Assuming the expression for radius of the orbit, derive an expression for total energy of an electron in hydrogen atom.

Explanation

Consider an electron of mass m and charge -e revolving round the nucleus of an atom of atomic number Z in the `n^(th)` orbit of radius .r.. Let v be the velocity of the electron. The electron possesses potential energy , because , it is in the electrostatic field of the nucleus . The electron also possesses kinetic energy by virtue of its motion.
Potential energy of the electron is given by , `E_p`=(potential at a distance r from the nucleus )(-e)
`=1/(4piepsilon_0)[(Ze)/r](-e)`
`E_p=-(Ze^2)/(4piepsilon_0r)`...(1)
kinetic energy of the electron is given by , `E_k=1/2 mv^2`...(2)
From Bohr.s postulate,
`(mv^2)/r=1/(4piepsilon_0)[(Ze^2)/r^2]`
`therefore mv^2=(Ze^2)/(4piepsilon_0r)`
Substituting this value of `mv^2` in equation (2)
`E_k=1/2 ((Ze^2)/(4piepsilon_0r))`...(3)
Total energy of the electron revolving in the `n^(th)` orbit is given by `E_n=E_p+E_k`
`E_n=-(Ze^2)/(4piepsilon_0r)+1/2 ((Ze^2)/(4piepsilon_0r))`
Using (1) and (2)
`=(Ze^2)/(4piepsilon_0r)[(-1)/1+1/2]`
`=(Ze^2)/ (4piepsilon_0r)[-1/2]`
`therefore E_n=-(Ze^2)/(8piepsilon_0r)`