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Question

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


Solution

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. $$F_{centrifugal} = F_{coulomb}$$
$$\therefore$$ $$\dfrac{mv_n^2}{r_n} = \dfrac{ke^2}{r_n^2}$$ where $$k = \dfrac{1}{4\pi \epsilon_o} = 9\times 10^9$$
$$\implies$$ $$r_n = \dfrac{ke^2}{mv_n^2}$$
Also we use $$mv_nr_n = \dfrac{nh}{2\pi}$$
Eliminating $$v_n$$ from both equations, we get  $$r_n = \dfrac{ke^2}{m}.\dfrac{4\pi^2m^2 r_n^2}{n^2 h^2}$$
$$\implies$$ $$r_n = \dfrac{h^2 n^2}{4\pi^2 m ke^2} $$
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  $$r_n = 0.529$$ $$ n^2$$ $$A^o$$

658201_623497_ans_8674599dd41347069d62e9bbdb08dec9.png

Physics

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