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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. |
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Answer» Solution :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 butRydberg 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. |
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