Saved Bookmarks
| 1. |
In the spectrum of hydrogen, find the ratio of the longest wavelength in the Lyman series to the longest wavelength in the Balmer series |
|
Answer» For Lyman series, \(\frac{1}{\lambda_1}\) = R \((\frac{1}{1^2}-\frac{1}{2^2})\) = \(\frac{3}{4}\)R For Balmer series, \(\frac{1}{\lambda_1}\) = R \((\frac{1}{2^2}-\frac{1}{3^2})\) = \(\frac{5}{36}\)R Dividing above two equations, \(\frac{\lambda_1}{\lambda_2}\) = \(\frac{(5/36)R}{(3/4)R}\) = \(\frac{5}{27}\) |
|