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The first line is the Lyman series has w...

The first line is the Lyman series has wavelength `lambda`.The first line in Balmer series has wavelength

A

`27/5lambda`

B

`5/27lambda`

C

`9/2lambda`

D

`2/9lambda`

Text Solution

Verified by Experts

The correct Answer is:
a

Lyman series is Given by,`1/lambda=R[1/1^2-1/n^2]`
For first line, `n=2`br`implies 1/lambda=R[1/1-1/2^2]=(3R)/4`
`implies R=4/(3lambda)`
Foe Balmers series,`1/lambda_B=R[1/2^2-1/n^2],n=3`
`implies 1/lambda_B=R[1/4-1/9]=R[5/36]`
`implies 1/lambda_B=4/(3lambda)[5/36]`
`implies 1/lambda_B=5/27 lambda`
`therefore lambda_B=27/5lambda`
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