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

The correct option is 3.4 eV.

To understand why this is correct, we can look at the relationship between the total energy, kinetic energy, and potential energy of an electron orbiting in a hydrogen-like atom.

According to Bohr's model of the hydrogen atom, the total energy ($E$) of an electron in a given orbit is the sum of its kinetic energy ($KE$) and its potential energy ($PE$):

$E=KE+PE$

From the electrostatic force providing the necessary centripetal force for circular motion, we have:
$KE=\frac{1}{2}\frac{kZ{e}^2}{r}$
And the potential energy is:
$PE=-\frac{kZ{e}^2}{r}$

Substituting these into the total energy equation gives:
$E=\frac{1}{2}\frac{kZ{e}^2}{r}-\frac{kZ{e}^2}{r}=-\frac{1}{2}\frac{kZ{e}^2}{r}$

Comparing this with the expression for kinetic energy, we find a direct relation:
$KE=-E$

This equation states that the kinetic energy of the electron in any orbit is equal in magnitude to its total energy, but is positive in sign.

Given:
Total energy of the first excited state ($E$) = –3.4 eV

Therefore, the kinetic energy ($KE$) in the same orbit is:
$KE=-(-3.4\text{ eV})=3.4\text{ eV}$