For Hydrogen like atom, energy emitted is E=−13.6Z2(ni21−nf21)
⇒E=C(nf21−ni21)
⇒hν=C[nf21−ni21]
Therefore, the ratiov2v1=[nf21−ni21]3−1[nf21−ni21]2−1
=[11−91][11−41]=9843
=43×89
⇒v2v1=3227
⇒v2=2732v1=2732×3×1015Hz=932×1015Hz
Hence, x=32.