Obtain the equations of frequency of radiation and wave number when electron makes transits from the high energy state to the lower energy state in hydrogen atom.

Explanation

According to the third postulate of Bohr.s model when an atom makes a transition from the higher energy state with quantum number `n_(i)` to the lower energy state with quantum number ng`n_(f) (n_(i) gt n_(f))`. then a photon of energy is emitted equal to their energy difference.
State, Energy of electron in `n_(i)` state,
`E_(n_(i))=-(me^(4))/(8 epsi_(0)^(2)h^(2)n_(i)^(2)) " ".....(1)`
Energy of electron in `n_(f)` state,
`E_(n_(f))=-(me^(4))/(8epsi_(0)^(2)h^(2)n_(k)^(2)) " "...(2)`
`E_(n_(i)) gt E_(n_(f))`
`:. E_(n_(f))-E_(n_(f))=-(me^(4))/(8 epsi_(0)^(2)h^(2)n_(i)^(2))-(-(me^(4))/(8 epsi_(0)^(2)h^(2)n_(f)^(2))) hv_(if)= (me^(4))/(8 epsi_(0)^(2)h^(2))[(1)/(n_(f)^(2))-(1)/(n_(i)^(2))]` this is the Rydbergy formula for the spectrum of hydrogen atom
`:.v_(if)=(me^(4))/(8 epsi_(0)^(2)h^(3))[(1)/(n_(f)^(2))-(1)/(n_(i)^(2))]`
But `c= lambda_(if) v_(if)`
`:.v_(if)=(c)/(lambda_(if))`
`:.(c)/(lambda_((if)))=(me^(4))/(8 epsi_(0)^(2)h^(2)c)[(1)/(n_(f)^(2))-(1)/(n_(i)^(2))]`
`:.(1)/(lambda_(if))=(me^(4))/(8epsi_(0)^(2)h^(3)c) [(1)/(n_(f)^(2))-(1)/(n_(i)^(2))]`
which is formula for the wave number of emitted photon but Rydberg constant R is `(me^(4))/(8 epsi_(0)^(2)h^(3)c)` and putting the accepted values of each simplifying gives theoretical value `R=.03xx10^(7)m^(-1)`.
The value of R from Balmer experiment it is R= `1.097xx10^(7)m^(-7)`
`:. (1)/(lambda_(if))=R[(1)/(n_(r)^(2))-(1)/(n_(i)^(2))]`
where `n_(f)` = quantum number of lower state.
`n_(i)=` quantum number of higher state.
The similarity of the theoretical and experimental values of Rydberg constant supported of the Bohr.s model.