In hydrogen spectrum, the shortest wavelength in Balmer series is the `lambda`. The shortest wavelength in the Brackett series will be

Options

A. `2lambda`
B. `4lambda`
C. `9lambda`
D. `16lambda`

Answer

B

Explanation

`(1)/(lambda)=R[(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]`
For shortest wavelength in Balmer series,
`n_(1)=2, n_(2)=oo`
`(1)/(lambda)=R[(1)/(2^(2))-(1)/(oo^(2))] or lambda=(4)/(R )`
For shortest wavelength in Brackett series.
`(1)/(lambda.)=R[(1)/(4^(2))-(1)/(oo^(2))] rArr lambda.=(16)/(R )=4xx(4)/(R )=4lambda`