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Q.
In hydrogen atom, energy of first excited state is $-3.4\, eV$. Then, KE of same orbit of hydrogen atom :-
AIPMTAIPMT 2002Structure of Atom
Solution:
$\because$ Total energy $\left(E_{n}\right)= KE + PE$
In first excited state $=\frac{1}{2} m v^{2}+\left[-\frac{Z e^{2}}{r}\right]$
$=+\frac{1}{2} \frac{Z e^{2}}{r}-\frac{Z e^{2}}{r}$
Energy of first excited state is $3.4 eV$
$\therefore KE =\frac{1}{2} \frac{Z e^{2}}{r} $
$=+3.4 \,e V$