In the Bohr model of a hydrogen atom, the centripetal force is furnished by the Coulomb attraction between the proton and the electrons. If `a_(0)` is the radius of the ground state orbit, `m` is the mass and `e` is the charge on the electron and `e_(0)` is the vacuum permittivity, the speed of the electron is

Options

A. `(e ) /(sqrt(4piepsi_0a_0m))`
B. `(e ) /(sqrt(epsi_0a_0m))`
C. `(sqrt(4piepsi_0a_0m))/(e )`
D. zero

Answer

A

Explanation

Coulomb's force between the proton and the electron is
`(1)/(4piepsi_0) (e^2)/(r^2)`
and the C.P. force `= (mv^2)/(r )`
`therefore (mv^2)/(r ) = (1)/(4piepsi_0) (e^2)/(r^2)`
`therefore v^2 = (1)/(4piepsi_0) (e^2)/(mr)`
`therefore v = (e) /( sqrt(4piepsi_0mr))`
For the ground state of the hydrogen atom , `r=a_0`
`therefore v = (e)/(sqrt(4piepsi_0 ma_0))`