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Calculate the wavelength of the first line in the balmer series of hydrogen spectrum |
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Answer» For the first in Balmer series `n_(1) = 2 and n_(2) = 3` so `bar v R_(H)[(1)/(n_(1)^(2)) - (1)/(n_(2)^(2))] cm^(-1)` `= 109677[(1)/((2)^(2)) - (1)/((3)^(2))]cm^(-1)` `= 109677 xx (5)/(36) cm^(-1)` `= 15232.9 cm^(-1)` `or lambda = (1)/(bar v) = (1)/(15.232.9)cm = 6.565 xx 10^(-5) cm = 656.5 nm` |
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