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Calculate the energy emitted when electrons of `1.0 g`1 of hydrogen transition giving spectrum lines of the lowest in the visible regain of its atomic spectrum `R_(H) = 1.1 xx 10^(7) m^(-1) , c= 3 xx 10^(8) m s^(-1)` and `h = 6.62 xx 10^(-34) J s` |
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Answer» For visible line spectrum, i.e. Balmer series `n_(1) = 2` Also for minimum energy transion `n_(2) = 3` so `:. (1)/(lambda) = R_(H)((1)/(n_(1)^(2))- (1)/(n_(2)^(2)))` `therefore1/lambda=R_(H)(1/2^(2)-1/3^(2))` `= 1.1 xx 10^(7)((1)/(4)-(1)/(9)) = 1.1xx 10^(7) xx (5)/(36)m^(-1)` `:. lambda = 6.55 xx 10^(-7) m` `E = (hc)/(lambda)= (6.62 xx 3.0 xx 10^(8))/(6.55 xx 10^(7)) = 3.037 xx 10^(-19) J` If N elecytron show the transition in `1 g of H` atom , then energy released `= E xx N` `= 3.03 xx 10^(-19) xx 6.023 xx 10^(23)` `= 18.29 xx 10^(6) J = 182.5 J` |
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