Question

By assuming Bohr's postulates derive an expression for radius of $n^{th}$ orbit of electron, revolving round the nucleus of hydrogen atom.

Answer 1

Let the electron of mass $m$ revolves around a nucleus (H-atom) in an orbit of radius $r_n$ with linear velocity $v_n$.

As the electron revolves in an stationary orbit, thus centrifugal force acting on the electron is balanced by the coulombic force i.e. $F_{centrifugal}=F_{coulomb}$

$\therefore$ $\frac{m{v}_n^2}{r_n}=\frac{k{e}^2}{r_n^2}$ where $k=\frac{1}{4\pi {ϵ}_o}=9\times {10}^9$

$⟹$ $r_n=\frac{k{e}^2}{m{v}_n^2}$

Also we use $m{v}_n{r}_n=\frac{nh}{2\pi }$

Eliminating $v_n$ from both equations, we get $r_n=\frac{k{e}^2}{m}.\frac{4{\pi }^2{m}^2{r}_n^2}{n^2{h}^2}$

$⟹$ $r_n=\frac{h^2{n}^2}{4{\pi }^{2}mk{e}^2}$

Putting $m=9.1\times {10}^{-31}kg$, $h=6.626\times {10}^{-34}Js$ and $e=1.6\times {10}^{-19}C$

We get radius of $n^{th}$ Bohr orbit $r_n=0.529$ $n^2$ $A^o$