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

Question

In hydrogen atom, energy of first excited state is -3.4 eV. Then, KE of same orbit of hydrogen atom :- (A) + 3.4 eV (B) +6.8 eV (C) -13.6 eV (D) + 13.6 eV

Tardigrade Admin · Accepted Answer

∵ Total energy (En)= KE + PE In first excited state =(1/2) m v2+[-(Z e2/r)] =+(1/2) (Z e2/r)-(Z e2/r) Energy of first excited state is 3.4 eV ∴ KE =(1/2) (Z e2/r) =+3.4 e V

Q. In hydrogen atom, energy of first excited state is $- 3.4 e V$ . Then, KE of same orbit of hydrogen atom :-

+ 3.4 eV

+6.8 eV

-13.6 eV

+ 13.6 eV

$∵$ Total energy $(E_{n}) = K E + PE$
In first excited state $= \frac{1}{2} m v^{2} + [- \frac{Z e ^{2}}{r}]$
$= + \frac{1}{2} \frac{Z e ^{2}}{r} - \frac{Z e ^{2}}{r}$
Energy of first excited state is $3.4 e V$
$∴ K E = \frac{1}{2} \frac{Z e ^{2}}{r}$
$= + 3.4 e V$

Q. In hydrogen atom, energy of first excited state is −3.4eV. Then, KE of same orbit of hydrogen atom :-