Question

In hydrogen atom, if `lambda_(1),lambda_(2),lambda_(3)` are shortest wavelengths in Lyman, Balmer and Paschen series respectively, then `lambda_(1):lambda_(2):lambda_(3)`equals
A. `1:4:9`
B. `9:4:1`
C. `1:2:3`
D. `3:2:1`

Answer 1

Correct Answer - A
For hydrogen atom,`1/lambda=R((1)/n_(1)^(2)-(1)/n_(2)^(2)),n_(2)gtn_(1)`
For Lyman series,`n_(1)=1,n_(2)=infty`
Rightarrow`1/lambda_(1)=R`
For Balmer series, `n_(1)=2,n_(2)=infty`
Rightarrow`1/lambda_(2)=R/4`
For Paschen series,`n_(1)=3,n_(2)=infty`
Rightarrow`1/lambda_(3)=R/9`
So, `lambda_(1)=1/R,lambda_(2)=4/R,lambda_(3)=9/R`
`lambda_(1):lambda_(2):lambda_(3)=1:4:9`