InterviewSolution
Saved Bookmarks
| 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^(@)` |
|