Obtain an expression for the total energy of an electron in the `n^(th)` orbit of hydrogen atom in terms of absolute constants.

Explanation

The radius of `n^(th)` permitted orbit of the electron is given by `r=(epsilon_(0)n^(2)h^(2))/(pimZe^(2))` . Substituting this value of r in the equation, `E_(n)=-(Ze^(2))/(8piepsilon_(0)r)` we get,
`E_(n)=-(Ze^(2))/(8piepsilon_(0)r)xx(pimZe^(2))/(epsilon_(0)n^(2)h^(2))`
i.e., `E_(n)=(-Ze^(2))/(8epsilon_(0)^(2)n^(2)h^(2))` for hydrogen like atoms.
For hydrogen atom, put Z= 1
`:.` 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))=-13.6eV`