The ionisation energy of a hydrogen atom is 1 | vedclass

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(B) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like atom is given by the formula: $E_n = -\frac{13.6 Z^2}{n^2} \; eV$.For a hydroge

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$54.4$

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(B) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like atom is given by the formula: $E_n = -\frac{13.6 Z^2}{n^2} \; eV$.For a hydrogen atom,$Z = 1$ and $n = 1$,so the ground state energy is $E_1 = -13.6 \; eV$. The ionisation energy is the energy required to remove the electron,which is $13.6 \; eV$.For a singly ionised helium atom $(He^+)$,the atomic number $Z = 2$. The ground state energy is $E_1 = -\frac{13.6 	imes (2)^2}{(1)^2} = -13.6 	imes 4 = -54.4 \; eV$.Therefore,the ionisation energy required to remove the electron from the ground state of $He^+$ is $54.4 \; eV$.

The ionisation energy of a hydrogen atom is $13.6 \; eV$. The ionisation energy of a singly ionised helium atom would be ....... $eV$.

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