What would be the approximate quantum number, , for a circular orbit of hydrogen, cm in diameter? — Atomic Structure Chemistry Question

💡 Solution & Explanation

Radius $r = \frac{\text{diameter}}{2} = 0.5 \times 10^{-5} \text{ cm} = 5 \times 10^{-8} \text{ cm} = 5 \times 10^{-10} \text{ m} = 5$ Å. Using $r = 0.529 \times n^2$ Å, we have $n^2 = \frac{5}{0.529} \approx 9.45$. Thus $n \approx 3$. Wait, diameter is $1 \times 10^{-5}$ cm = $10^{-7}$ m = 1000 Å. Radius = 500 Å. $n^2 = \frac{500}{0.529} \approx 945 \implies n \approx 31$. Therefore, correct answer is A.