Derive an experession for the total energy of an electron in stationary state of hydrogen atom. Assuming the expression for the radius.

Explanation

Consider an electron of mass m and charge -e revolving around 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_0 r))` ...(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)` .
The radius of `n^(th)` permitted orbit of the electron is given by `r=(epsilon_0 n^2 h^2)/(pi m Ze^2) `
Substituting this value of r in the equation ,
`E_n = -(Ze^2)/(8 pi epsilon_0 r)`
we get , `E_n =-(Ze^2)/(8pi epsilon_0) xx(pimZe^2)/(epsilon_0 n^2 h^2)`
i.e., `E_n=(-Z^2 me^4)/(8epsilon_0^2n^2h^2)` for hydrogen like atoms.
For hydrogen atom , put Z=1
`therefore` Total energy of the electron in the `n^(th)` orbit of hydrogen atom is
`E_n=(-me^4)/(8epsilon_0^2 n^2 h^2)`
Note : The Total energy of the electron in the first orbit is
`E_1 = (-me^4)/(8epsilon_0^2 h^2)` =-13.6 eV