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Q. The one electron species having ionization energy of 54.4 eV is

CMC MedicalCMC Medical 2011

Solution:

$ {{E}_{n}}=\frac{-13.6\times {{(Z)}^{2}}}{{{n}^{2}}}eV $ If $ n=1 $ $ {{E}_{n}}=-13.6\,{{Z}^{2}} $ $ \therefore $ $ {{Z}^{2}}=\frac{-\,54.4}{-13.6} $ $ {{Z}^{2}}=4 $ $ Z=2 $ Thus, the species with atomic number 2 (i.e., $ H{{e}^{+}} $ ) has the ionisation energy 54.4 eV.