Chemistry - Atomic Structure Question with Solution | TestHub

First line in balmer series means transition from$\mathrm{n}=3$to$\mathrm{n}=2$. Now,$9^{\mathrm{t}\mathrm{h}}$orbit of${\mathrm{L}\mathrm{i}}^{2+}$has same energy as third orbit of$\mathrm{\prime }\mathrm{H}\mathrm{\prime }$atom &$6^{\mathrm{t}\mathrm{h}}$orbit of${\mathrm{L}\mathrm{i}}^{2+}$has same energy as second orbit of$\mathrm{\prime }\mathrm{H}\mathrm{\prime }$atom. So$\mathrm{n}=9$to$\mathrm{n}=6$transition in${\mathrm{L}\mathrm{i}}^{2+}$would have same wavelength as first line in balmer series of

$\mathrm{H}$-spectrum.