As the orbit number increases, the K.E. and P.E. for an electron: — Atomic Structure Chemistry Question

💡 Solution & Explanation

$K.E. \propto \frac{1}{n^2}$ and $P.E. \propto -\frac{1}{n^2}$. As $n$ increases, the magnitude of both decreases. Since $P.E.$ is negative, becoming less negative means it increases. Thus, P.E. increases and K.E. decreases. Therefore, correct answer is D.