Derive an expression for the radius of `n^(th)` Bohr's orbit of hydrogen atom hence write the expression for the radius of first orbit of hydrogen atom.

Explanation

Consider an atom of atomic number Z. The nucleus on its nucleus is +Ze . Let an electron of mass m and charge '-e' revolve round the nucleus in a circular orbit.
Let an Let v be its velocity .
The coulomb's electrostatic force of attraction between the electron and the nucleus is
`=(1)/(4piepsilon_(0))[(Ze^(2))/(r^(2))]`
This provides the centripetal force `(mv^(2))/(r)` needed for the orbital motion of the electron .
`(mv^(2))/(r)=1/(4piepsilon_(0))[(Ze^(2))/(r^(2))]`
`mv^(2)r=(Ze^(2))/(4piepsilon_(0))" "...(1)`
From Bohr's Quantization rule, for the `n^(th)` orbit,
we have, `mvr=(nh)/(2pi) " "...(2)`
squaring equation (2)
`m^(2)v^(2)r^(2)=(n^(2)h^(2))/(4pi^(2))`
Dividing equation (3) by equation(1)
`(m^(2)v^(2)r^(2))/(mv^(2)r)=(n^(2)h^(2))/(piZe^(2))xx(4piepsilon_(0))/(Ze^(2))`
i.e., `mr=(epsilon_(0)n^(2)h^(2))/(piZe^(2))`
`:. r = (epsilon _(0)n^(2)h^(2))/(pimZe^(2))`.
For `H_(2)` atom put Z = 1 and `r=(epsilon_(0)n^(2)h^(2))/(pime^(2))` For I orbit , put n = 1
i.e., `r=epsilon_(0)h^(2)//pime^(2)` for the I orbit of hydrogen atom.