We know:
\[ r = 0.529 \frac{n^2}{Z} \implies 8.48 = 0.529 \frac{n^2}{1} \]
\[ n^2 = 16 \implies n = 4 \]
We also know:
\[ E \propto \frac{1}{n^2} \]
\[ E_n = \frac{E}{16} \]
Thus, $x = 16$.
Match the following:
Which of the following is the correct electronic configuration for \( \text{Oxygen (O)} \)?
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is: