InterviewSolution
Saved Bookmarks
| 1. |
N 3.1 = 0m0 d)n = 3,1 = 2, m = 2,£1,013. The wave number of the shortest wavelength transition in Balmer series of atomic hydrogen willa) 42154 b) 1437 A )3942 A d) 3647 A |
| Answer» | |