The longest wavelength of He^(+) in Paschen series in 'm'. Then what will be the shortest wavelength of Be^(3+) in Paschen series in terms of m ?

The longest wavelength of He^(+) in Paschen series in 'm'. Then what w

1.	The longest wavelength of He^(+) in Paschen series in 'm'. Then what will be the shortest wavelength of Be^(3+) in Paschen series in terms of m ?
Answer» Solution :`(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`