Question

In hydrogen atom, energy of first excited state is $–3.4\text{eV}$. Then find out K.E. of same orbit of hydrogen atom.

• A. $+3.4\text{eV}$
• B. $+6.8\text{eV}$
• C. $-13.6\text{eV}$
• D. $+13.6\text{eV}$

Answer 1

The correct option is A $+3.4\text{eV}$

Kinetic energy $=\frac{1}{2}m{v}^2=(\frac{\pi e^2}{nh}{)}^2\times 2m$
$(\text{As}v=\frac{2\pi e^2}{nh})$
$\text{Total energy}{E}_n=-\frac{2{\pi }^{2}m{e}^4}{n^2{h}^2}=-(\frac{\pi e^2}{nh}{)}^2\times 2m=-K.E.$
$\therefore K.E.=-E_n$
Energy of first excited state is $–3.4\text{eV}$
$\therefore$
Kinetic energy of same orbit $(n=2)$ will be $+3.4\text{eV}$.