The maximum wavelength of Brackett series of hydrogen atom is ...... `[R=1.097xx10^(7)m-1]`

Options

A. `35,890Å`
B. `14,440Å`
C. `62,160Å`
D. `40,400Å`

Answer

D

Explanation

The wavelength of Brackett series,
`(1)/(lamda)=R[(1)/(4^(2))-(1)/(n^(2))]`
For maximum wavelength we have to take n = 5
`:.(1)/(lamda_(max))=R[(1)/(4^(2))-(1)/(5^(2))]`
`:.(1)/(lamda_(max))=R[(1)/(16)-(1)/(25)]`
`:.(1)/(lamda_(max))=R[(25-16)/(25xx16)]=(Rxx9)/(25xx16)`
`:.lamda_(max)=(25xx16)/(9R)=(400)/(9xx1.097xx10^(7))`
`=40514xx10^(-10)m`
`=40,400Å` (Approximate value)