a hydrogen atom is ?13.6 eV. What will be the energy of the atom corresponding to \[n=2\] [Pb. CET 2000]

Joint Entrance Exam - Main (JEE Main) Chemistry in Joint Entrance Exam - Main (JEE Main) . 9 months ago

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ENERGY of an ELECTRON \[E=\frac{-{{E}_{0}}}{{{N}^{2}}}\] For energy level \[(n=2)\]      \[E=-\frac{13.6}{{{(2)}^{2}}}=\frac{-13.6}{4}=-3.4eV\].
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