Question

In hydrogen atom, energy of first excited state is -3.4eV. Then find out KE of same orbit of hydrogen atom

Answer 1

KE is the energy with which the electron moves in the orbit hence The energy is same but positive so answer is 3.4eV

Answer 2

Answer : The kinetic energy of same orbit of hydrogen atom is 3.4 eV.

Explanation :

For hydrogen atom,

The kinetic energy is equal to the negative of the total energy.

And the potential energy is equal to the twice of the total energy.

The first excited state energy of orbital = -3.4 eV

and The kinetic energy of same orbital = -(-3.4 eV) = 3.4 eV

Therefore, the kinetic energy of same orbit of hydrogen atom is 3.4 eV.