Question

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

• A. +3.4 eV
• B. + 6.8 eV
• C. -13.6 eV
• D. +13.6 eV

Answer 1

Answer: (A)

+3.4 eV

Answer 2

$\because$ Total energy $\left(E_{n}\right)= KE + PE$ In first excited state, $=\frac{1}{2} m v^{2}+\left[-\frac{Z e^{2}}{r}\right]$ $=+\frac{1}{2} \frac{Z e^{2}}{r}-\frac{Z e^{2}}{r}$ $-3.4\, eV =-\frac{1}{2} \frac{Z e^{2}}{r}$ $\therefore KE =\frac{1}{2} \frac{Z e^{2}}{r}=+3.4\, eV$