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Last line of Lyman series for `H-` atom has wavelength `lambda_(1) A,2^(nd)` line of Balmer series has wavelength `lambda_(2)A` thenA. `(16)/(lambda_(1))=(9)/(lambda_(2))`B. `(16)/(lambda_(2))=(3)/(lambda_(1))`C. `(4)/(lambda_(1))=(1)/(lambda_(2))`D. `(16)/(lambda_(1))=(3)/(lambda_(2))` |
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Answer» Correct Answer - 2 `(1)/(lambda_(1))=R(1)^(2)[(1)/(1^(2))-(1)/(oo^(2))]" "` and `" "(1)/(lambda_(2))=R(1)^(2)[(1)/(2^(2))-(1)/(4^(2))]` `:. " "lambda_(1)=(1)/(R)" "`and `" "lambda_(2)=(16)/(3R)" ":." "(16)/(lambda_(2))=(3)/(lambda_(1))` |
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