Question
Using Bohr’s postulates, derive the expression for the orbital period of the electron moving in the $n^{th}$ orbit of hydrogen atom.
✓
Answer
$mvr=\frac{nh}{2\pi}\text{ }\text{ }\text{ }\text{ }\text{ }\dots\text{Bohr postulate}$
Also, $\frac{mv^2}{r}=\frac{1}{4\pi\in_0}\frac{e^2}{r^2}$
$\Leftrightarrow \text{ }mv^2r=\frac{e^2}{4\pi\in_0}$
$\therefore\text{ }v=\frac{e^2}{4\pi\in_0}\times\frac{2\pi}{nh}=\frac{e^2}{2\in_0nh}$
$T=\frac{2\pi r}{v}=\frac{2\pi mvr}{mv^2}$
$=\frac{2\pi\big(\frac{nh}{2\pi}\big)}{m\big(\frac{e^2}{2\in_0nh}\big)^2}$
$=\frac{4n^3h^3\in_{0}^2}{me^4}$
Also, $\frac{mv^2}{r}=\frac{1}{4\pi\in_0}\frac{e^2}{r^2}$
$\Leftrightarrow \text{ }mv^2r=\frac{e^2}{4\pi\in_0}$
$\therefore\text{ }v=\frac{e^2}{4\pi\in_0}\times\frac{2\pi}{nh}=\frac{e^2}{2\in_0nh}$
$T=\frac{2\pi r}{v}=\frac{2\pi mvr}{mv^2}$
$=\frac{2\pi\big(\frac{nh}{2\pi}\big)}{m\big(\frac{e^2}{2\in_0nh}\big)^2}$
$=\frac{4n^3h^3\in_{0}^2}{me^4}$
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