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If the first excitation energy of a hydrogen-like atom is 27 . 3 eV , then the ionization energy of this atom will be

Correct answer

36.4

Step-by-step solution

First excitation energy = Z 2 Rhc 1 1 2 − 1 2 2 = Z 2 Rhc 3 4 ∴ 3 4 Rhc = 27.3 eV ∴ Z 2 Rhc = 4 3 27.3 eV = 36.4 eV

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