What is the derivation of the radius of the n^{th} orbit in bohr's model?

the formula is R_{n} = (4πε_{0})n^{2}h^{2 }/ 4π^{2}me^{2}

According to Bohr’s model, the electron revolves revolve in stationary orbits where the angular momentum of electron is an integral multiple of h/2π.

mvr = nh/2π ------------------(1)

Here, h is Planck's constant.

Now, when an electron jumps from an orbit of higher energy E2 to an orbit of lower energy E1, it emits a photon. The energy of the photon is E2-E1.The relation between wavelength of the emitted radiation and energy of photon is given by the Einstein - Planck equation.

E2-E1= hν = hc/λ -------------(2)

For an electron of hydrogen moving with a constant speed v along a circle of radius R with the center at the nucleus, the force acting on the electron according to Coulomb’s law is:

F = e^{2}/4πε_{0}R^{2}

The acceleration of the electron is given by v^{2}/r. If m is the mass of the electron, then according to Newton’s law:

e^{2}/4πε_{0}r^{2} = mv^{2}/R ---------(3)

mv = [(1/4πε_{0}) (m e^{2}/R)]^{1/2} ----------------(4)

From equation (1) and (4) we get,

{[(1/4πε_{0}) (m e^{2}/R)]^{1/2}}^{2} x R^{2} = n^{2}h^{2}/2^{2}π^{2}

R = (4πε_{0})n^{2}h^{2 }/ 4π^{2}me^{2}

For nth orbit,

R_{n} = ε_{0 }n^{2}h^{2 }/ πme^{2}

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