To solve this problem, we need to find the wave number of the last line of the Balmer series in the hydrogen spectrum.$\\$The wave number $\nu$ is given by the Rydberg formula:$\\ \nu = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \\$For the Balmer series, $n_1 = 2$ and $n_2$ is the principal quantum number of the final state.$\\$The last line of the Balmer series corresponds to the transition from $n_2 = \infty$ to $n_1 = 2. \\$Substituting these values into the Rydberg formula:$\\ \nu = R \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) \\$Since $\frac{1}{\infty^2} = 0,$ the equation simplifies to:$\\ \nu = R \left( \frac{1}{4} \right) \\$Given that the Rydberg constant $R = 10^7 \,$m$^{-1},$ we have:$\\ \nu = 10^7 \times \frac{1}{4} \\ \nu = 0.25 \times 10^7 \,$m$^{-1} \\$Therefore, the wave number of the last line of the Balmer series is $0.25 \times 10^7 \,$m$^{-1}. \\$This corresponds to Option 1.