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The `lambda` of `H_(alpha)` line of the Balmer series is `6500 Å` What is the `lambda` of `H_(beta)` line of the Balmer series |
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Answer» For `H_(alpha)`line of the balmer series `n_(1) = 2,n_(2) = 3` For `H_(beta)`line of the balmer series `n_(1) = 2,n_(2) = 4` `1/(lambda_(H_(alpha))) = R_(H) [(1)/(2^(2)) - (1)/(3^(2))]`……..(i) ` and `1/(lambda_(H_(beta))) = R_(H) [(1)/(2^(2)) - (1)/(4^(2))]`……..(ii) By equation (i) and (ii) we get `:. (lambda_(beta))/(lambda_(alpha))= ((1)/(4) - (1)/(9))/((1)/(4) - (1)/(16))` `:. lambda_(beta) = lambda_(alpha) xx [(80)/(108)] = 6500 xx (80)/(108) = 4814.8 Å` |
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