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:

The energy of the nth orbit is $E_n = -\frac{13.6}{n^2} eV$. For $n=3$, $E_3 = -1.51 eV$, and for $n=4$, $E_4 = -0.85 eV$. The energy difference is $\Delta E = E_4 - E_3 = -0.85 - (-1.51) = 0.66 eV$.