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The energy of a hydrogen atom in the ground state is -13.6 eV . The energy of a He ⁺ ion in the first excited state will be

Options

  1. A-13.6 eV
  2. B-27.2 eV
  3. C-54.4 eV
  4. D-6.8 eV

Correct answer

A. -13.6 eV

Step-by-step solution

Energy E of an atom with principal quantum number n is given by E= -13.6 n ^2 Z^2 for first excited state n =2 and for He ⁺ Z =2 aligned E & = -136 (2)^2 (2)^2 & =-13.6 eV aligned

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