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Calculate the shortest and longest wavelength of Balmer series of hydrogen atom. Given R=1.097 xx 10^(7)m^(-1). |
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Answer» Solution :Shortest wavelength(HIGHEST frequency) is for the last line in the series. For last line in the BALMER.s series,`n_(1)=2,n_(2)=00` We have `(1)/(LAMBDA)=R((1)/(n_(1)^(2))-(1)/(n_(2)^(2)))` `(1)/(lambda)=1.097xx10^(7)[(1)/(2)^(2)-(1)/(00)]` `=1.097xx10^(7)((1)/(4))` `lambda=(4)/(1.097xx10)^(7)=3.646xx10^(-7)` Longest wavelength (smallest frequency) is for the first line in the series. For first line in the Balmer series, `n_(1) = 2, n_(2) =3`. we have `(1)/(lambda)=R((1)/(n_(1)^(2))-(1)/(n_(2)^(2)))=1.097xx10^(7)((1)/(2^(2))-(1)/(3^(2)))` `=1.097xx10^(7)((1)/(4)-(1)/(9))` `=1.097xx10^(7)((9-4)/(36))` `lambda=(1)/(0.1523xx10^(7))` `lambda=6.566xx10^(-7)` `lambda=6566A^(@)` |
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