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When an electric discharge is passed through hydrogen gas, the hydrogen molecules dissociate to produce excited hydrogen atoms. These excited atoms emit electromagnetic radiation of discrete frequencies which can be given by the general formula bar(v) = 109677 [(1)/(n_(i)^(2))- (1)/(n_(2)^(f))] What points of Bohr's model of an atom can be usedto arrive at this formula ? Based on these points derive the above formula giving description of each step and each term. |
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Answer» Solution :Two important points of Bohr's model that can be used to derive the given formula are: (i) Electrons revolve AROUND the nucleus in stationary states, i.e., orbits with fixed values of energy (quantized values). (ii) Energy is EMITTED or absorbed only when electron jumps from one orbit to another Derivation. Step (i) The energy of the electron in the nth stationary state is given by `E_(n) = - (2pi^(2) me^(4))/(n^(2)h^(2))` where m = mass of electron, e= charge on electron, h = Planck's constant Step (ii). When th electron jumps from outer stationay state `n_(2)` to inner stationary state `n_(1)`, the different of energy `(Delta E)` is emitted, i.e., `Delta E = E_(2) - E_(1) = - (2pi^(2) me^(4))/(n_(2)^(2) h^(2)) - (-(2pi^(2) me^(4))/(n_(1)^(2) h^(2))) = (2pi^(2) me^(4))/(h^(2)) ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))` Step (iii), `Delta E = hv = h(c)/(LAMDA) = hc BAR(v)` `:. bar(v) = (Delta E)/(hc) = (2pi^(2) me^(4))/(ch^(3)) ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))` Substituting the value of the constant, `pi, m, e, c and h` in CGS units, we get `bar(v) = 109677 cm^(-1) ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))` |
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