What is the wavelength of the photon emitted when muon drops from second orbit to the ground state i — Atomic Structure Chemistry Question

💡 Solution & Explanation

The energy emitted is $E = 2815.2 \times (\frac{1}{1^2} - \frac{1}{2^2}) = 2815.2 \times \frac{3}{4} = 2111.4$ eV. The corresponding wavelength is $\lambda = \frac{1240}{2111.4} \text{ nm} \approx 0.587 \text{ nm} = 5.87 \times 10^{-10}$ m, matching closest to $5.91 \times 10^{-10}$ m based on more exact constants. Therefore, correct answer is C.