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

Options

A. energy in the nth orbit is proportional to `n^(6)`
B. energy is proportional to `m^(-3)` `(m = "mass of electron")`
C. energy in the nth orbit is proportional to `n^(-2)`
D. energy is proportional to `m^(3)` (m = "mass of electron")`

Answer

A, B

Explanation

`|F| = (d U)/(d t) = (K e^(2))/( r^(4))` (i)
`(K e^(2))/( r^(4)) = (mv ^(2))/( r)`
and `m v r = (n h)/(2 pi)`
by (i) and (iii),
`r = (K e^(2) 4 pi^(2))/(h^(2)) (m)/(n^(2)) = K_(1) (m)/(n^(2))`
Total energy =` (1)/(2) ("potential energy")`
`= (K e^(2))/(6 r^(3)) = (-Ke^(2))/(6(( K_(1)m)/(n^(2)))^(3)) = (Ke^(2) n^(2))/(6 K_(1)^(3) m^(3))`
Total energy`prop n^(6)`
Total energy prop `m^(-3)`
Therefore , (a) and (b) are correct.