A) 40.8 eV
B) 27.2 eV
C) 54.4 eV
D) 13.6 eV
Correct Answer: C
Solution :
The excitation energy in the first excited state is \[E=Rhc{{Z}^{2}}\left( \frac{1}{{{1}^{2}}}-\frac{1}{{{2}^{2}}} \right)\] \[=(13.6eV)\times {{Z}^{2}}\times \frac{3}{4}\] \[\therefore \] \[40.8=13.6\times {{Z}^{2}}\times \frac{3}{4}\] \[\Rightarrow \] \[Z=2\] So, the ion in problem is \[H{{e}^{+}}.\]The energy of the ion in the ground state is \[E=-\frac{Rhc{{Z}^{2}}}{{{1}^{2}}}=-13.6\times 4=-54.4eV\] Hence, 54.4 eV is required to remove the electron from the ion.
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