Prove that `E_(n)=-13.6xxz^(2)/n^(2) eV//atom` for `n^(th)` orbit in single electron species.

Answer

The total energy of an electron revolving in a particular orbit is
`T.E.=K.E. + P.E.`
where `P.E.`=Potential energy, `K. E.` =Kinetic energy, `T.E.`=Total energy
The `K.E.` of an electron`=1/2 mv^(2)`
and the `P.E.` of an electron `=-(KZe^(2))/(r )`
Hence, `T.E. =1/2 mv^(2)-(KZe^(2))/(r )`
we know that, `(mv^(2))/(r )=(KZe^(2))/(r^(2))` or `mv^(2)=(KZe^(2))/(r )`
Substituting the value of `mv^(2)` in the above equation:
`T.E. =(KZe^(2))/(2r)-(KZe^(2))/(r )=-(KZe^(2))/(2r)`
So, `T.E. =-(KZe^(2))/(2r)`
substituting the value of 'r' in the equation of `T.E.`
Then `T.E.=-(KZe^(2))/(2)xx(4pi^(2)Ze^(2)m)/(n^(2)h^(2))=-(2pi^(2)Z^(2)e^(4)mK^(2))/(n^(2)h^(2))`
Thus, the total energy of an electrons in `n^(th)` orbit is given by
`T.E.=E_(n)=-(2pi^(2)me^(4)k^(2))/(h^(2))(z^(2)/n^(2)) ...(iv)`
Putting the value of `m, e, h` and `pi` we get the expression of total energy
`E_(n)=-13.6 Z^(2)/n^(2) eV//atom n uarr T.E. uarr ; Z uarr T.E. darr`