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The ratio of maximum wavelength of Lyman and Balmer series in hydrogen emission spectra will be ...... |
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Answer» `(9)/(31)` `(1)/(lambda_(L))=R[(1)/(1^(2))-(1)/(N^(2))]` `(1)/(lambda_(L))=R[(1)/(1^(2))-(1)/(2^(2))]`For maximum wavelength n=3 `(1)/(lambda_(L))=R[(1)/(1)-(1)/(4)]=(3R)/(4)` `:.lambda_(L)=(4)/(3R)....(1)` For Balmer series `(1)/(lambda_(B))=R [(1)/(2^(2))-(1)/(3^(2))]` For maximum wavelength n = 3 `(1)/(lambda_(B))=R[(1)/(4)-(1)/(9)]=(5R)/(36)` `:.lambda_(B)=(36)/(5R) .....(2)` `:.(lambda_(L))/(lambda_(B))=(4)/(3R)XX(5R)/(36)=(5)/(27)` |
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