InterviewSolution
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InterviewSolution
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What hydrogen-like ion has the wavelength difference the first lines of Balmer and Lyman series equal to `59.3 nm`? `R_(H) = 109.678 cm^(-1)` |
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Answer» We have `(1)/(lambda) = R_(H)Z^(2) [(1)/(n_(1)^(2)) - (1)/(n_(2)^(2))]`For first line of Balmer series `(1)/(lambda_(B)) =R_(H).Z^(2) [(1)/(2^(2)) - (1)/(3^(2))] = (5)/(36) xx R_(H) xx Z^(2)` or `lambda_(B) = (36)/(5.R_(H).Z^(2))`.......(i) For first line of Lyman series `(1)/(lambda_(L)) =R_(H).Z^(2) [(1)/(1^(2)) - (1)/(2^(2))] = (3)/(4) xx R_(H) xx Z^(2)` or `lambda_(L) = (4)/(3R_(H).Z^(2))`.....(ii) Given `lambda_(B) - lambda_(L) = 59.3 xx 10^(-7) cm ` or `(36)/(5R_(H).Z^(2)) -(4)/(3R_(H).Z^(2))= 59.8 xx 10^(-7)` `(1)/(5R_(H).Z^(2))[7.2 - 1.333] = 59.3 xx 10^(-7)` or `Z^(2) = (5.867)/(R_(H) xx 59.3 xx 10^(-7))` `= (5.867)/(109678 xx 59.8 xx 10^(-7))` `:. Z = 3` `:. "H-like atom is" Li^(2+)` |
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