Atomic Structure: Practice Problem & Solution

Solution Explained:

To solve this problem, we apply the core principles of Atomic Structure. Understanding the underlying formula is key to arriving at the correct answer below:

Energy of incident photon $E = \frac{hc}{\lambda} = \frac{12400}{975} eV \approx 12.75 eV$. The atom reaches an energy of $-13.6 + 12.75 = -0.85 eV$, which corresponds to $n = 4$ state since $-13.6/4^2 = -0.85 eV$. Number of spectral lines is $n(n-1)/2 = 4(3)/2 = 6$.