Question

The kinetic energy of an electron in the $2^{nd}$ orbit of bohr's model of a hydrogen atom is: ($a_0$ is Bohr radius)

• A. $\frac{h^2}{4{\pi }^{2}m{a}_0^2}$
• B. $\frac{h^2}{16{\pi }^{2}m{a}_0^2}$
• C. $\frac{h^2}{32{\pi }^{2}m{a}_0^2}$
• D. $\frac{h^2}{64{\pi }^{2}m{a}_0^2}$

Answer 1

The correct option is C $\frac{h^2}{32{\pi }^{2}m{a}_0^2}$

The nuclear attraction on the electron $=$ Centrifugal force

So,$\frac{kZ{e}^2}{r^2}=\frac{m{v}^2}{r}$

$v^2=\frac{kZ{e}^2}{mr}$ ...(i)

Angular momentum, $mvr=n\frac{h}{2\pi }$

$v^2=\frac{n^2{h}^2}{4{\pi }^2{m}^2{r}^2}$ ...(ii)

Comparing (i) and (ii)

$\frac{kZ{e}^2}{mr}=\frac{n^2{h}^2}{4{\pi }^2{m}^2{r}^2}$

$r=\frac{n^2}{Z}\left[\frac{h^2}{4{\pi }^{2}mk{e}^2}\right]=\frac{n^2{a}_0}{Z}$

$\therefore$ $v_n=\frac{hZ}{2\pi a_{0}n}$

$KE=\frac{1}{2}m{v}_n^2=\frac{h^2{Z}^2}{8{\pi }^{2}m{a}_0^2{n}^2}$

$KE(forn=2,Z=1)=\frac{h^2}{32{\pi }^{2}m{a}_0^2}$.

Hence, the correct option is $\text{C}$