As an electron jumps from the fourth orbit to the second orbit in Be^3+ ion, its — Atomic Structure Chemistry Question

💡 Solution & Explanation

When an electron transitions from a higher orbit (n = 4) to a lower orbit (n = 2), the radius of revolution decreases. Let's analyze the properties: 1) K.E. is ∝ 1/n^2, so K.E. increases. 2) Speed v is ∝ 1/n, so speed increases. 3) Frequency of revolution f is ∝ 1/n^3, so frequency increases. 4) P.E. is negative and ∝ -1/n^2, so P.E. becomes more negative, meaning it decreases. Thus, all statements are correct.