Using Bohr's postulates of the atomic model. Derive the expression for radius of nth electron orbit, thus obtaining the expression for Bohr's radius.

Explanation

The electrostatic force of attraction `Fe` between the revolving electrons and the nuculeus provide requisite centripetal force `(Fe)` to keep them in their orbits
`Fc = Fe`
`(m V^(2))/(r ) = (1)/(4 pi epsilon_(0)) (e^(2))/(r^(2))`
Thus the relation between the orbit radius and electron velocity is
`R = (e^(2))/(4 pi epsilon_(0) m V^(2))`
`K.E = (1)/(2) m v^(2) = (e^(2))/(8 pi epsilon_(0) R)`
`U = -(e^(2))/(4pi epsilon_(0) R)`
The `-ve` sign in `U` signifies that the lectrostatics Force is in the `-r` direction
Total energy `E = K+U = (e^(2))/(8 pi epsilon_(0) r) - (e^(2))/(4 pi epsilon_(0) r)` `= (-e^(2))/(8pi epsilon_(0) r)`
Total enerfy is negative. This implies that electron is bound to the atom.
OR
`F = (Ze^(2))/(4pi epsilon_(0) r^(2))`
`F = (Ze^(2))/(4pi epsilon_(0) r^(2)) = (m V^(2))/(r )` (centripetal force)
`R = (Ze^(2))/(4 pi epsilon_(0) m V^(2))` ____________(i)
From, Bohr's qunantization rule
`m v r = (n h)/(2pi)` __________(ii)
Eliminating `r` from (i) and (ii) we get,
`V = (ze^(2))/(2 epsilon_(0) h n)`
Substituing this in (ii)
`r = (epsilon_(0) h^(2) n^(2))/(pi m Z e^(2))`
For Bohr's radius `n = 1 , Z = 1`
`r_(0) = 0.529 Å`