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प्रश्न
The first line of Balmer series has wavelength 6563 A. What will be the wavelength of the first member of Lyman series?
विकल्प
1215.4 Å
2500 Å
7500 Å
600 Å
MCQ
उत्तर
1215.4 Å
Explanation:
`1/lambda_"Balmer" = "R"_"H" [1/2^2 - 1/3^2] = (5"R"_"H")/36`
`therefore 1/(lambda_"L""Balmer") = 3"R"_"H"/4`
`therefore 1/(lambda_"L""Balmer") = lambda_"Balmer" xx 5/27 = 1215 .4 Å`
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