To solve this problem, we need to determine the number of spectral lines emitted when a hydrogen atom is excited and then returns to lower energy levels.$\\$Given:$\\$• Wavelength of the exciting radiation $\lambda = 975$ Å$\\$First, we calculate the energy of the photon using the formula:$\\ E = \frac{hc}{\lambda} \\$Where:$\\$• $h = 6.626 \times 10^{-34}$ J s (Planck's constant)$\\$• $c = 3 \times 10^8$ m/s (speed of light)$\\$• $\lambda = 975 \times 10^{-10}$ m (convert Å to meters)$\\ E = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{975 \times 10^{-10}} \\ E = \frac{19.878 \times 10^{-26}}{975 \times 10^{-10}} \\ E = 2.04 \times 10^{-18}$ J$\\$Convert energy from joules to electron volts (1 eV = 1.602 \times 10^{-19$J): \\ E = \frac{2.04 \times 10^{-18}}{1.602 \times 10^{-19}}$ eV$\\ E \approx 12.74$ eV$\\$The energy levels of a hydrogen atom are given by:$\\ E_n = -13.6 \frac{1}{n^2}$ eV$\\$The ground state energy (n=1) is $-13.6$ eV.$\\$The energy difference between the ground state and the excited state is $12.74$ eV.$\\$Calculate the excited state level:$\\ E_{excited} = -13.6 + 12.74 = -0.86$ eV$\\$Find $n$ such that $E_n = -0.86$ eV:$\\ -13.6 \frac{1}{n^2} = -0.86 \\ \frac{1}{n^2} = \frac{0.86}{13.6} \\ n^2 = \frac{13.6}{0.86} \\ n^2 \approx 15.81 \\ n \approx 4 \\$The atom is excited to the $n=4$ level.$\\$The number of spectral lines emitted when the electron returns to lower energy levels is given by:$\\$Number of lines$= \frac{n(n-1)}{2} \\$For $n=4: \\$Number of lines$= \frac{4(4-1)}{2} \\$Number of lines$= \frac{4 \times 3}{2} \\$Number of lines$= 6 \\$Therefore, the number of spectral lines emitted is $6$, which corresponds to Option 3.