To solve this problem, we need to use the Rydberg formula for hydrogen spectral lines.$\\$The Rydberg formula is given by:$\\ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \\$where:$\\ • \lambda$ is the wavelength of the emitted light,$\\ • R_H$ is the Rydberg constant for hydrogen $(1.097 \times 10^7 \,$m$^{-1}), \\ • n_1$ and $n_2$ are the principal quantum numbers with $n_2 > n_1. \\$Let's calculate the wavelengths for each transition:$\\$A. $n_2 = 3$ to $n_1 = 2: \\ \frac{1}{\lambda} = R_H \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \\ = 1.097 \times 10^7 \left( \frac{1}{4} - \frac{1}{9} \right) \\ = 1.097 \times 10^7 \left( \frac{5}{36} \right) \\ = 1.523 \times 10^6 \,$m$^{-1} \\ \lambda = \frac{1}{1.523 \times 10^6} \approx 656.3 \,$nm$\\$B. $n_2 = 4$ to $n_1 = 2: \\ \frac{1}{\lambda} = R_H \left( \frac{1}{2^2} - \frac{1}{4^2} \right) \\ = 1.097 \times 10^7 \left( \frac{1}{4} - \frac{1}{16} \right) \\ = 1.097 \times 10^7 \left( \frac{3}{16} \right) \\ = 2.056 \times 10^6 \,$m$^{-1} \\ \lambda = \frac{1}{2.056 \times 10^6} \approx 486.1 \,$nm$\\$C. $n_2 = 5$ to $n_1 = 2: \\ \frac{1}{\lambda} = R_H \left( \frac{1}{2^2} - \frac{1}{5^2} \right) \\ = 1.097 \times 10^7 \left( \frac{1}{4} - \frac{1}{25} \right) \\ = 1.097 \times 10^7 \left( \frac{21}{100} \right) \\ = 2.303 \times 10^6 \,$m$^{-1} \\ \lambda = \frac{1}{2.303 \times 10^6} \approx 434.1 \,$nm$\\$D. $n_2 = 6$ to $n_1 = 2: \\ \frac{1}{\lambda} = R_H \left( \frac{1}{2^2} - \frac{1}{6^2} \right) \\ = 1.097 \times 10^7 \left( \frac{1}{4} - \frac{1}{36} \right) \\ = 1.097 \times 10^7 \left( \frac{32}{144} \right) \\ = 2.466 \times 10^6 \,$m$^{-1} \\ \lambda = \frac{1}{2.466 \times 10^6} \approx 410.2 \,$nm$\\$Now, match the calculated wavelengths with the given options:$\\$A. $656.3 \,$nm$\rightarrow$III$\\$B. $486.1 \,$nm$\rightarrow$IV$\\$C. $434.1 \,$nm$\rightarrow$II$\\$D. $410.2 \,$nm$\rightarrow$I$\\$Thus, the correct option is:$\\$Option 4: A->III, B->IV, C->II, D->I