Supose the potential energy between electron and proton at a distance r is given by `-(Ke^(2))/(3r^(3))` . Applicatiojn of Bohr's theroy of hydrogen atom in this case shows that

Answer

`E_(n) = (-n^(2) h^(2))/(128 K^(2) e^(4) m^(3))`

Explanation

According to gives situation for an electron revolving in an `n^(th)` orbit, the potential energy is given as
`U = - (ke^(2))/(3 r_(n)^(3))` (i)
The contripeletal force on the electron at speed `v_(n)` then we have
`F = - (dU)/(dr) = (ke^(2))/(r_(n)^(3))` (ii)
If an `n^(th)` orbit electron revolve at speed `v_(n)` then we have
`mv_(H)^(2))/(r_(n) = (ke^(2))/(r_(n)^(4)) or mv_(n)^(2) = (ke^(2))/(r_(n)^(3))` (iii)
From Eqs. (iii) and (iv), we have `v_(n) = (nh)/(2 pi mr_(n)) and m ((nh)/(2 pi mr_(n))) ^(2) = (ke^(2))/(r_(n)^(3))`
or `r_(n) = (4 pi^(2) Ke^(2) m)/(n^(2) h^(2))` and `v_(n) = (n^(3) h^(3))/(8 pi^(2) Km^(2) e^(2))`
Now energy in `n^(th)` orbit is equal to negative of `KE`.
`E_(n) = (-1)/(2)mv_(n)^(2) = - (ke^(2))/(2 r_(n)^(3))`
` = - (1)/(2) ke^(2) ((n^(2) h^(2))/(4 pi^(2) Ke^(2) m)) ^(3)`
`= (-n^(2) h^(2))/(128 pi^(2) K^(2) e^(2) m)`