What is maximum wavelength of line of Balmer series of Hydrogen spectrum ($R_H=1.1\times {10}^7{m}^{-1}$) :

The correct option is B 654 $nm$
For Balmer series, lower orbit number $n_1$ = 2 and maximum wavelength can be obtained for that transition for which energy difference is minimum. So, here for the transition from 3 to 2 $\Delta$$E$ is minimum.
Hence, wavelength for this transition can be obtained by using Rydberg formula as,
$\frac{1}{\lambda }$ = $1.1\times {10}^7{m}^{-1}$ $\times Z^2$ ($\frac{1}{n_1^2}$ - $\frac{1}{n_2^2}$)
where $n_1$ = ground state orbit $n_2$ = higher state orbit
Z = atomic number
$\frac{1}{\lambda }$ = $1.1\times {10}^7{m}^{-1}$ $\times 1^2$ ($\frac{1}{2^2}$ - $\frac{1}{3^2}$)
$\lambda$ = 654 $nm$