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NEET2019PhysicsAtomic PhysicsActual

15 eV is given to e⁻ in 4^ th orbit, then find its final energy when it comes out of H -atom.

Options

  1. A14.15 eV
  2. B13.6 eV
  3. C12.08 eV
  4. D15.85 eV

Correct answer

A. 14.15 eV

Step-by-step solution

Energy of 4^ th orbit of H -atom =-13.6 1 16 =-0.85 eV So energy released = total energy - ionization energy of 4^ th orbit =15-0.85=14.15 eV

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