\(\frac{1}{{{\lambda _B}}}\,\, = \,\,{Z^2}{R_H}\left( {\frac{1}{{{2^2}}} - \frac{1}{{{3^2}}}} \right)\,\,\,{\lambda _B}\,\, = \,\,\frac{{36}}{{5{R_H}{Z^2}}}\)

લાયમન શ્રેણીની \(1^{st}\) રેખાની તરંગલંબાઈ

\(\frac{1}{{{\lambda _L}}}\,\, = \,\,{Z^2}{R_H}\left( {\frac{1}{{{1^2}}} - \frac{1}{{{2^2}}}} \right)\,\,\,\,\,\,{\lambda _L}\,\, = \,\,\frac{4}{{3{R_H}{Z^2}}}\)

તફાવત \({\lambda _B} - {\lambda _L}\,\, = \,\,59.3 \times {10^{ - 7}}\,\, = \,\,\frac{{36}}{{5{R_H}{Z^2}}} - \frac{4}{{3{R_H}{Z^2}}}\)

ઉકેલતા \(Z = 3\) મળે.  તેથી \(Li^{2+}\)