The ionazation potential of the hydrogen is 13.6ev .The energy needed to ionise a hydrogen atom which is in its second exited state is about
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According to the question, the Ionisation potential of hydrogen atom is 13.6eV.
Therefore, the change in energy involved in removing the electron from n=2 is given by
the formula E(n)=−13.6n2 eV/atom
However, in this case the electron is completely removed from n=2, therefore
ΔE=E(∞)−E(2)
Therefore,
ΔE=0−(−13.622)
ΔE=+13.64
ΔE=3.4eV
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