1.

The shortest wavelength in the Lyman series is 911.6Å. Then the longest wavelength in Lyman series is ......

Answer»

`1215Å`
`oo`
`2430Å`
`600Å`

Solution :For LYMAN series `(1)/(lamda)=R[(1)/(l^(2))-(1)/(n^(2))]`
`:.(1)/(lamda_(min))=R[(1)/(1^(2))-(1)/(oo^(2))]`
`:.(1)/(lamda_(min))=R""......(1)`
`[:.(1)/(oo^(2))=0]`
For maximum wavelength n = 2,
`(1)/(lamda_(min))=R[(1)/(l^(2))-(1)/(2^(2))]`
`=R[(1)/(l)-(1)/(4)]`
`(1)/(lamda_(max))=(3R)/(4)""......(2)`
`:.(lamda_(max))/(lamda_(min))=(R)/((3R)/(4))`
Taking RATIO of equation (1) and (2),
`:.lamda_(max)=(4)/(3)xxlamda_(min)`
`=(4)/(3)xx911.6Å`
`=4xx303.9=1215.6Å`
`:.lamda_(max)=1215Å`
(Taking near by value in integer)


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