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In a hydrogen atom, the energy of the first excited state is -3.4 eV. Then, find out the K.E. of the same orbit of the hydrogen atom.

NEET UG 2002 — Chemistry Physical Chemistry

2002mcqeasy

In a hydrogen atom, the energy of the first excited state is −3.4eV-3.4 \mathrm{eV}−3.4eV. Then, find out the K.E. of the same orbit of the hydrogen atom.

Official previous-year question

Held on 30 Apr 2002 · Verified 9 Jul 2026.

Options

  1. A

    +3.4eV+3.4 \mathrm{eV}+3.4eV

  2. B

    +6.8eV+6.8 \mathrm{eV}+6.8eV

  3. C

    −13.6eV-13.6 \mathrm{eV}−13.6eV

  4. D

    +13.6eV+13.6 \mathrm{eV}+13.6eV

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