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If R_(H) represents Rydberg constant, then the energy of the electron in the ground state of hydrogen atom is |
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Answer» `- (hc)/(R_(H))` `bar(v) = R_(H) ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))`....(i) According to Bohr model, for an electronic TRANSITION, `Delta E = - (2pi^(2) me^(4))/(h^(2)) ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))` But `Delta E = hv = (hc)/(LAMDA) = hc bar(v)` `:. hc bar(v) = - (2pi^(2) me^(4))/(h^(2)) ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))` or `bar(v) = - (2pi^(2) me^(4))/(ch^(3)) ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))`...(ii) COMPARING eqns. (i) and (ii) `R_(H) = - (2pi^(2) me^(4))/(ch^(3)) :. E_(1) = - (2pi^(2) me^(4))/(h^(2)) = - R_(H) ch` |
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