Using Rautherfold model of the atom, derive the expression for the total electron in hydrogen atom. What is the significance of total negative energy possedssed by the electron?
Using Bohr's postulates of the atomic model, derive the expression for radius of `n^(th)` electron orbit. Hence obtain the expression for Bohr's radius
This therory developed is applicable to Hydrogen atoms and ions having just one electron. Let us assume that the nucleus has a positive charge `Ze` (i.e. ther are `Z` protons in it) and electrons moves with a consant speed `V` along a circle of radius `r` with the centre at nucleus By Coulomb's law

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`