1.

The second member of Lyman series in hydrogen spectrum has wavelength `5400 Å`. Find the wavelength of first member.

Answer» `(1)/(lamda)=R((1)/(n_(1)^(2))-(1)/(n_(2)^(2)))`
For second member of Lyman series,
`(1)/(5400)=R((1)/(1^(2))-(1)/(3^(2)))rArr(1)/(5400)=(8R)/(9)rarr(1)`
For first member of Lyman series,
`(1)/(lamda^(1))=R((1)/(1^(2))-(1)/(2^(2)))`
`(1)/(lamda^(1))=(3R)/(4)" "rarr(2)`
`((1))/((2))rArr(lamda^(1))/(5400)=(8R)/(9)xx(4)/(3R)`
`therefore lamda^(1)=(32)/(27)xx5400=6400 Å`.


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