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| 1. |
What are the wavelengths of two longest line (in nm) in Lymington series of Hydrogen atom |
| Answer» According to Rydberg Balmer equation{tex}\\frac{1}{\\lambda}{/tex} = {tex}R\\left[\\frac{1}{n_{1}^{2}}-\\frac{1}{n_{2}^{2}}\\right]=R\\left[\\frac{1}{1^{2}}-\\frac{1}{n_{2}^{2}}\\right]{/tex}The wavelength {tex}(\\lambda){/tex}will be longest when n2\xa0is the smallest i.e. n = 2 and 3 for two longest wavelength lines.For : n2 = 2 ,\xa0{tex}\\frac{1}{\\lambda}=\\left(1.097 \\times 10^{-2} \\mathrm{nm}^{-1}\\right)\\left[\\frac{1}{1^{2}}-\\frac{1}{2^{2}}\\right]{/tex}{tex}\\left(1.097 \\times 10^{-2} \\mathrm{nm}^{-1}\\right) \\times \\frac{3}{4}=8.228 \\times 10^{-3} \\mathrm{nm}^{-1}{/tex} or {tex}\\lambda=121.54 \\mathrm{nm}{/tex}For: n2 = 3; {tex}\\frac{1}{\\lambda}=\\left(1.097 \\times 10^{-2} \\mathrm{nm}^{-1}\\right){/tex}= {tex}\\left(1.097 \\times 10^{-2} \\mathrm{nm}^{-1}\\right) \\times(8 / 9)=9.75 \\times 10^{-3} \\mathrm{nm}^{-1}{/tex}; {tex}\\lambda=102.56 \\mathrm{nm}{/tex} | |