The first line of Balmer series has wvaelength `6563 Å`. What will be the wavelength of the ifrst member of Lyman series?A. `1215.4 Å`B. `2500 Å`C. `7500 Å`D. `600 Å`

The first line of Balmer series has wvaelength `6563 Å`. What will be

1.	The first line of Balmer series has wvaelength `6563 Å`. What will be the wavelength of the ifrst member of Lyman series?A. `1215.4 Å`B. `2500 Å`C. `7500 Å`D. `600 Å`
Answer» Correct Answer - A `(1)/(lambda_("Balmer")) = R [(1)/(2^(2)) - (1)/(3^(2))] = (5R)/(36), (1)/(lambda_("Lyman")) = R [(1)/(1^(2)) - (1)/(2^(2))] = (3R)/(4)` `:. Lambda_("Lyman") = lambda_("Balmer") xx (5)/(27) = 1215.4 Å`

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The first line of Balmer series has wvaelength `6563 Å`. What will be the wavelength of the ifrst member of Lyman series?A. `1215.4 Å`B. `2500 Å`C. `7500 Å`D. `600 Å`

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