Derive the expression for the radius of the orbit of the electron and its velocity using Bohr atom model.

Explanation

(i) Consider an atom which contains the nucleus at rest and an electron revolving around the nucleus in a circular orbit of radius `r_n`. as shown in Figure.
(ii) Nucleus is made up of protons and neutrons. Since proton is positively charged and neutron is electrically neutral, the charge of a nucleus is purely the total charge of protons.

(iii) Let Z be the atomic number of the atom, then +Ze is the charge of the nucleus. Let -e be the charge of the electron. From Coulomb.s law, the force of attraction between the nucleus and the electron is
`vecF_("coloumb") = 1/(4 piepsi_(0)) ((+Z3)(-e))/r_n^2`
`=- 1/(4 piepsi_(0))(Ze^2)/r_n^2hatr`
(iv) This force provides necessary centripetal force `vecF_("centripetal") = (mv_n^2)/r_n hatr`
(v) where m be the mass of the electron that moves with a velocity `v_n`, in a circular orbit. Therefore,
`|vecF_("coloumb")|=|vecF_("centripetal")|`
`1/(4 piepsi_(0))(Ze^2)/r_n^2 = (mv_m^2)/r_n`,
`r_n = (4 piepsi_(0)(mv_nr_n)^2)/(Zme^2)`
From Bohr.s ssumption, the angular momentum quantization condition
`mv_n r_n = nh`
`:." "r_n = (4pi epsi_0(mv_nr_n)^2)/(Zme^2)`
`r_n = (4pi epsi_0(nh)^2)/(Zme^2) = (4piepsi_0 n^2h^2)/(Zme^2)`
`r_n = ((epsi_0 h^2)/(pime^2))n^2/Z" "( :.h= h/(2pi))`
Where n`inNN`. Since, `epsi_0`, h, e and `pi` are constants. Therefore, the radius of the orbit becomes
`r_n = a_0 n^2/Z`
where `a_0=(epsi_0 h^2)/(pime^2) = 0.529`.