A H e + ion is in its first excited state. Its ionization energy is:

A He+ ion is in its first excited state. Its ionization energy is:
  1. 13.60 eV
  2. 48.36 eV
  3. 54.40 eV
  4. 6.04 eV

Solution

E=-13.6z2n2
=-13.62222=-13.60 eV
Total energy of He+ ion is in its first excited state is -13.60 eV . Hence, its ionization energy will be 13.60 eV 

Asked in: JEE Main 2019 (09 Apr Shift 2)

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