What will be the longest wavelength line in the Balmer series of spectrum of H atom?

Using the formula:

$\frac{1}{\lambda }=R{Z}^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

Value of R is $1.097\times {10}^{-2} {\mathrm{nm}}^{-1}$ and for hydrogen, Z=1.

n₁ is 2 for Balmer series and for determining longest wavelength, minimum wavelength is considered because energy is inversely proportional to the wavelength.

$$\begin{gathered} \frac{1}{\mathrm{\lambda }}=1.097\times {10}^{-2} {\mathrm{nm}}^{-1}\times 1^2\times \left(\frac{1}{2^2}-\frac{1}{3^2}\right) \\ \mathrm{\lambda }=656 \mathrm{nm} \end{gathered}$$