What is the angular speed of an electron revolving in the third orbit of ion? — Atomic Structure Chemistry Question

💡 Solution & Explanation

Angular speed $\omega = \frac{v}{r}$. Using $v = 2.18 \times 10^6 \frac{Z}{n}$ m/s and $r = 0.529 \times 10^{-10} \frac{n^2}{Z}$ m. For $\text{He}^+$ ($Z=2, n=3$), $\omega = \frac{2.18 \times 10^6 \times (2/3)}{0.529 \times 10^{-10} \times (9/2)} \approx 6.1 \times 10^{15} \text{ s}^{-1}$. Therefore, correct answer is A.