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In hydrogen atom, energy of first excited state is -3.4 eV. Find out KE of the same orbit of Hydrogen atom.

1. +3.4 eV
2. +6.8 eV
3. -13.6 eV
4. +13.6 eV

Correct answer: +3.4 eV

Solution

In the first excited state of a hydrogen atom, the total energy (E) is given as -3.4 eV. The kinetic energy (KE) is equal to the magnitude of the total energy but positive, as KE = -E. Thus, KE = +3.4 eV.

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