Derive the energy expression for hydrogen atom 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/(4piepsi_(0))((+Ze)(-e))/(r_(n)^(2))hatr`
`=-1/(4piepsi_(0))(Ze^(2))/(r_(n)^(2))hatr`
(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/(4piepsi_(0))(Ze^(2))/(r_(n)^(2))=(mv_(n)^(2))/r_(n)`
`r_(n)=(4piepsi_(0)(mv_(n)r_(n))^(2))/(Zme^(2))`
From Bohr.s assumption, the angular momentum quantization condition
`mv_(n)r_(n)=nh`
`thereforer_(n)=(4piepsi_(0)(mv_(n)r_(n))^(2))/(Zme^(2))`
`r_(n)=(4piepsi_(0)(nh)^(2))/(Zme^(2))=(4piepsi_(0)n^(2)h^(2))/(Zme^(2))`
`r_(n)=((epsi_(0)h^(2))/(pime^(2)))n^(2)/Z" "(thereforeh=h/(2pi))`
where `ninNN`. Since, `epsi_(0),h,eandpi` 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))=0.529Å`.