For which orbit in He ion, the circumference is 26.5 Å? — Atomic Structure Chemistry Question

💡 Solution & Explanation

Circumference $= 2\pi r$. Thus, $2\pi (0.529 \frac{n^2}{Z}) = 26.5$. For He$^+$, $Z=2$. $2 \times 3.1415 \times 0.529 \times \frac{n^2}{2} = 26.5 \implies 1.661 \times n^2 = 26.5 \implies n^2 \approx 15.95 \implies n = 4$. Therefore, correct answer is C.