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Q.
Excitation energy of a hydrogen like ion in its first excitation state is $40.8\, eV$. Energy needed to remove the electron from the ion in ground state is
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Solution:
Excitation energy
$\Delta E=E_{2}-E_{1}=13.6Z^{2}\left[\frac{1}{1^{2}}-\frac{1}{2^{2}}\right]$
$40.8=13.6Z^{2}\times\frac{3}{4}$
or $Z = 2$
So, energy required to remove the electron from ground state
$=+\frac{13.6Z^{2}}{\left(1\right)^{2}}=13.6\left(2\right)^{2}$
$= 54.4\, eV$