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The energy required to ionize an unknown hydrogen-like species from its first excited state is exactly 13.6 eV . Identify the species.

Options

  1. AH atom
  2. BHe ⁺ ion
  3. CLi ²⁺ ion
  4. DBe ³⁺ ion

Correct answer

B. He ⁺ ion

Step-by-step solution

The ionization energy of a hydrogen-like species from a state with principal quantum number n is given by E = 13.6 Z^2 n^2 eV , where Z is the atomic number. The first excited state corresponds to n = 2 . Given that the ionization energy from this state is 13.6 eV , we can write: 13.6 = 13.6 Z^2 2^2 Z^2 4 = 1 Z^2 = 4 Z = 2 The species with atomic number Z = 2 is the helium ion ( He ⁺ ). Answer: He ⁺ ion

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