Question
Using Bohr’s postulates of the atomic model, derive the expression for radius of $n^{th}$ electron orbit. Hence obtain the expression for Bohr’s radius.
✓
Answer
For the electron, we have
Bohr’s Postulate $(\text{mvr} = \frac{\text{nh}}{2\pi})$
$\frac{\text{mv}^{2}}{\text{r}} = \frac{1}{4\pi\in_{o}}\frac{\text{ze}^{2}}{\text{r}^{2}}$
and mvr $ = \frac{\text{nh}}{2\pi}$
$\therefore\text{m}^{2}\text{v}^{2}\text{r}^{2} = \frac{\text{n}^{2}\text{h}^{2}}{4\pi^{2}}$
and $mv^2r = \frac{1}{4\pi\in_{o}}\text{ze}^{2}$
$\therefore\text{r} = \frac{\in_{o}\text{n}^{2}\text{h}^{2}}{\pi\text{ze}^{2}\text{m}}$
Bohr’s radius $($for $n = 1) = \in_{o } \text{h}^{2}/\pi\text{ze}^{2}\text{m}.$
Bohr’s Postulate $(\text{mvr} = \frac{\text{nh}}{2\pi})$
$\frac{\text{mv}^{2}}{\text{r}} = \frac{1}{4\pi\in_{o}}\frac{\text{ze}^{2}}{\text{r}^{2}}$
and mvr $ = \frac{\text{nh}}{2\pi}$
$\therefore\text{m}^{2}\text{v}^{2}\text{r}^{2} = \frac{\text{n}^{2}\text{h}^{2}}{4\pi^{2}}$
and $mv^2r = \frac{1}{4\pi\in_{o}}\text{ze}^{2}$
$\therefore\text{r} = \frac{\in_{o}\text{n}^{2}\text{h}^{2}}{\pi\text{ze}^{2}\text{m}}$
Bohr’s radius $($for $n = 1) = \in_{o } \text{h}^{2}/\pi\text{ze}^{2}\text{m}.$
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