1.

The ratio of the largest to shortest wavelength in Balmer series of hydrogen spectra is :

Answer»

`(25)/(9)`
`(17)/6`
`9/5`
`5/4`

Solution :`1/lambda ALPHA [1/n_(F)^(2)-1/n_(i)^(2)]`
`(lambda_("longest"))/(lambda_("shortest"))=((1/4-1/oo))/((1/4-1/9))=9/5`


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