\(\frac{1}{\lambda}= R \left(1-\frac{1}{4}\right)= R \left(\frac{3}{4}\right)\)

\(\quad \lambda =\frac{4}{3 R } \ldots(1)\)

\(3^{\text {rd }} \text { line(Paschen) }\)

\(\frac{1}{\lambda_{3}}= R \left(\frac{1}{3^{2}}-\frac{1}{6^{2}}\right)=\frac{ R }{9} \times \frac{3}{4}\)

\(2^{nd}\) line(Balmer)

\(\frac{1}{\lambda_{2}}= R \left(\frac{1}{2^{2}}-\frac{1}{4^{2}}\right)=\frac{ R }{4} \times \frac{3}{4}\)

Thus a \(\lambda=\lambda_{3}-\lambda_{2}=\frac{12}{R}-\frac{16}{3 R}=\frac{20}{3 R}\) putting

\(a\left(\frac{4}{3 R}\right)=\frac{20}{3 R} \Rightarrow a=5\)