To solve this problem, we need to understand the hydrogen spectral series and the formula for the wavelength of emitted photons.$\\$The wavelength of light emitted in a hydrogen atom transition is given by the Rydberg formula:$\\ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \\$where $R$ is the Rydberg constant, $n_1$ and $n_2$ are the principal quantum numbers,$\\$and $n_2 > n_1. \\$For the Balmer series, $n_1 = 2$ and $n_2 \geq 3. \\$The shortest wavelength in the Balmer series corresponds to the transition from $n_2 = \infty$ to $n_1 = 2. \\$Thus, the shortest wavelength $\lambda$ in the Balmer series is:$\\ \frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) = R \left( \frac{1}{4} \right) \\$For the Brackett series, $n_1 = 4$ and $n_2 \geq 5. \\$The shortest wavelength in the Brackett series corresponds to the transition from $n_2 = \infty$ to $n_1 = 4. \\$Thus, the shortest wavelength $\lambda'$ in the Brackett series is:$\\ \frac{1}{\lambda'} = R \left( \frac{1}{4^2} - \frac{1}{\infty^2} \right) = R \left( \frac{1}{16} \right) \\$Now, we compare the two wavelengths:$\\ \frac{1}{\lambda} = \frac{R}{4} \quad$and$\quad \frac{1}{\lambda'} = \frac{R}{16} \\$Taking the ratio of the two equations:$\\ \frac{\lambda'}{\lambda} = \frac{R/16}{R/4} = \frac{1/16}{1/4} = \frac{1}{4} \\$Thus, $\lambda' = 4\lambda. \\$Therefore, the shortest wavelength in the Brackett series is $4\lambda,$ which corresponds to Option 2.