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=((nh)/(2pi))(1)/(6(Ke^(2))^(2)m^(3))`

Explanation

`F=(delU)/(delr)=(3Ke^(2)r^(-4))/(3)=Ke^(2)r^(-4)`
This froce provide centipetal force so
`F=(mv^(2))/(r )=(Ke^(2))/(r^(4))`
`K=(1)/(2)mv^(2)=(1)/(2)(Ke^(2))/(r^(3))`
`E_("total")=("Potential"+"Kinetic")=U+K=(-Ke^(2))/(3r^(3))+(Ke^(2))/(2r^(3))=(Ke^(2))/(6r^(3))`.....(i)
Now, `mv^(2)=(Ke^(2))/(r^(3))`
or `m^(2)v^(2)=(Kme^(2))/(r^(3))`
`mvr=sqrt((Kme^(2)r^(2))/(r^(3)))`
`mvr=sqrt((Kme^(2))/(r ))`
Now, quantization of angular momentum gives
`L=(nh)/(2pi)=mvr`
`(nh)/(2pi)=sqrt((Kme^(2))/(r ))`
`r=(Kme^(2))/((nh//2pi)^(2))`
`r^(3)=((Kme^(2))^(3))/((nh//2pi)^(6))`
putting `r^(3)` in equ. (i)
`T=(Ke^(2))/(6)((nh)/(2pi))^(6)((1)/(Ke^(2)m))^(3)`