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calculate the longest wavelength in Balmer series and the series limit . (Given R=1.097xx10^(7)m^(-1) ) |
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Answer» Solution :`overline(v)=(1)/(lamda)=R((1)/(n_(1)^(2))-(1)/(n_(2)^(2)))` For shortest WAVELENGTH `n_(1)=2,n_(2)=oo` Substitution with final answer `lamda=3646Å or lamda=3.646xx10^(-7)m` For longest wavelength `n_(1)=2,n_(2)=3` Substitution with final answer `lamda=6563Å or lamda=6.563xx10^(-7)m` Important Note: Any other relevant or ALTERNATE answer may be considered. Detailed Answer: The lines of BALMER series is given by `(1)/(lamda)=R[(1)/(2^(2))-(1)/(n_(2)^(2))]` For short wavelength, `n=-oo` `(1)/(lamda)=R[(1)/(4)-(1)/((-oo)^(2))]` `(1)/(lamda)=(R)/(4)` `Rlamda=4` `lamda=(4)/(1.097xx10^(7))` `lamda=(4)/(1.097)xx10^(-7)` `lamda=3646Å`, For LONG wavelength `(1)/(lamda_("long"))=R[(1)/((2)^(2))-(1)/((3)^(2))]` `(1)/(lamda_("long"))=R[(1)/(4)-(1)/(9)]` `(1)/(lamda_("long"))=R[(9-4)/(36)]`. `(1)/(lamda_("long")=R[(5)/(36)]` `lamda(5R)=36` `lamda=(36)/(5R)xx10^(-7)=(36)/(5xx1.097xx10^(+7))` `lamda_("long")=6.563xx10^(-7)=6563`Å |
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