K.E = 1/2 mv²

= ${\left( {{{\pi {e^2}} \over {nh}}} \right)^2} \times 2m$ [ $\because$ v = ${{2\pi {e^2}} \over {nh}}$]

$\therefore$ Total energy = E_n = ${{ - 2{\pi ^2}m{e^4}} \over {{n^2}{h^2}}}$

= -${\left( {{{\pi {e^2}} \over {nh}}} \right)^2}$ $\times$ 2m = -K.E

$\therefore$ K.E = - E_n

Energy of first excited state is -3.4 eV

$\therefore$ Kinetic energy of the same orbit (n = 2) will be +3.4 ev