Correct Answer - B
`n_(2)+n_(1)=4` . . . (i)
`n_(2)^(2)-n_(1)^(2)=8`
i.e., `(n_(2)+n_(1))(n_(2)-n_(1))=8`
`4(n_(2)-n_(1))=8`
`(n_(2)-n_(1))=2`
From eqns. (i) and (ii) `n_(2)=3,n_(1)=1`
`(1)/(lamda)=R_(H)Z^(2)[(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]`
`=R_(H)xx2^(2)[(1)/(1^(2))-(1)/(3^(2))]`
`(1)/(lamda)=R_(H)xx(32)/(9)` ltbr. `thereforelamda=(9)/(32R_(H))`.