Question

Derivation for radius of the nth orbit of hydrogen atom

Answer 1

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 = e2/4πε0R2

The acceleration of the electron is given by v2/r. If m is the mass of the electron, then according to Newton’s law:

e2/4πε0r2 = mv2/R  ---------(3)

mv = [(1/4πε0) (m e2/R)]1/2  ----------------(4)

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

{[(1/4πε0) (m e2/R)]1/2}2 x R2 = n2h2/22π2

R = (4πε0)n2h2 / 4π2me2

For nth orbit,

Rn = ε0 n2h2 / πme2

Put in the values for hydrogen, you will get your desired answer!