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Taking the wavelength of first Balmer li...

Taking the wavelength of first Balmer line in hydrogen spectrum (`n = 3` to `n = 2`) as 660 nm, the wavelength of the `2^(nd)` Balmer line (`n = 4` to `n = 2`) will be:

A

`488.9` nm

B

`388.9` nm

C

`889.2`nm

D

`642.7` nm

Text Solution

Verified by Experts

The correct Answer is:
A
`1/(lambda_(1))=R[1/(2^(2))-1/(3^(2))]=5/36 R`
`1/(lambda_(2))=R [1/(2^(2))-1/(4^(2))] =R [3/16]`
`(lambda_(2))/(lambda_(1)) = (5R//36)/(3R//16)=5/36 xx16/3` .
`lambda_(2) = 5/36 xx16/3 xx 660 = 488.9 nm `.
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