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CBSE Class 12-science Answered

The wavelength of the first member of Balmer series in the hydrogen spectrum is 6563 A. Calculate the wavelength of the first member of lyman series in the same spectrum.
Asked by rutujasarangmehta | 15 Mar, 2016, 07:00: PM
answered-by-expert Expert Answer
begin mathsize 12px style The space wavelength space of space the space spectral space lines space of space hydrogen space spectrum space are space given space by space the space formula colon
1 over straight lambda equals straight R open parentheses fraction numerator 1 over denominator straight n subscript straight f superscript 2 end fraction minus fraction numerator 1 over denominator straight n subscript straight i superscript 2 end fraction close parentheses
where space straight R space equals space Rydberg space constant
For space the space first space member space of space the space Balmer space series comma space straight n subscript straight f space equals space 2 space and space straight n subscript straight i space equals space 3
therefore 1 over straight lambda subscript 1 equals straight R open parentheses 1 over 2 squared minus 1 over 3 squared close parentheses
therefore 1 over straight lambda subscript 1 equals fraction numerator 5 straight R over denominator 36 end fraction space......... space left parenthesis space Equation space 1 right parenthesis
For space the space first space member space of space the space Lyman space series comma space straight n subscript straight f space equals space 1 space and space straight n subscript straight i space equals space 2
therefore fraction numerator 1 over denominator straight lambda subscript 1 apostrophe end fraction equals straight R open parentheses 1 over 1 squared minus 1 over 2 squared close parentheses
therefore fraction numerator 1 over denominator straight lambda subscript 1 apostrophe end fraction equals fraction numerator 3 straight R over denominator 4 end fraction space space......... space left parenthesis space Equation space 2 right parenthesis space
From space equation space 1 space and space 2 comma space we space get
fraction numerator straight lambda subscript 1 over denominator straight lambda subscript 1 apostrophe end fraction space equals space fraction numerator 5 straight R over denominator 36 end fraction space cross times space fraction numerator 4 over denominator 3 straight R end fraction
therefore fraction numerator straight lambda subscript 1 over denominator straight lambda subscript 1 apostrophe end fraction space equals 5 over 27
therefore straight lambda subscript 1 equals 5 over 27 straight lambda subscript 1 apostrophe
Given space that space straight lambda subscript 1 apostrophe space equals space 6563
therefore straight lambda subscript 1 space end subscript equals space 5 over 27 space cross times space 6563 space
therefore straight lambda subscript 1 space end subscript equals space 1215 space straight A with degree on top
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Answered by Yashvanti Jain | 16 Mar, 2016, 12:14: PM
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