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If the wavelength of the first line of the Balmer series of hydrogen atom is `656.1 nm `the wavelngth of the second line of this series would beA. `218.7 nm`B. `328.0 nm`C. `486.nm`D. `640.0nm` |
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Answer» Correct Answer - C `bar v_(1) = (1)/(lambda_(1)) = R ((1)/(2^(2)) - (1)/(3^(2))) = (5R)/(36)` `lambda = (36)/(5R)` `bar v_(2) = (1)/(lambda_(2)) = R ((1)/(2^(2)) - (1)/(4^(2))) = (3R)/(16)` `lambda_(2) = (16)/(3R)` When `(36)/(5R) rarr 656.1 nm` ` (16)/(3R) rArr (656.1 xx 5R xx 16)/(36 xx 3R) = 486 nm` |
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