Use Bohr's postulates hydrogen atom to deduce the expression for the kinetic energy (K.E.) of the electron revolving in the $\mathrm{n}^{\text {th }}$ orbit and show that, K.E. $\frac{\mathrm{e}^2}{8 \pi \varepsilon_0 e_n}$, where $r_n$ is the radius of the $n^{\text {th }}$ orbit. How is the orbit. How is the potential energy in the orbit related to the orbital radius orbit related to the orbital radius $r_n$ ?
CBSE OUTSIDE DELHI - SET 2 GUHAWATI 2015
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For an electron (mass ‘m’ and charge ‘e’) revolving in $n^{th} $ stable circular orbit of radius $'r_n’$, with velocity $v_{n,}$ in the hydrogen atom (z=1), we have
$\frac{\text{mv}^{2}_{n}}{\text{r}_{n}} = \frac{1}{4\pi\varepsilon_{0}}\frac{\text{e}^{2}}{\text{r}_{n}^{2}}$
$\therefore = \text{E}_{k} = \frac{1}{2}\text{mv}^{2}_{n} = \frac{\text{e}^{2}}{8\pi\in_{0}\text{r}_{n}}$
$\therefore\text{E}_{p} = \frac{1}{4\pi\varepsilon_{0}}\frac{(+\text{e})\times( - \text{e})}{\text{r}_{n}}$
$ =- \frac{\text{e}^{2}}{4\pi\varepsilon_{0}\text{r}_{n}}$
$\frac{\text{mv}^{2}_{n}}{\text{r}_{n}} = \frac{1}{4\pi\varepsilon_{0}}\frac{\text{e}^{2}}{\text{r}_{n}^{2}}$
$\therefore = \text{E}_{k} = \frac{1}{2}\text{mv}^{2}_{n} = \frac{\text{e}^{2}}{8\pi\in_{0}\text{r}_{n}}$
$\therefore\text{E}_{p} = \frac{1}{4\pi\varepsilon_{0}}\frac{(+\text{e})\times( - \text{e})}{\text{r}_{n}}$
$ =- \frac{\text{e}^{2}}{4\pi\varepsilon_{0}\text{r}_{n}}$
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