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| 1. |
Calculate the shortest wavelength in helium spectrum of lyman series |
| Answer» Lyman series, n = 2, 3, 4... to n= 1For short wavelength, n ={tex}\\infty{/tex}\xa0to n = 1we know that ,\xa0{tex}E = \\frac { 12375 } { \\lambda ( \\mathop A\\limits_{}^o ) } = \\frac { 12375 } { 9134 } \\mathrm { eV } = 13.54 \\mathrm { eV }{/tex}Energy of nth\xa0orbit, E = 13.54/n2So, energy of n = 1, energy level = 13.54eVEnergy of n = 2, energy level= 13.54/22 = 3.387 eVSo, short wavelength of Balmer series =\xa0{tex}\\frac { 12375 } { 3.387 }{/tex}\xa0= 3653\xa0{tex}\\mathop A\\limits_{}^o {/tex} | |