In a hydrogen spectum if electron moves from `6^(th)` to `2^(nd)` by transition in multi steps then find out the number of lines in spectum

Explanation

(a) Calculation of total number of lines :
`=((n_(2)-n_(1))[(n_(2)-n_(1))+1])/(2)=((6-3)[(6-3)+1])/(2)=(3xx4)/(2)=6`
(b) Calculation of number lines present in `U.V`. region.
`e^(-)` moves from `6^(th)` to `3^(rd)` orbit in multisteps.
For `U.V.` region, `e^(-)` should becomes into `1^(st)` shell. So the number of lines in `U.V`. region zero.
(c ) Calculation of toal number of lines in visible region.
For visible region, `e^(-)` should becomes into `2^(nd)` shell, so the number of lines in visible region zero.
(d) Calculation of total number of lines in I.R. region.
In I.R. region, Paschen, Bracket and Pfund series are present.
Number of lines in Paschen series`= n_(2)-3`
`= 6-3`
`=3`
Number of lines in Bracket series ` = n_(2)-4`
`6-4`
`=2`
Number of lines in Pfund series ` = n_(2)-5`
`=6-5`
`=1`
So total number of lines `= 3+2+1 =6`