Question

The magnitude of potential energy of an electron in the second Bohr orbit of a hydrogen atom is:

[$a_0$ is Bohr radius].

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

Answer 1

The correct option is B $\frac{h^2}{16{\pi }^{2}ma{}^2{}_0}$

$\because mvr=\frac{2h}{2\Pi }.....[n=2$ for $2nd$ orbit$]$

$\therefore V=\frac{h}{m\Pi r}$

$\Rightarrow KE=\frac{1}{2}m{v}^2=\frac{1}{2}\times m\times \frac{h^2}{{\Pi }^2{m}^2{r}^2}=\frac{h^2}{2{\Pi }^{2}m{r}^2}$

$\Rightarrow [P.E.]=2KE=\frac{h^2}{{\Pi }^{2}m{r}^2}....[PE=\frac{K{e}^2}{r}$ & $\frac{m{v}^2}{r}=\frac{K{e}^2}{r^2}]$ $\therefore PE=2KE$

$r$= radius of $2nd$ orbit,

$\because r=a_o\times \frac{n^2}{z}=a_o\times \frac{(2{)}^2}{1}=4a_o$

$\therefore \left[PE\right]=\frac{h^2}{16{\Pi }^{2}m{a}_o^2}$