The relation between the orbit radius and the electron velocity for a dynamically stable orbit in a hydrogen atom is (where, all notations have their usual meanings)A. `v=sqrt((4piepsilon_0)/(me^2r))`B. `r=sqrt((e^2)/(4piepsilon_0v))`C. `v=sqrt((e^2)/(4piepsilon_0mr))`D. `r=sqrt((ve^2)/(4piepsilon_0m))`

The relation between the orbit radius and the electron velocity for a

1.	The relation between the orbit radius and the electron velocity for a dynamically stable orbit in a hydrogen atom is (where, all notations have their usual meanings)A. `v=sqrt((4piepsilon_0)/(me^2r))`B. `r=sqrt((e^2)/(4piepsilon_0v))`C. `v=sqrt((e^2)/(4piepsilon_0mr))`D. `r=sqrt((ve^2)/(4piepsilon_0m))`
Answer» Correct Answer - C In hydrogen atom electrostatic force of attraction `(F_e)` between the revolving electrons and the nucleus provides the requisite centripetal force `(F_c)` to keep them in their orbits. Thus, `F_e=F_(c) therefore (mv^2)/(r)=(1)/(4piepsilon_0)(e^2)/(r^2)` or `v^2=(e^2)/(4piepsilon_0mr)rArrv=sqrt((e^2)/(4piepsilon_0mr))`