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

polonijempuf

polonijempuf

Answered question

2022-12-18

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

Answer & Explanation

Ainsley Kirk

Ainsley Kirk

Beginner2022-12-19Added 8 answers

The ground state energy of hydrogen atom is -13.6 eV.
Ground state energy of hydrogen atom, E = -13.6 eV
This is the total energy of a hydrogen atom. Kinetic energy is equal to the negative of the total energy.
Kinetic energy =−E=−(−13.6)=13.6eV
Potential energy is equal to the negative of two times of kinetic energy.
Potential energy =−2×(13.6)=−27.2eV.

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