InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
lambda_(1) and lambda_(2) are the wavelength of the first numbers of the Lyman and Paschen series respectively, then find lambda_(1): lambda_(2) |
|
Answer» Solution : Wavelength of first LINE in Lyman series `lambda_(1)` is, `:.(1)/(lambda_(1))=R[(1)/(1^(2))-(1)/(n_(i)^(2))] ` [ `:.` for first line `n_(1)=2]` `:.(1)/(lambda_(1))=R[(1)/(1^(2))-(1)/(r^(2))]=R[1-(1)/(4)]` `:. (1)/(lambda_(1))=(3R)/(4)....(1)` `rArr` Wavelength of first line in PASCHEN series `lambda_(2)`, `:.(1)/(lambda_(1))=R[(1)/(3^(2))-(1)/(4^(2))] [ :.` for first line `n_(i)=4]` `:.(1)/(lambda_(2)) [(1)/(9)-(1)/(16)]=R[(7)/(14)]` `:.(1)/(lambda_(2))=(7R)/(144) ...(2)` `rArr` Taking ratio of equation (2) and equation (1), `(lambda_(1))/(lambda_(2))=(7R)/(144)xx(4)/(3R)` `:. (lambda_(1))/(lambda_(2))=(7)/(108)` |
|