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

`(1)/(2)Ke (1+2ln((nh)/(2pisqrt(kema^(2)))))`

Explanation

`|F|=(delU)/(deltar)=(Ke)/(r )=(mv^(2))/(r )`
`rArr mv^(2)=ke`.....(i)
Also `mvr=(nh)/(2pi)`....(20
`:. c=sqrt((Ke)/(m))` and `r=(nh)/(2pim)sqrt((m)/(Ke))`
`E_(n)=KE+PE`
`(1)/(2)mv^(2)+U=(1)/(2)ke+ke ln((nh)/(2pia)sqrt((1)/(kem)))`
`=(1)/(2)ke (1+2l n(nh)/(2pisqrt(kema^(2))))`