InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
Find the longest and shortest wavelengths in the Lyman series for hydrogen. In what region of the electromagnetic spectrum does each series lie? |
|
Answer» The transition equation for Lyman series is given by `(1)/(lambda) = R ((1)/(1^(2)) - (1)/(n^(2)))`. N = 2,3` …. The largest wavelength is corresponds to `n = 2` `:. (1)/(lambda_(max)) = 1.097 xx 10^(7)) ((1)/(1) - (1)/(4))` `= 0.823 xx 10^(7)` `implies lambda_(max) = 1.2154 xx 10^(7) m = 1215 Å` The shortest wavelength corresponds to `n = oo` `:. (1)/(lambda_(min)) = 1.097 xx 10^(7)) ((1)/(1) - (1)/(oo))` or ` lambda_(min) = 0.911 xx 10^(-7) m = 911 Å` Both of these wavelength lie in the ultraviolet `(UV)` region of electromagnetic spectrum. S |
|