A hydrogen atom initially in the n = 6 state decays to the n = 2 state. The emitted photon is detected in a photographic plate. What is the wavelength of the detected photon? The lowest level energy state of hydrogen is -13.6 eV. (h = 6.626 × 10^-34 J ∙ s, 1 eV = 1.60 × 10^-19 J, c = 3.00 × 10⁸ m/s) A) 410 nm B) 93.8 nm C) 1090 nm D) 93.1 nm

Question

Quizplus · Accepted Answer

The energy difference between the n=6 and n=2 states is calculated using the formula E = -13.6 eV (1/n2^2 - 1/n1^2). Converting this energy to joules and using E = hc/λ, we find the wavelength to be approximately 410 nm.

A Hydrogen Atom Initially in the N = 6 State