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

Options

  1. A13.60 e V
  2. B48.36 e V
  3. C54.40 e V
  4. D6.04 e V

Correct answer

A. 13.60 e V

Step-by-step solution

E = - 13.6 z 2 n 2 = - 13.6 2 2 2 2 = - 13.60   e V Total energy of H e + ion is in its first excited state is - 13.60   e V . Hence, its ionization energy will be 13.60   e V

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