Using the formula for the radius of nth orbit `r_(n)=(n^(2)h^(2)epsi_(0))/(pi mZe^(2))` derive an expression for the total energy of electron in `n^(th)` Bohr's orbit.

Explanation

The atomic model of Bohr is shown in the figure.

Let mass of electron m, charge e, linear speed in `n^(th)` orbit `v_(n)` and orbital radius `r_(n)`.
Positive charge on nucleus Ze, where Z = atomic number of element.
The necessary centripetal force is provided by Colombian attractive force between an electron and the positive charge of the nucleus.
`:.(mv_(n)^(2))/(r_(n))=((Ze)(e))/(4pi epsi_(0)r_(n)^(2))`
`:.(1)/(2)mv_(n)^(2)=(Ze^(2))/(8 pi epsi_(0)r_(n))`
`:.` Kinetic energy `K=(Ze^(2))/(8pi epsi_(0)r_(n)) " "....(1)`
And potential energy.
`U=(kq_(1)q_(2))/(r_(n))`
`:.U=-((Ze)(e))/(4pi epsi_(0) r_(n))=-(Ze^(2))/(4pi epsi_(0)r_(n))....(2)`
`rArr` Total energy of electron,
`E_(n)` = kinetic energy K + potential energy
`=(Ze^(2))/(8 pi epsi_(0)r_(n))-(Ze^(2))/(4pi epsi_(0)r_(n))`[ `:.` from equation (1) and (2)]
`=-(Ze^(2))/(8pi epsi_(0)r_(n))`
Now putting `r_(n)=(n^(2)h^(2)epsi_(0))/(pi m Ze^(2))`
`E_(n)=-(Ze^(2))/(8pi epsi_(0))xx(pi m Ze^(3))/(n^(2)h^(2)epsi_(0))`
`:. E_(n)=(mZ^(2)e^(4))/(8n^(2)h^(2)epsi_(0)^(2)) " "...(3)`
`:. E_(n) prop -(Z^(2))/(n^(2))`[ `:.` All other terms are contant]
For hydrogen atom Z=1.
`E_(n)=(me^(4))/(8 n^(2)h^(2) epsi_(0)^(2)) " "...(4)`
which is the total energy of electron in the `n^(th)` orbit of atom. It is assumed that deriving this formula the electronic orbits are circular.
Putting the accepted value of m, e, h and €, in equation (4),
`E_(n)=(2.18xx10^(-18))/(n^(2))J`
but `1.6xx10^(-19)KJ=1eV`
`:.E_(n)=-(2.18xx10^(-18))/(n^(2)xx1.6xx10^(-19))eV`
`=-(13.6)/(n^(2))eV [ "where" (me^(4))/(8h^(2) epsi_(0)^(2))=3.6eV]`
The negative sign of the total energy indicates that electron is bound with nucleus.