A photon of 2.55 eV is emitted out by an electronic transition in hydrogen atom. The change in de-Br — Atomic Structure Chemistry Question

💡 Solution & Explanation

An energy difference of 2.55 eV in H atom corresponds to a transition between $n=4$ (-0.85 eV) and $n=2$ (-3.4 eV). The de-Broglie wavelength in orbit $n$ is $\lambda_n = \frac{2\pi r_n}{n} = \frac{2\pi (0.529 n^2)}{n} = n (2\pi \times 0.529) = n \times 3.32$ \AA. The change $\Delta\lambda = (4 - 2) \times 3.32 = 6.64$ \AA. Therefore, correct answer is C.