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The maximum wavelength of Brackett series of hydrogen atom is ...... [R=1.097xx10^(7)m-1] |
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Answer» `35,890Å` `(1)/(lamda)=R[(1)/(4^(2))-(1)/(n^(2))]` For maximum wavelength we have to take n = 5 `:.(1)/(lamda_(max))=R[(1)/(4^(2))-(1)/(5^(2))]` `:.(1)/(lamda_(max))=R[(1)/(16)-(1)/(25)]` `:.(1)/(lamda_(max))=R[(25-16)/(25xx16)]=(Rxx9)/(25xx16)` `:.lamda_(max)=(25xx16)/(9R)=(400)/(9xx1.097xx10^(7))` `=40514xx10^(-10)m` `=40,400Å` (Approximate value) |
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