1.

The maximum wavelength of Lyman series is ....

Answer»

`(4xx1.097xx10^(7))/(3)m`
`(3)/(4xx1.097xx10^(7))m`
`(4)/(4xx1.097xx10^(7))m`
`(3)/(4)xx1.097xx10^(7)m`

SOLUTION :Wave NUMBER of Lyman series
`(1)/(lambda)=R[(1)/(1^(2))-(1)/(n^(2))]`
but for MAXIMUM wavelength n=2
`:. lambda_(1)/(lambda_(max))=1.097xx10^(7)[(1)/(1)-(1)/(2^(2))]`
`:. lambda_(1)/(lambda_(max))=1.097xx10^(7)[1-(1)/(2^(2))]`
`:. lambda_(1)/(lambda_(max))=(3)/(4)xx1.097xx10^(7)m^(-1)`
`:. lambda_(max)=(4)/(3xx1.097xx10^(7))`


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