For Balmer series, \(\mathrm{n}_1=2\).
Hence, \(\bar{v}=R\left(\frac{1}{2^2}-\frac{1}{n_2^2}\right)\) \(\bar{v}=\frac{1}{\lambda}\), For \(\lambda\) to be longest (maximum) \(\bar{v}\) should be minimum. This can be so when \(\mathrm{n}_2\) is minimum, i.e., \(\mathrm{n}_2=3\). Hence,
\(\bar{v}=\left(1.097 \times 10^7 \mathrm{~m}^{-1}\right)\left(\frac{1}{2^2}-\frac{1}{3^2}\right)\)
\(=1.097 \times 10^7 \times \frac{5}{36} \mathrm{~m}^{-1}=1.523 \times 10^6 \mathrm{~m}^{-1}\)