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

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 $Li^{2+}$