If the shortest wavelength of the spectral line of H - atom in Lyman series is x, then the longest wavelength of the line in Balmer series of Li2+ is
4x5
vmax=1λmin=RHZ2n21(n2=∞)
For H - atom, Z = 1 : For Lyman series, n1=1
Hence, 1x=RH
For largest wavelength in the Balmerseries of Li2+,Z=3,n1=2;n2=3
Hence, 1λmax′RHZ2(1n21−1n22)
=1x×32(122−132)=9x[536]=54x
or λmax=4x5