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`
Text Solution
Verified by Experts
The correct Answer is:
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`
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