The circumference of the third orbit of ion is m. The de-Broglie wavelength of electron revolving in — Atomic Structure Chemistry Question

💡 Solution & Explanation

By the de-Broglie relation applied to Bohr orbits, the circumference is an integer multiple of the wavelength: $2\pi r = n\lambda$. For the third orbit, $n = 3$. Given the circumference is $x$, we have $x = 3\lambda \implies \lambda = \frac{x}{3}$. Therefore, correct answer is A.