The longest wavelength of ` He^(+)` in paschen series is "m", then shortest wavelenght of `Be^(+3)` in Pacchen series is( in terms of m):

Explanation

`(1)/(lamda_(He^(+))) = RZ^(2) ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))`
`(1)/(m) = R xx4 ((1)/(3^(2) ) - (1)/(4^(2))) = 4R xx (7)/(144) = (7R)/(36)`...(i)
(`:'` For Paschen series, `n_(1) = 3` and for longest wavelength, `n_(2) = 4`. For He Z = 2)
`(1)/(lamda_(Be^(3+))) = R xx 16 ((1)/(3^(2)) - (1)/(oo)) = (16)/(R)`...(ii)
( `:'` For shortest wavelength, `n_(2) = oo` and for Be, Z = 4)
Dividing eqn. (i) by eqn. (ii), we get
`:. (lamda_(Be^(3+)))/(m) = (7 xx 9)/(16 xx 36) or lamda_(Be^(3+)) = (7)/(64) m`