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The wavelength of the first line in the hbalmer series is `656 nm ` .Calculate the wavelength of the second line and the limeting line in the Ralmerseries |
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Answer» Wavelength of the second line According to Rydbery formula `(1)/(lambda) = R_(H)((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))` For the first line `n_(1) = 2,n_(2)= 3 and lambda = 656 nm` `(1)/(656) = R_(H) ((1)/(2^(2)) - (1)/(3^(2)))` ` R_(H) = ((1)/(4) - (1)/(9)) = R_(H) xx (5)/(36) = (5R_(H))/(36)`....(i) For the second line `n_(1) = 2,n_(2) = 4` `(1)/(lambda) = R_(H) ((1)/(2^(2)) - (1)/(4^(2))) = R_(H)((1)/(4) - (1)/(16))` `=R_(H) xx(3)/(16) = (3R_(H))/(16)`.....(ii) Dividing equation (i) by equation (ii),we get `(lambda)/(656) = (5)/(36) xx (16)/(3)`M `or lambda = (656 xx 5xx 16)/(36 xx 3) = (52450)/(108) = 485.9 m` IN balmer series the limiting line coreesponding to transition from `n = oo "to" n = 2` therefore `n_(1) = 2 and n_(2) = oo` `therefore (1)/(lambda) = R_(H)[(1)/(n_(1)^(2)) - (1)/(n_(2)^(2))]` `(1)/(lambda) = 109677 [(1)/(2^(2)) - (1)/(oo^(2))] = 109677 xx (1)/(4) cm^(-1) = 27419.25 cm^(-1)` Wavelength lambda = `(1)/(27419.25) = 3.47 xx 10^(-5) = 3647 Å` |
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