1.

The shortest wavelength of `He^(+)` in Balmer series is `x`. Then longest wavelength in the Paschene series of `Li^(+2)` is :-A. `(36x)/(5)`B. `(16x)/(7)`C. `(9x)/(5)`D. `(5x)/(9)`

Answer» Correct Answer - B
`(IE)_(Li^(2+))=(IE)_(H)xxƶ^(2)`
`21.8xx10^(-19)xx9J//"atom"`
`lambda = (h)/sqrt(2ME)`
`lambda=(6.62xx10^(-34))/sqrt(2xx9.1xx10^(-31)xx2.18xx10^(-9)xx9)`
`lambda=1.17A^(@)`


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