An electron is moving round the nucleus of a hydrogen atom in a circular orbit of radius `r`. The coulomb force `vec(F)` between the two is (where `k=(1)/(4piepsilon_(0)))`

Options

A. `K(e^(2))/(r^(2))hatr`
B. `-K(e^(2))/(r^(2))hatr`
C. `k(e^(2))/r r`
D. `-k(e^(2))/r r`

Answer

B

Explanation

Let charges on an electron and hydrogen nucleus are `q_(1)` and `q_(2)`. The Coulomb.s force between them at a distance r is
`F=1/ (4pi epsilon_(0)) (q_(1)q_(2))/(r^(2)) r`
putting `1/( 4 pi epsilon_(0)) =k ` (given)
`F=-k(q_(1)q_(2))/(r^(2))r`
Since the nucleus of hydrogen atom has one proton, so charge on nucleus is e i.e. `q_(2)=e` also `q_(1)=e` for electron.
So `F=-k(e.e.)/(r^(2)) r = -k(e^(2))/(r^(2)) r`
but `r=r/(|r|)=r/r`
Hence `F=-k(e^(2))/(r^(2)) . r/r =-k(e^(2))/(r^(3)) . r`